Transitive Closure of Symmetric Relation is Symmetric

Let $S$ be a set. Let $\mathcal R$ be a symmetric relation on $S$. Let $\mathcal T$ be the transitive closure of $\mathcal R$. The $\mathcal T$ is symmetric.


Transitive Closure of Reflexive Relation is Reflexive

Let $S$ be a set. Let $\mathcal R$ be a reflexive relation on $S$. Let $\mathcal T$ be the transitive closure of $\mathcal R$. Then $\mathcal T$ is reflexive.


Transitive Closure of Reflexive Symmetric Relation is Equivalence

Let $S$ be a set. Let $\mathcal R$ be a symmetric and reflexive relation on $S$. Then the transitive closure of $\mathcal R$ is an equivalence relation.


Union of Overlapping Convex Sets in Toset is Convex

Let $\left({ S, \preceq }\right)$ be a totally ordered set. Let $U$ and $V$ be convex sets in $S$. Let $U \cap V \ne \varnothing$. Then $U \cup V$ is also convex.


Rule of Material Equivalence

==== Formulation 1 ==== {{:Rule of Material Equivalence/Formulation 1}} ==== Formulation 2 ==== {{:Rule of Material Equivalence/Formulation 2}}


Union of Overlapping Convex Sets in Toset is Convex/Infinite Union

Let $\left({S, \preceq}\right)$ be a totally ordered set. Let $\mathcal A$ be a set of convex subsets of $S$. For any $P, Q \in \mathcal A$, let there be elements $C_0, \dotsc, C_n \in \mathcal A$ such that: :$C_0 = P$ :$C_n = Q$ :For $k = 0, \dots, n-1: C_k \cap C_{k+1} \ne \varnothing$ Then $\bigcup \mathcal A$ is convex in $S$.


Rule of Material Equivalence/Formulation 2

:$\vdash \left({p \iff q}\right) \iff \left({\left({p \implies q}\right) \land \left({q \implies p}\right)}\right)$


==== Formulation 1 ==== {{:Non-Equivalence as Disjunction of Conjunctions/Formulation 1}} ==== Formulation 2 ==== {{:Non-Equivalence as Disjunction of Conjunctions/Formulation 2}}


Continuous Involution is Homeomorphism

Let $\struct {S, \tau}$ be a topological space. Let $f: S \to S$ be a continuous involution. Then $f$ is a homeomorphism.


Involution is Permutation

Let $S$ be a set. Let $f: S \to S$ be an involution. Then $f$ is a permutation.


Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Forward Implication/Proof

: $\neg \left ({p \iff q}\right) \vdash \left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)$


Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Reverse Implication/Proof

: $\left({\neg p \land q}\right) \lor \left({p \land \neg q}\right) \vdash \neg \left ({p \iff q}\right)$


Non-Equivalence as Disjunction of Conjunctions/Formulation 2

:$\vdash \paren {\neg \paren {p \iff q} } \iff \paren {\paren {\neg p \land q} \lor \paren {p \land \neg q} }$


Non-Equivalence as Equivalence with Negation/Formulation 1/Forward Implication

:$\neg \paren {p \iff q} \vdash \paren {p \iff \neg q}$


Non-Equivalence as Equivalence with Negation/Formulation 1

:$\neg \paren {p \iff q} \dashv \vdash \paren {p \iff \neg q}$


Sign of Function Matches Sign of Definite Integral

Let $f$ be a real function continuous on some closed interval $\closedint a b$, where $a < b$. Then: :If $\forall x \in \closedint a b: \map f x \ge 0$ then $\displaystyle \int_a^b \map f x \rd x \ge 0$ :If $\forall x \in \closedint a b: \map f x > 0$ then $\displaystyle \int_a^b \map f x \rd x > 0$ :If $\forall x \in \closedint a b: \map f x \le 0$ then $\displaystyle \int_a^b \map f x \rd x \le 0$ :If $\forall x \in \closedint a b: \map f x < 0$ then $\displaystyle \int_a^b \map f x \rd x < 0$


Non-Equivalence as Equivalence with Negation/Formulation 2

: $\vdash \neg \left ({p \iff q}\right) \iff \left({p \iff \neg q}\right)$


Symmetric Closure of Relation Compatible with Operation is Compatible

Let $\struct {S, \circ}$ be a magma. Let $\RR$ be a relation compatible with $\circ$. Let $\RR^\leftrightarrow$ be the symmetric closure of $\RR$. Then $\RR^\leftrightarrow$ is compatible with $\circ$.


Singleton is Convex Set (Order Theory)

Let $\left({S, \preceq}\right)$ be an ordered set. Let $x \in S$. Then the singleton $\left\{{x}\right\}$ is convex.


Rule of Transposition/Variant 1/Formulation 2

:$\vdash \paren {p \implies \neg q} \iff \paren {q \implies \neg p}$


Rule of Transposition/Variant 1

==== Formulation 1 ==== {{:Rule of Transposition/Variant 1/Formulation 1}} ==== Formulation 2 ==== {{:Rule of Transposition/Variant 1/Formulation 2}}


Rule of Transposition/Variant 1/Formulation 2/Reverse Implication

: $\vdash \left({q \implies \neg p}\right) \implies \left({p \implies \neg q}\right)$


Rule of Transposition/Variant 1/Formulation 1

:$p \implies \neg q \dashv \vdash q \implies \neg p$


Rule of Transposition/Variant 2

==== Formulation 1 ==== {{:Rule of Transposition/Variant 2/Formulation 1}} ==== Formulation 2 ==== {{:Rule of Transposition/Variant 2/Formulation 2}}


Rule of Transposition/Variant 2/Formulation 1

: $\neg p \implies q \dashv \vdash \neg q \implies p$


Rule of Transposition/Variant 2/Formulation 1/Proof 2

:$\neg p \implies q \dashv \vdash \neg q \implies p$


Rule of Transposition/Variant 2/Formulation 2

: $\vdash \left({\neg p \implies q}\right) \iff \left({\neg q \implies p}\right)$


Boundary of Polygon is Topological Boundary

Let $P$ be a polygon embedded in $\R^2$. Denote the boundary of $P$ as $\partial P$. Let $\Int P$ and $\Ext P$ denote the interior and exterior of $\partial P$, where $\partial P$ is considered as a Jordan curve. Then the topological boundary of $\Int P$ is equal to $\partial P$, and the topological boundary of $\Ext P$ is equal to $\partial P$.


Jordan Curve and Jordan Arc form Two Jordan Curves

Let $\closedint a b$ denote the closed real interval between $a \in \R, b \in \R: a \le b$. Let $\gamma: \closedint a b \to \R^2$ be a Jordan curve. Let the interior of $\gamma$ be denoted $\Int \gamma$. Let the image of $\gamma$ be denoted $\Img \gamma$. Let $\sigma: \closedint c d \to \R^2$ be a Jordan arc such that: :$\map \sigma c \ne \map \sigma d$ :$\map \sigma c, \map \sigma d \in \Img \gamma$ and: :$\forall t \in \openint c d: \map \sigma t \in \Int \gamma$ Let $t_1 = \map {\gamma^{-1} } {\map \sigma c}$. Let $t_2 = \map {\gamma^{-1} } {\map \sigma d}$. Let $t_1 < t_2$. Define: :$-\sigma: \closedint c d \to \Img \sigma$ by $-\map \sigma t = \map \sigma {c + d - t}$ Let $*$ denote concatenation of paths. Let $\gamma \restriction_{\closedint a {t_1} }$ denote the restriction of $\gamma$ to $\closedint a {t_1}$. Define: :$\gamma_1 = \gamma {\closedint a {t_1} } * \sigma * \gamma {\restriction_{\closedint {t_2} b} }$ Define: :$\gamma_2 = \gamma {\closedint a {t_1} } * \paren {-\sigma}$ Then $\gamma_1$ and $\gamma_2$ are Jordan curves such that: :$\Int {\gamma_1} \subseteq \Int \gamma$ and: :$\Int {\gamma_2} \subseteq \Int \gamma$


Restriction of Total Ordering is Total Ordering

Let $\struct {S, \preceq}$ be a total ordering. Let $T \subseteq S$. Let $\preceq \restriction_T$ be the restriction of $\preceq$ to $T$. Then $\preceq \restriction_T$ is a total ordering of $T$.


Ordering Cycle implies Equality/General Case

Let $\left({S,\preceq}\right)$ be an ordered set. Let $x_0, x_1, \dots, x_n \in S$. Suppose that for $k = 0, 1, \dots, n - 1: x_k \preceq x_{k+1}$. Suppose also that $x_n \preceq x_0$. Then $x_0 = x_1 = \dots = x_n$.


Finite Non-Empty Subset of Ordered Set has Maximal and Minimal Elements

Let $\left({S, \preceq}\right)$ be an ordered set. Let $T \subseteq S$ be a finite, non-empty subset of $S$. Then $T$ has a maximal element and a minimal element.


Star Convex Set is Path-Connected

Let $A$ be a star convex subset of a vector space $V$ over $\R$ or $\C$. Then $A$ is path-connected.


Minimal Element of Chain is Smallest Element

Let $\left({S, \preceq}\right)$ be an ordered set. Let $C$ be a chain in $S$. Let $m$ be a minimal element of $C$. Then $m$ is the smallest element of $C$.


Maximal Element of Chain is Greatest Element

Let $\left({S, \preceq}\right)$ be an ordered set. Let $C$ be a chain in $S$. Let $m$ be a maximal element of $C$. Then $m$ is the greatest element of $C$.


Modus Ponendo Ponens/Variant 3

:$\vdash \paren {\paren {p \implies q} \land p} \implies q$


Rule of Material Equivalence/Formulation 1/Proof 2

:$p \iff q \dashv \vdash \left({p \implies q}\right) \land \left({q \implies p}\right)$


Rule of Material Equivalence/Formulation 2/Proof 1

:$\vdash \left({p \iff q}\right) \iff \left({\left({p \implies q}\right) \land \left({q \implies p}\right)}\right)$


Rule of Material Equivalence/Formulation 2/Proof 2


Modus Tollendo Ponens/Variant/Formulation 2/Proof 2

: $\vdash \left({p \lor q}\right) \iff \left({\neg p \implies q}\right)$


Rule of Transposition/Formulation 2/Proof 2

: $\vdash \left({p \implies q}\right) \iff \left({\neg q \implies \neg p}\right)$


Convex Set Characterization (Order Theory)

Let $\left({S, \preceq, \tau}\right)$ be an ordered set. Let $C \subseteq S$. {{TFAE}} {{begin-axiom}} {{axiom | n = 1         | t = $C$ is convex. }} {{axiom | n = 2         | t = $C$ is the intersection of an upper set with a lower set. }} {{axiom | n = 3         | t = $C$ is the intersection of its upper closure with its lower closure. }} {{end-axiom}}


Constructive Dilemma/Formulation 3

:$\paren {p \implies q} \land \paren {r \implies s}, p \lor r \vdash q \lor s$


Upper Closure is Upper Set

Let $(S, \preceq, \tau)$ be an ordered set. Let $T$ be a subset of $S$. Let $U$ be the upper closure of $T$. Then $U$ is an upper set.


Destructive Dilemma/Formulation 1/Proof 3

:$p \implies q, r \implies s \vdash \neg q \lor \neg s \implies \neg p \lor \neg r$


Open Ray is Open in GO-Space/Definition 1

Let $\struct {S, \preceq, \tau}$ be a generalized ordered space. Let $p \in S$. Then: :$p^\prec$ and $p^\succ$ are $\tau$-open where: :$p^\prec$ is the strict lower closure of $p$ :$p^\succ$ is the strict upper closure of $p$.


Upper and Lower Closures of Open Set in GO-Space are Open

Let $\left({X, \preceq, \tau}\right)$ be a Generalized Ordered Space/Definition 1. Let $A$ be open in $X$. Then the upper and lower closures of $A$ are open.


Equivalence of Definitions of Generalized Ordered Space/Definition 1 implies Definition 3

Let $\struct {S, \preceq, \tau}$ be a generalized ordered space by Definition 1: {{:Definition:Generalized Ordered Space/Definition 1}} Then $\struct {S, \preceq, \tau}$ is a generalized ordered space by Definition 3: {{:Definition:Generalized Ordered Space/Definition 3}}


Upper Set is Convex

Let $\left({S, \preceq}\right)$ be an ordered set. Let $T \subseteq S$ be an upper set. Then $T$ is convex in $S$.


Lower Set is Convex

Let $\left({S, \preceq}\right)$ be an ordered set. Let $T \subseteq S$ be a lower set. Then $T$ is convex in $S$.


GO-Space Embeds Densely into Linearly Ordered Space

Let $\left({Y, \preceq, \tau}\right)$ be a generalized ordered space (GO-space) by Definition 3. That is: : let $\left({Y, \tau}\right)$ be a Hausdorff space and: : let $\tau$ have a sub-basis consisting of upper sets and lower sets relative to $\preceq$. Then $\left({Y, \preceq, \tau}\right)$ is a GO-space by Definition 2. That is, there is a linearly ordered space $\left({X, \preceq', \tau'}\right)$ and a mapping from $Y$ to $X$ which is a order embedding and a topological embedding.


Upper Set with no Minimal Element

Let $\struct {S, \preceq}$ be an ordered set. Let $U \subseteq S$. Then: :$U$ is an upper set in $S$ with no minimal element {{iff}}: :$\displaystyle U = \bigcup \set {u^\succ: u \in U}$ where $u^\succ$ is the strict upper closure of $u$.


Lower Sets in Totally Ordered Set form Nest

Let $\left({S, \preceq}\right)$ be a totally ordered set. Let $\mathcal L$ be a set of lower sets in $S$. Then $\mathcal L$ is a nest. That is, $\mathcal L$ is totally ordered by $\subseteq$.


Exclusive Or is Self-Inverse

:$\paren {p \oplus q} \oplus q \dashv \vdash p$ where $\oplus$ denotes the exclusive or operator.


Conjunction has no Inverse

Let $\land$ denote the conjunction operation of propositional logic. Then there exists no binary logical connective $\circ$ such that: :$(1): \quad \forall p, q \in \left\{{T, F}\right\}: \left({p \land q}\right) \circ q = p$


Disjunction has no Inverse

Let $\lor$ denote the disjunction operation of propositional logic. Then there exists no binary logical connective $\circ$ such that: :$(1): \quad \forall p, q \in \left\{{T, F}\right\}: \left({p \lor q}\right) \circ q = p$


Binary Logical Connectives with Inverse

Let $\circ$ be a binary logical connective. Then there exists another binary logical connective $*$ such that: :$\forall p, q \in \set {\F, \T}: \paren {p \circ q} * q \dashv \vdash p \dashv \vdash q * \paren {p \circ q}$ {{iff}} $\circ$ is either: :$(1): \quad$ the exclusive or operator or: :$(2): \quad$ the biconditional operator. That is, the only truth functions that have an inverse operation are the exclusive or and the biconditional.


Biconditional is Self-Inverse

:$\paren {p \iff q} \iff q \dashv \vdash p$ where $\iff$ denotes the biconditional operator.


Finite Chain is Order-Isomorphic to Finite Ordinal

Let $\left({S, \preceq}\right)$ be an ordered set. Let $C$ be a finite chain in $S$. Then for some finite ordinal $\mathbf n$: :$\left({C, {\preceq \restriction_C} }\right)$ is order-isomorphic to $\mathbf n$. That is: :$\left({C, {\preceq \restriction_C} }\right)$ is order-isomorphic to $\N_n$ where $\N_n$ is the initial segment of $\N$ determined by $n$: :$\N_n = \left\{ {k \in \N: k < n}\right\} = \left\{ {0, 1, \ldots, n - 1}\right\}$


Complete Linearly Ordered Space is Compact

Let $\left({X, \preceq, \tau}\right)$ be a linearly ordered space. Let $\left({X, \preceq}\right)$ be a complete lattice. Then $\left({X, \tau}\right)$ is compact.


Condition for Well-Foundedness/Reverse Implication

Let $\left({S, \preceq}\right)$ be an ordered set. Suppose that there is no infinite sequence $\left \langle {a_n}\right \rangle$ of elements of $S$ such that $\forall n \in \N: a_{n+1} \prec a_n$. Then $\left({S, \preceq}\right)$ is well-founded.


Condition for Well-Foundedness/Reverse Implication/Proof 1

Let $\struct {S, \preceq}$ be an ordered set. Suppose that there is no infinite sequence $\sequence {a_n}$ of elements of $S$ such that $\forall n \in \N: a_{n + 1} \prec a_n$. Then $\struct {S, \preceq}$ is well-founded.


Inversion Mapping on Ordered Group is Dual Order-Isomorphism

Let $\struct {G, \circ, \preceq}$ be an ordered group. Let $\iota: G \to G$ be the inversion mapping, defined by $\map \phi x = x^{-1}$. Then $\iota$ is a dual order-isomorphism.


Properties of Ordered Group/OG2/Proof 2

Let $\left({G, \circ, \preceq}\right)$ be an ordered group with identity $e$. Let $\prec$ be the reflexive reduction of $\preceq$. Let $x, y \in G$. Then the following equivalences hold: :$(\operatorname{OG}2.1):\quad x \preceq y \iff e \preceq y \circ x^{-1}$ :$(\operatorname{OG}2.2):\quad x \preceq y \iff e \preceq x^{-1} \circ y$ :$(\operatorname{OG}2.3):\quad x \preceq y \iff x \circ y^{-1} \preceq e$ :$(\operatorname{OG}2.4):\quad x \preceq y \iff y^{-1} \circ x \preceq e$ :$(\operatorname{OG}2.1'):\quad x \prec y \iff e \prec y \circ x^{-1}$ :$(\operatorname{OG}2.2'):\quad x \prec y \iff e \prec x^{-1} \circ y$ :$(\operatorname{OG}2.3'):\quad x \prec y \iff x \circ y^{-1} \prec e$ :$(\operatorname{OG}2.4'):\quad x \prec y \iff y^{-1} \circ x \prec e$


Properties of Ordered Group/OG4/Proof 1

Let $\struct {G, \circ, \preceq}$ be an ordered group with identity $e$. Let $x \in G$. Then the following equivalences hold: {{begin-axiom}} {{axiom | n = \operatorname {OG} 4.1         | m = x \preceq e \iff e \preceq x^{-1} }} {{axiom | n = \operatorname {OG} 4.2         | m = e \preceq x \iff x^{-1} \preceq e }} {{axiom | n = \operatorname {OG} 4.1'         | m = x \prec e \iff e \prec x^{-1} }} {{axiom | n = \operatorname {OG} 4.2'         | m = e \prec x \iff x^{-1} \prec e }} {{end-axiom}}


Properties of Ordered Group/OG4/Proof 2

Let $\left({G, \circ, \preceq}\right)$ be an ordered group with identity $e$. Let $x \in G$. Then the following equivalences hold: {{begin-axiom}} {{axiom|n = \operatorname{OG}4.1        |m = x \preceq e \iff e \preceq x^{-1} }} {{axiom|n = \operatorname{OG}4.2        |m = e \preceq x \iff x^{-1} \preceq e }} {{axiom|n = \operatorname{OG}4.1'        |m = x \prec e \iff e \prec x^{-1} }} {{axiom|n = \operatorname{OG}4.2'        |m = e \prec x \iff x^{-1} \prec e }} {{end-axiom}}


Properties of Ordered Group/OG5/Proof 1

Let $\left({G, \circ, \preceq}\right)$ be an ordered group with identity $e$. Let $\prec$ be the reflexive reduction of $\preceq$. Let $x \in G$. Let $n \in \N_{>0}$ be a strictly positive integer. Then the following hold: :$x \preceq e \implies x^n \preceq e$ :$e \preceq x \implies e \preceq x^n$ :$x \prec e \implies x^n \prec e$ :$e \prec x \implies e \prec x^n$


Properties of Ordered Group/OG5/Proof 2


Mapping from Totally Ordered Set is Dual Order Embedding iff Strictly Decreasing

Let $\left({S, \preceq_1}\right)$ be a totally ordered set. Let $\left({T, \preceq_2}\right)$ be an ordered set. Let $\phi: S \to T$ be a mapping. Then $\phi$ is a dual order embedding {{iff}} $\phi$ is strictly decreasing. That is: :$\forall x, y \in S: x \preceq_1 y \iff \phi \left({y}\right) \preceq_2 \phi \left({x}\right)$ {{iff}} :$\forall x, y \in S: x \prec_1 y \implies \phi \left({y}\right) \prec_2 \phi \left({x}\right)$


Let $\struct {S, \preceq_1}$ be a totally ordered set and let $\struct {T, \preceq_2}$ be an ordered set. Let $\phi: S \to T$ be a strictly increasing mapping. Then $\phi$ is an order embedding. <section notitle="true" name="proofs"> == Proof 1 == {{:Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing/Reverse Implication/Proof 1}} == Proof 2 == {{:Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing/Reverse Implication/Proof 2}} </section> Category:Order Embeddings Category:Total Orderings ip0v3n07lm6pfuiq8dh8hmlp58or3cs


Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing/Reverse Implication/Proof 2

Let $\struct {S, \preceq_1}$ be a totally ordered set and let $\struct {T, \preceq_2}$ be an ordered set. Let $\phi: S \to T$ be a strictly increasing mapping. Then $\phi$ is an order embedding.


Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing/Reverse Implication/Proof 1

Let $\struct {S, \preceq_1}$ be a totally ordered set. Let $\struct {T, \preceq_2}$ be an ordered set. Let $\phi: S \to T$ be a strictly increasing mapping. Then $\phi$ is an order embedding .


Foundational Relation is Antireflexive

Let $\mathcal R$ be a foundational relation on a set or class $A$. Then $\mathcal R$ is antireflexive.


Foundational Relation is Asymmetric

Let $\struct {S, \RR}$ be a relational structure, where $S$ is a set or a proper class. Let $\RR$ be a foundational relation. Then $\RR$ is asymmetric.


Upper Set with no Smallest Element is Open in GO-Space

Let $\struct {S, \preceq, \tau}$ be a generalized ordered space. Let $U$ be an upper set in $S$ with no smallest element. Then $U$ is open in $\struct {S, \preceq, \tau}$.


Lower Set with no Maximal Element

Let $\left({S, \preceq}\right)$ be an ordered set. Let $L \subseteq S$. Then: : $L$ is a lower set in $S$ with no maximal element {{iff}}: : $\displaystyle L = \bigcup \left\{{l^\prec: l \in L }\right\}$ where $l^\prec$ is the strict lower closure of $l$.


Lower Set with no Greatest Element is Open in GO-Space

Let $\struct {S, \preceq, \tau}$ be a generalized ordered space. Let $L$ be a lower set in $S$ with no greatest element. Then $L$ is open in $\struct {S, \preceq, \tau}$.


Lower Set is Dual to Upper Set

Let $\left({S, \preceq}\right)$ be an ordered set. Let $T \subseteq S$. The following are dual statements: :$T$ is a lower set in $S$ :$T$ is an upper set in $S$


Open Ray is Dual to Open Ray

Let $\struct {S, \preceq}$ be a totally ordered set. Let $R$ be an open ray in $\struct {S, \preceq}$. Then $R$ is an open ray in $\struct {S, \succeq}$, where $\succeq$ is the dual ordering of $\preceq$.


Topologies on Set form Complete Lattice

Let $X$ be a non-empty set. Let $\mathcal L$ be the set of topologies on $X$. Then $\left({\mathcal L, \subseteq}\right)$ is a complete lattice.


Complement of Lower Set is Upper Set

Let $\left({S, \preceq}\right)$ be an ordered set. Let $L$ be a lower set. Then $S \setminus L$ is an upper set.


GO-Space Embeds as Closed Subspace of Linearly Ordered Space

Let $(X, \preceq_X, \tau_X)$ be a generalized ordered space. Then there is a linearly ordered space $(Y, \preceq_Y, \tau_Y)$ and a mapping $\phi: X \to Y$ such that $\phi$ is a topological embedding and an order embedding, and $\phi(X)$ is closed in $Y$.


Union of Total Ordering with Lower Sets is Total Ordering

Let $\left({Y, \preceq}\right)$ be a totally ordered set. Let $X$ be the disjoint union of $Y$ with the set of lower sets of $Y$. Define a relation $\preceq'$ on $X$ extending $\preceq$ by letting: :$y_1 \preceq' y_2 \iff y_1 \preceq y_2$ :$y \preceq' L \iff y \in L$ :$L_1 \preceq' L_2 \iff L_1 \subseteq L_2$ :$L \preceq' y \iff y \in Y \setminus L$ Then $\preceq'$ is a total ordering.


Lower Closure is Lower Set

Let $\left({S, \preceq}\right)$ be an ordered set. Let $T$ be a subset of $S$. Let $L$ be the lower closure of $T$. Then $L$ is a lower set.


Ordered Set is Lower Set in Itself

Let $(S, \preceq)$ be an ordered set. Then $S$ is a lower set in $S$.


Ordered Set is Convex in Itself

Let $\struct {S, \preceq}$ be an ordered set. Then $S$ is a convex set in $S$.


Equivalence of Definitions of Generalized Ordered Space/Definition 3 implies Definition 1

Let $\left({S, \preceq, \tau}\right)$ be a generalized ordered space by Definition 3: {{:Definition:Generalized Ordered Space/Definition 3}} Then $\left({S, \preceq, \tau}\right)$ is a generalized ordered space by Definition 1: {{:Definition:Generalized Ordered Space/Definition 1}}


Strict Upper Closure is Upper Set

Let $(S, \preceq)$ be an ordered set. Let $p \in S$. Then $p^\succ$, the strict upper closure of $p$, is an upper set.


Strict Lower Closure is Lower Set

Let $\struct {S, \preceq}$ be an ordered set. Let $p \in S$. Then $p^\prec$, the strict lower closure of $p$, is a lower set.


Topology Discrete iff All Singletons Open

Let $(X, \tau)$ be a topological space. Then $\tau$ is the discrete topology on $X$ iff: : For all $x \in X$: $\{ x \} \in \tau$ That is, iff every singleton of $X$ is $\tau$-open.


Characteristic Function of Universe

Let $S$ be a set. Let $\chi_S: S \to \left\{ {0, 1}\right\}$ be its characteristic function (in itself). Then: :$\chi_S = f_1$ where $f_1: S \to \left\{ {0, 1}\right\}$ is the constant mapping with value $1$.


Faltings' Theorem

Let $C$ be a curve over $\Q$ of genus $g > 1$. Then $C$ has only finitely many rational points.


Supremum of Lower Closure of Set

Let $\left({S, \preceq}\right)$ be an ordered set. Let $T \subseteq S$. Let $L = T^\preceq$ be the lower closure of $T$ in $S$. Let $s \in S$ Then $s$ is the supremum of $T$ {{iff}} it is the supremum of $L$.


Upper Closure is Smallest Containing Upper Set

Let $\struct {S, \preceq}$ be an ordered set. Let $T \subseteq S$. Let $U = T^\succeq$ be the upper closure of $T$. Then $U$ is the smallest upper set containing $T$ as a subset.


Upper Closure is Closure Operator

Let $\left({S, \preceq}\right)$ be an ordered set. Let $T^\succeq$ be the upper closure of $T$ for each $T \subseteq S$. Then $\cdot^\succeq$ is a closure operator.


Equivalence of Definitions of Upper Set

Let $\left({S, \preceq}\right)$ be an ordered set. Let $U \subseteq S$. {{TFAE|def = Upper Set}}


Topological Closure is Closure Operator

The topological closure operator is a closure operator.


Reflexive Closure is Closure Operator

Let $S$ be a set. Let $R$ be the set of all endorelations on $S$. Then the reflexive closure operator on $R$ is a closure operator.


Set Closure is Smallest Closed Set/Closure Operator

Let $S$ be a set. Let $\cl: \powerset S \to \powerset S$ be a closure operator. Let $T \subseteq S$. Then $\map \cl T$ is the smallest closed set (with respect to $\cl$) containing $T$ as a subset.


Equivalence of Definitions of Lower Set

{{TFAE|def = Lower Set}} Let $\left({S, \preceq}\right)$ be an ordered set. Let $U \subseteq S$. Then the following are equivalent: {{begin-axiom}} {{axiom | n = 1         | m = \forall u \in U: \forall s \in S: s \preceq u \implies s \in U }} {{axiom | n = 2         | m = U^\preceq \subseteq U }} {{axiom | n = 3         | m = U^\preceq = U }} {{end-axiom}} where $U^\preceq$ is the lower closure of $U$.


Zero and One are the only Consecutive Perfect Squares/Proof 1

{{:Zero and One are the only Consecutive Perfect Squares}} <section title="Proof" name="proof"> Let $x$ and $h$ be integers such that $x^2 + 1 = \paren {x - h}^2$ {{begin-eqn}} {{eqn | l = x^2 + 1       | r = \left({x - h}\right)^2 }} {{eqn | l = 1       | r = -2 x h + h^2 }} {{eqn | l = 2 x h       | r = h^2 - 1 }} {{eqn | l = 2 x h       | r = \paren {h - 1} \paren {h + 1} }} {{end-eqn}} We have that Consecutive Integers are Coprime. However, both sides must have the same unique prime factorization by the Fundamental Theorem of Arithmetic Therefore $h$ cannot have any prime factors since they cannot be shared by $\paren {h - 1} \paren {h + 1}$. This leaves $h = -1$, $h = 0$, or $h = 1$ as the only possibilities since they are the only integers with no prime factors. If $h = -1$ then $h + 1 = 0$, so $2 x h = 0$. It follows that $x = 0$. If $h = 1$ then $h - 1 = 0$, so $2 x h = 0$. It follows that $x = 0$. If $h = 0$, then $2 x \cdot 0 = \paren {-1} \paren 1$, which is a contradiction. Therefore the only pairs of consecutive perfect squares are: :$0^2 = 0$ and $\paren {0 + \paren {-1} }^2 = \paren {-1}^2 = 1$ and: :$0^2 = 0$ and $\paren {0 + 1}^2 = 1^2 = 1$ {{qed}} </section> Category:Zero and One are the only Consecutive Perfect Squares htz0vf4du4yfpb36uyojnh9tye85uv2


Zero and One are the only Consecutive Perfect Squares/Proof 2

{{:Zero and One are the only Consecutive Perfect Squares}} <section title="Proof" name="proof"> Suppose that $k, l \in \Z$ are such that their squares are consecutive, i.e.: :$l^2 - k^2 = 1$ Then we can factor the left-hand side as: :$l^2 - k^2 = \left({l + k}\right) \left({l - k}\right)$ By Invertible Integers under Multiplication, it follows that: :$l + k = \pm 1 = l - k$ Therefore, it must be that: :$\left({l + k}\right) - \left({l - k}\right) = 0$ That is, $2 k = 0$, from which we conclude $k = 0$. So if $n$ and $n + 1$ are squares, then necessarily $n = 0$. The result follows. {{qed}} </section> Category:Zero and One are the only Consecutive Perfect Squares k8g8dzfo5fx677zrb5d7aaqbvi1ouhp


Convergent Series of Natural Numbers

Let $\left({a_n}\right)_{n \in \N}$ be a sequence of natural numbers. Then the following are equivalent: $(1): \quad \displaystyle \sum_{n \mathop = 1}^\infty a_n$ converges $(2): \quad \exists N \in \N: \forall n \ge N: a_n = 0$ That is, $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ converges {{iff}} only finitely many of the $a_n$ are non-zero.


Closure Operator from Closed Sets

Let $S$ be a set. Let $\CC$ be a set of subsets of $S$. Let $\CC$ be closed under arbitrary intersections: :$\forall \KK \in \powerset \CC: \bigcap \KK \in \CC$ where $\bigcap \O$ is taken to be $S$. Define $\cl: \powerset S \to \CC$ by letting: :$\map \cl T = \bigcap \set {C \in \CC: T \subseteq C}$ Then $\cl$ is a closure operator whose closed sets are the elements of $\CC$.


Intersection is Decreasing

Let $U$ be a set. Let $\mathcal F$ and $\mathcal G$ be sets of subsets of $U$. Then $\mathcal F \subseteq \mathcal G \implies \bigcap \mathcal G \subseteq \bigcap \mathcal F$, where by convention $\bigcap \varnothing = U$. That is, $\bigcap$ is a decreasing mapping from $(\mathcal P(\mathcal P(U)), \subseteq)$ to $(\mathcal P(U), \subseteq)$, where $\mathcal P(U)$ is the power set of $U$.


Open Ray is Open in GO-Space

Let $\left({S, \preceq, \tau}\right)$ be a generalized ordered space. Let $p \in S$. Then: : $p^\prec$ and $p^\succ$ are $\tau$-open where: : $p^\prec$ is the strict lower closure of $p$ : $p^\succ$ is the strict upper closure of $p$.


Union is Increasing

Let $U$ be a set. Let $\mathcal F$ and $\mathcal G$ be sets of subsets of $U$. Then $\mathcal F \subseteq \mathcal G \implies \bigcup \mathcal F \subseteq \bigcup \mathcal G$. That is, $\bigcup$ is an increasing mapping from $(\mathcal P(\mathcal P(U)), \subseteq)$ to $(\mathcal P(U), \subseteq)$, where $\mathcal P(U)$ is the power set of $U$.


Intersection is Idempotent/Indexed Family

Let $\family {F_i}_{i \mathop \in I}$ be a non-empty indexed family of sets. Suppose that all the sets in the $\family {F_i}_{i \mathop \in I}$ are the same. That is, suppose that for some set $S$: :$\forall i \in I: F_i = S$ Then: :$\displaystyle \bigcap_{i \mathop \in I} F_i = S$ where $\displaystyle \bigcap_{i \mathop \in I} F_i$ is the intersection of $\family {F_i}_{i \mathop \in I}$.


Factor Principles/Conjunction on Left/Formulation 1/Proof 2

:$p \implies q \vdash \paren {r \land p} \implies \paren {r \land q}$


Factor Principles/Conjunction on Left/Formulation 1/Proof 1

: $p \implies q \vdash \paren {r \land p} \implies \paren {r \land q}$


Constructive Dilemma for Join Semilattices

Let $\struct {S, \vee, \preceq}$ be a join semilattice. Let $a, b, c, d \in S$. Let $a \preceq b$. Let $c \preceq d$. Then $\paren {a \vee c} \preceq \paren {b \vee d}$.


Praeclarum Theorema for Meet Semilattices

Let $(S, \wedge, \preceq)$ be a meet semilattice. Let $a, b, c, d \in S$. Let $a \preceq b$. Let $c \preceq d$. Then $(a \wedge c) \preceq (b \wedge d)$.


Supremum is Increasing relative to Product Ordering

Let $(S, \preceq)$ be an ordered set. Let $I$ be a set. Let $f, g: I \to S$. Let $f \left[{I}\right]$ denote the image of $I$ under $f$. Let: :$\forall i \in I: f \left({i}\right) \preceq g \left({i}\right)$ That is, let $f \preceq g$ in the product ordering. Let $f \left[{I}\right]$ and $g \left[{I}\right]$ admit suprema. Then: : $\sup f \left[{I}\right] \preceq \sup g \left[{I}\right]$


Reflexive Closure of Strict Ordering is Ordering

Let $S$ be a set. Let $\prec$ be a strict ordering on $S$. Let $\preceq$ be the reflexive closure of $\prec$. Then $\preceq$ is an ordering.


Reflexive Closure is Reflexive

Let $\mathcal R$ be a relation on a set $S$. Then $\mathcal R^=$, the reflexive closure of $\mathcal R$, is reflexive.


Reflexive Closure is Inflationary

Let $S$ be a set. Let $R$ denote the set of all endorelations on $S$. Then the reflexive closure operator is an inflationary mapping on $R$.


Reflexive Closure is Order Preserving

Let $S$ be a set. Let $R$ denote the set of all endorelations on $S$. Then the reflexive closure operator is an order preserving mapping on $R$. That is: :$\forall \RR, \SS \in R: \RR \subseteq \SS \implies \mathcal R^= \subseteq \SS^=$ where $\RR^=$ and $\SS^=$ denote the reflexive closure of $\RR$ and $\SS$ respectively.


Reflexive Closure is Idempotent

Let $S$ be a set. Let $R$ denote the set of all endorelations on $S$. Then the reflexive closure operator is an idempotent mapping on $R$. That is: :$\forall \RR \in R: \RR^= = \paren {\RR^=}^=$ where $\RR^=$ denotes the reflexive closure of $\RR$.


Transitive Closure Always Exists (Relation Theory)/Proof 1

Let $\mathcal R$ be a relation on a set $S$. Then the transitive closure $\mathcal R^+$ of $\mathcal R$ always exists.


Transitive Closure Always Exists (Relation Theory)/Proof 2


Equivalence of Definitions of Transitive Closure (Relation Theory)/Union of Compositions is Smallest

Let $\RR$ be a relation on a set $S$. Let: :$\RR^n := \begin{cases} \RR & : n = 0 \\ \RR^{n - 1} \circ \RR & : n > 0 \end{cases}$ where $\circ$ denotes composition of relations. {{explain|Really? I would have thought $\RR^1 {{=}} \RR$, not $\RR^0 {{=}} \RR$. If anything, the diagonal relation $\Delta_S$ should be $\RR^0$.}} Finally, let: :$\displaystyle \RR^+ = \bigcup_{i \mathop \in \N} \RR^i$ Then $\RR^+$ is the smallest transitive relation on $S$ that contains $\RR$.


Equivalence of Definitions of Transitive Closure (Relation Theory)/Finite Chain Equivalent to Union of Compositions

The finite chain and union of compositions definitions of '''transitive closure''' are equivalent.


Chasles' Relation

Let $\mathcal E$ be an affine space. Let $p, q, r \in \mathcal E$ be points. Then: :$\vec{p q} = \vec{p r} + \vec{r q}$


Affine Coordinates are Well-Defined

Let $\mathcal E$ be an affine space with difference space $V$ over a field $k$. Let $\mathcal R = \left({p_0, e_1, \ldots, e_n}\right)$ be an affine frame in $\mathcal E$. Define a mapping $\Theta_{\mathcal R} : k^n \to \mathcal E$ by: :$\displaystyle \Theta_\mathcal R \left({\lambda_1, \ldots, \lambda_n}\right) = p_0 + \sum_{i \mathop = 1}^n \lambda_i e_i$ Then $\Theta_\mathcal R$ is a bijection.


Transitive Chaining

Let $\mathcal R$ be a transitive relation on a set $S$. Let $n \in \N$ be a natural number. Let $n \ge 2$. Let $\langle x_k \rangle_{k \in \left\{ {1, 2, \dots, n}\right\} }$ be a sequence of $n$ terms. For each $k \in \left\{ {1, 2, \dots, n-1}\right\}$, let $x_k \mathrel {\mathcal R} x_{k+1}$. That is, let $x_1 \mathrel {\mathcal R} x_2$, $x_2 \mathrel {\mathcal R} x_3, \dotsc, x_n-1 \mathrel {\mathcal R} x_n$. Then $x_1 \mathrel {\mathcal R} x_n$


Szpilrajn Extension Theorem

Let $\struct {S, \prec}$ be a strictly ordered set. {{Disambiguate|Definition:Strictly Ordered Set}} Then there is a strict total ordering on $S$ of which $\prec$ is a subset.


Strict Ordering can be Expanded to Compare Additional Pair

Let $\left({S, \prec}\right)$ be an ordered set. Let $a$ and $b$ be distinct, $\prec$-incomparable elements of $S$. That is, let: :$a \not \prec b$ and $b \not \prec a$. Let ${\prec'} = {\prec} \cup \left\{ {\left({a, b}\right)} \right\}$. Define a relation $\prec'^+$ by letting $p \prec'^+ q$ {{iff}}: : $p \prec q$ or: : $p \preceq a$ and $b \preceq q$ where $\preceq$ is the reflexive closure of $\prec$. Then: : $\prec'^+$ is a strict ordering : $\prec^+$ is the transitive closure of $\prec'$.


Characterization of Interior of Triangle

Let $\triangle$ be a triangle embedded in $\R^2$. Denote the vertices of $\triangle$ as $A_1, A_2, A_3$. For $i \in \set {1, 2, 3}$, put $j = i \bmod 3 + 1$, $k = \paren {i + 1} \bmod 3 + 1$, and: :$U_i = \set {A_i + s t \paren {A_j - A_i} + \paren {1 - s} t \paren {A_k - A_i} : s \in \openint 0 1, t \in \R_{>0} }$ Then: :$\displaystyle \Int \triangle = \bigcap_{i \mathop = 1}^3 U_i$ where $\Int \triangle$ denotes the interior of the boundary of $\triangle$.


Equivalence of Definitions of Reflexive Transitive Closure

Let $\mathcal R$ be a relation on a set $S$. {{TFAE|def = Reflexive Transitive Closure}}


Intersection of Relation with Inverse is Symmetric Relation

Let $\mathcal R$ be a relation on a set $S$. Then $\mathcal R \cap \mathcal R^{-1}$, the intersection of $\mathcal R$ with its inverse, is symmetric.


Intersection of Closed Sets is Closed/Closure Operator

Let $S$ be a set. Let $f: \powerset S \to \powerset S$ be a closure operator on $S$. Let $\CC$ be the set of all subsets of $S$ that are closed with respect to $f$. Let $\AA \subseteq \CC$. Then $\bigcap \AA \in \CC$.


Closure is Closed/Power Set

Let $S$ be a set. Let $\powerset S$ be the power set of $S$. Let $\cl: \powerset S \to \powerset S$ be a closure operator. Let $T \subseteq S$. Then $\map \cl T$ is a closed set with respect to $\cl$.


Relation Intersection Inverse is Greatest Symmetric Subset of Relation

Let $\RR$ be a relation on a set $S$. Let $\powerset \RR$ be the power set of $\RR$. By definition, $\powerset \RR$ is the set of all relations on $S$ that are subsets of $\RR$. Then the greatest element of $\powerset \RR$ that is symmetric is: :$\RR \cap \RR^{-1}$


Composition of Compatible Closure Operators

Let $S$ be a set. Let $f, g: \mathcal P \left({S}\right) \to \mathcal P \left({S}\right)$ be closure operators on $S$. Let $\mathcal C_f$ and $\mathcal C_g$ be the sets of closed sets of $S$ with respect to $f$ and $g$ respectively. For each subset $T$ of $S$, let the following hold: : $(1): \quad$ If $T$ is closed with respect to $g$, then $f \left({T}\right)$ is closed with respect to $g$. :: That is, if $T \in \mathcal C_g$ then $f \left({T}\right) \in \mathcal C_g$. : $(2): \quad$ If $T$ is closed with respect to $f$, then $g \left({T}\right)$ is closed with respect to $f$. :: That is, if $T \in \mathcal C_f$ then $g \left({T}\right) \in \mathcal C_f$. Let $\mathcal C_h = \mathcal C_f \cap \mathcal C_g$. Then: : $\mathcal C_h$ induces a closure operator $h$ on $S$ : $f \circ g = g \circ f = h$, where $\circ$ represents composition of mappings.


Closure Operator from Closed Elements

Let $\struct {S, \preceq}$ be an ordered set. Let $C \subseteq S$. Suppose that $C$ is a subset of $S$ with the property that every element of $S$ has a smallest successor in $C$. Let $\cl: S \to S$ be defined as follows: For $x \in S$: :$\map \cl x = \map \min {C \cap x^\succeq}$ where $x^\succeq$ is the upper closure of $x$. That is, let $\map \cl x$ be the smallest successor of $x$ in $C$. Then: :$\cl$ is a closure operator on $S$ :The closed elements of $\cl$ are precisely the elements of $C$.


Product of Affine Spaces is Affine Space

Let $\mathcal E, \mathcal F$ be affine spaces. Let $\mathcal G = \mathcal E \times \mathcal F$ be the product of $\mathcal E$ and $\mathcal F$. Then $\mathcal G$ is an affine space.


Intersection of Complete Meet Subsemilattices invokes Closure Operator

Let $\struct {S, \preccurlyeq}$ be an ordered set. Let $f_i$ be a closure operator on $S$ for each $i \in I$. Let $C_i = \map {f_i} S$ be the set of closed elements with respect to $f_i$ for each $i \in I$. Suppose that for each $i \in I$, $C_i$ is a '''complete meet subsemilattice''' of $S$ in the following sense: :For each $D \subseteq C_i$, $D$ has an infimum in $S$ such that $\inf D \in C_i$. Then $C = \displaystyle \bigcap_{i \mathop \in I} C_i$ induces a closure operator on $S$.


Reflexive Reduction of Ordering is Strict Ordering

Let $\mathcal R$ be an ordering on a set $S$. Let $\mathcal R^\ne$ be the reflexive reduction of $\mathcal R$. Then $\mathcal R^\ne$ is a strict ordering on $S$.


Vectorialization of Affine Space is Vector Space

Let $\mathcal E$ be an affine space over a field $k$ with difference space $E$. Let $\mathcal R = \tuple {p_0, e_1, \ldots, e_n}$ be an affine frame in $\mathcal E$. Let $\struct {\mathcal E, +, \cdot}$ be the vectorialization of $\mathcal E$. Then $\struct {\mathcal E, +, \cdot}$ is a vector space.


Reflexive Reduction is Antireflexive

Let $\mathcal R$ be a relation on a set $S$. Let $\mathcal R^\ne$, be the reflexive reduction of $\mathcal R$. Then $\mathcal R^\ne$ is antireflexive.


Subband iff Idempotent under Induced Operation

Let $\left({S, \circ}\right)$ be a band. Let $\left({\mathcal P \left({S}\right), \circ_\mathcal P}\right)$ be the algebraic structure consisting of the power set of $S$ and the operation induced on $\mathcal P \left({S}\right)$ by $\circ$. Let $X \in \mathcal P \left({S}\right)$. Then $X$ is idempotent {{iff}} $\left({X, \circ}\right)$ is a subband of $\left({S, \circ}\right)$.


Restriction of Idempotent Operation is Idempotent

Let $\left({S, \circ}\right)$ be an algebraic structure. Let $T \subseteq S$. Let the operation $\circ$ be idempotent. Then $\circ$ is also idempotent upon restriction  to $\left({T, \circ \restriction_T}\right)$.


Composition of Commuting Idempotent Mappings is Idempotent

Let $S$ be a set. Let $f, g: S \to S$ be idempotent mappings from $S$ to $S$. Let: :$f \circ g = g \circ f$ where $\circ$ denotes composition. Then $f \circ g$ is idempotent.


Compositions of Closure Operators are both Closure Operators iff Operators Commute

Let $\left({S, \preceq}\right)$ be an ordered set. Let $f$ and $g$ be closure operators on $S$. Then the following are equivalent: :$(1): \quad f \circ g$ and $g \circ f$ are both closure operators. :$(2): \quad f$ and $g$ commute (that is, $f \circ g = g \circ f$). :$(3): \quad \operatorname{img}\left({f \circ g}\right) = \operatorname{img}\left({g \circ f}\right)$ where $\operatorname{img}$ represents the image of a mapping.


Composition of Increasing Mappings is Increasing

Let $\struct {S, \preceq_S}$, $\struct {T, \preceq_T}$ and $\struct {U, \preceq_U}$ be ordered sets. Let $g: S \to T$ and $f: T \to U$ be increasing mappings. Then their composition $f \circ g: S \to U$ is also increasing.


Composition of Inflationary Mappings is Inflationary

Let $\left({S, \preceq}\right)$ be an ordered set. Let $f, g: S \to S$ be inflationary mappings. Then $f \circ g$, the composition of $f$ and $g$, is also inflationary.


Fixed Point of Idempotent Mapping

Let $S$ be a set. Let $f: S \to S$ be an idempotent mapping. Let $f \left[{S}\right]$ be the image of $S$ under $f$. Let $x \in S$. Then $x$ is a fixed point of $f$ {{iff}} $x \in f \left[{S}\right]$.


Symmetric Closure of Ordering may not be Transitive

Let $\left({S, \preceq}\right)$ be an ordered set. Let $\preceq^\leftrightarrow$ be the symmetric closure of $\preceq$. Then it is not necessarily the case that $\preceq^\leftrightarrow$ is transitive.


Composition of Idempotent Mappings

Let $S$ be a set. Let $f, g: S \to S$ be idempotent mappings. Suppose that $f \circ g$ and $g \circ f$ have the same images. That is, suppose that $f \sqbrk {g \sqbrk S} = g \sqbrk {f \sqbrk S}$. Then $f \circ g$ and $g \circ f$ are idempotent.


Composition of Inflationary and Idempotent Mappings

Let $\struct {S, \preceq}$ be an ordered set. Let $f$ and $g$ be inflationary and idempotent mappings on $S$. Then the following are equivalent: :$(1): \quad f \circ g$ and $g \circ f$ are both idempotent :$(2): \quad f$ and $g$ commute (that is, $f \circ g = g \circ f$) :$(3): \quad \Img {f \circ g} = \Img {g \circ f}$ where: :$\circ$ represents composition :$\Img f$ represents the image of a mapping $f$.


Closure is Smallest Closed Successor

Let $\struct {S, \preceq}$ be an ordered set. Let $f: S \to S$ be a closure operator on $S$. Let $x \in S$. Then $\map f x$ is the smallest closed element that succeeds $x$.


Closed Elements Uniquely Determine Closure Operator

Let $\left({S, \preceq}\right)$ be an ordered set. Let $f, g: S \to S$ be closure operators on $S$. Suppose that $f$ and $g$ have the same closed elements. Then $f = g$.


Square of Number Always Exists

Let $x$ be a number. Then its square $x^2$ is guaranteed to exist.


Schröder Rule

Let $A$, $B$ and $C$ be relations on a set $S$. Then the following are equivalent statements: :$(1): \quad A \circ B \subseteq C$ :$(2): \quad A^{-1} \circ \overline C \subseteq \overline B$ :$(3): \quad \overline C \circ B^{-1} \subseteq \overline A$ where: : $\circ$ denotes relation composition : $A^{-1}$ denotes the inverse of $A$ : $\overline A$ denotes the complement of $A$.


Equivalence of Definitions of Dual Relation

{{TFAE|def = Dual Relation}} Let $\RR \subseteq S \times T$ be a relation.


Trivial Gradation is Gradation

Let $\left({R, +, \circ}\right)$ be a ring. Let $\left({M, e,\cdot}\right)$ be a monoid. Let :$\displaystyle R = \bigoplus_{m \mathop \in M} R_m$ be the trivial $M$-gradation on $R$. This is a gradation on $R$.


Fixed Point of Mappings is Fixed Point of Composition

Let $S$ be a set. Let $f, g: S \to S$ be mappings. Let $x \in S$ be a fixed point of both $f$ and $g$. Then $x$ is also a fixed point of $f \circ g$, the composition of $f$ and $g$.


Fixed Point of Mappings is Fixed Point of Composition/General Result

Let $S$ be a set. Let $n \in \N$ be a strictly positive integer. Let $\N_n$ be the initial segment of $n$ in $\N$. That is, let $\N_n = \left\{{0, 1, \dots, n-1}\right\}$. For each $i \in \N_n$, let $f_i: S \to S$ be a mapping. Let $x \in S$ be a fixed point of $f_i$ for each $i \in \N_n$. Let $g = f_0 \circ f_1 \circ \dots \circ f_{n-1}$ be the composition of all the $f_i$s. Then $x$ is a fixed point of $g$.


Polynomial has Integer Coefficients iff Content is Integer

$f$ has integer coefficients {{iff}} $\cont f$ is an integer.


Content of Scalar Multiple

:$\cont {q f} = q \cont f$


Fixed Point of Composition of Inflationary Mappings

Let $\left({S, \preceq}\right)$ be a ordered set. Let $f, g: S \to S$ be inflationary mappings. Let $x \in S$. Then: : $x$ is a fixed point of $f \circ g$ {{iff}}: : $x$ is a fixed point of both $f$ and $g$.


Group of Units is Group

Let $\struct {R, +, \circ}$ be a ring with unity. Then the set of units of $\struct {R, +, \circ}$ forms a group under $\circ$. Hence the justification for referring to the group of units of $\struct {R, +, \circ}$.


Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal/Forward Implication

Let $\struct {D, +, \circ}$ be a principal ideal domain. Let $p$ be an irreducible element of $D$. Let $\ideal p$ be the principal ideal of $D$ generated by $p$. Then $\ideal p$ is a maximal ideal of $D$.


Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal/Reverse Implication

Let $\left({D, +, \circ}\right)$ be a principal ideal domain. Let $\left({p}\right)$ be the principal ideal of $D$ generated by $p$. Let $\left({p}\right)$ be a maximal ideal of $D$. Then $p$ is irreducible.


Subring of Polynomials over Integral Domain Contains that Domain

Let $\struct {R, +, \circ}$ be a commutative ring. Let $\struct {D, +, \circ}$ be an integral subdomain of $R$. Let $x \in R$. Let $D \sqbrk x$ denote the ring of polynomials in $x$ over $D$. Then $D \sqbrk x$ contains $D$ as a subring and $x$ as an element.


Subring of Polynomials over Integral Domain is Smallest Subring containing Element and Domain

Let $\struct {R, +, \circ}$ be a commutative ring. Let $\struct {D, +, \circ}$ be an integral subdomain of $R$. Let $x \in R$. Let $D \sqbrk x$ denote the ring of polynomials in $x$ over $D$. Then $D \sqbrk x$ is the smallest subring of $R$ which contains $D$ as a subring and $x$ as an element.


Cantor-Bernstein-Schröder Theorem/Lemma

Let $S$ be a set. Let $T \subseteq S$. Suppose that $f: S \to T$ is an injection. Then there is a bijection $g: S \to T$.


Cantor-Bernstein-Schröder Theorem/Proof 5

Let $S$ and $T$ be sets. Let $f: S \to T$ and $g: T \to S$ be injections. Then there exists a bijection $\phi: S \to T$.


Quotient Space of Real Line may be Indiscrete

Let $\struct {\R, \tau}$ denote the real number line with the usual (Euclidean) topology. Let $\Q$ be the rational numbers. Let $\mathbb I$ be the irrational numbers. Then $\set {\Q, \mathbb I}$ is a partition of $\R$. Let $\sim$ be the equivalence relation induced on $\R$ by $\set {\Q, \mathbb I}$. Let $T_\sim := \struct {\R / {\sim}, \tau_\sim} $ be the quotient space of $\R$ by $\sim$. Then $T_\sim$ is an indiscrete space.


Quotient Space of Real Line may not be Kolmogorov

Let $\struct {\R, \tau}$ denote the real number line with the usual (Euclidean) topology. Then there exists an equivalence relation $\sim$ on $\R$ such that the quotient space $\struct {\R / {\sim}, \tau_\sim}$ is not Kolmogorov.


Quotient Space of Real Line may be Kolmogorov but not Fréchet

Let $\struct {\R, \tau}$ denote the real number line with the usual (Euclidean) topology. Define an equivalence relation $\sim$ by letting $x \sim y$ {{iff}} either: :$x = y$ or: :$x, y \in \Q$ Let $\struct {\R / {\sim}, \tau_\sim}$ be the quotient space of $\R$ by $\sim$. Then $\struct {\R / {\sim}, \tau_\sim}$ is a Kolmogorov space but not a Fréchet space.


Unique Representation in Polynomial Forms/General Result

Let $f$ be a polynomial form in the indeterminates $\set {X_j: j \in J}$ such that $f: \mathbf X^k \mapsto a_k$. For $r \in \R$, $\mathbf X^k \in M$, let $r \mathbf X^k$ denote the polynomial form that takes the value $r$ on $\mathbf X^k$ and zero on all other monomials. Let $Z$ denote the set of all multiindices indexed by $J$. Then the sum representation: :$\ds \hat f = \sum_{k \mathop \in Z} a_k \mathbf X^k$ has only finitely many non-zero terms. Moreover it is everywhere equal to $f$, and is the unique such sum.


Law of Excluded Middle for Two Variables

:$\vdash (p \land q) \lor (\lnot p \land q) \lor (p \land \lnot q) \lor (\lnot p \land \lnot q)$


Ring of Polynomial Forms is Commutative Ring with Unity

Let $\struct {R, +, \circ}$ be a commutative ring with unity. Let $A = R \sqbrk {\set {X_j: j \in J} }$ be the set of all polynomial forms over $R$ in the indeterminates $\set {X_j: j \in J}$. Then $\struct {A, +, \circ}$ is a commutative ring with unity.


Idempotent Elements form Subsemigroup of Commutative Semigroup

Let $\struct {S, \circ}$ be a semigroup such that $\circ$ is commutative. Let $I$ be the set of all elements of $S$ that are idempotent under $\circ$. That is: :$I = \set {x \in S: x \circ x = x}$ Then $\struct {I, \circ}$ is a subsemigroup of $\struct {S, \circ}$.


Product of Commuting Idempotent Elements is Idempotent

Let $\struct {S, \circ}$ be a semigroup. Let $a, b \in S$ be idempotent elements of $S$. Let $a$ and $b$ commute: :$a \circ b = b \circ a$ Then $a \circ b$ is idempotent.


Set of All Self-Maps is Semigroup

Let $S$ be a set. Let $S^S$ be the set of all mappings from $S$ to itself. Let the operation $\circ$ represent composition of mappings. Then the algebraic structure $\struct {S^S, \circ}$ is a semigroup.


Integers form Subdomain of Reals

The integral domain of integers $\struct {\Z, +, \times}$ forms a subdomain of the field of real numbers.


Identity Element is Idempotent

Let $\struct {S, \circ}$ be an algebraic structure. Let $e \in S$ be an identity with respect to $\circ$. Then $e$ is idempotent under $\circ$.


Identity of Algebraic Structure is Preserved in Substructure

Let $\struct {S, \circ}$ be an algebraic structure with identity $e$. Let $\struct {T, \circ}$ be a algebraic substructure of $\struct {S, \circ}$. That is, let $T \subseteq S$. Let $e \in T$. Then $e$ is an identity of $\struct {T, \circ}$.


Idempotent Elements form Submonoid of Commutative Monoid

Let $\left({S, \circ}\right)$ be a commutative monoid. Let $e \in S$ be the identity element of $\left({S, \circ}\right)$. Let $I$ be the set of all elements of $S$ that are idempotent under $\circ$. That is: :$I = \left\{{x \in S: x \circ x = x}\right\}$ Then $\left({I, \circ}\right)$ is a submonoid of $\left({S, \circ}\right)$ with identity $e$.


Inverse Image under Order Embedding of Strict Upper Closure of Image of Point

Let $\struct {S, \preceq}$ and $\struct {T, \preceq'}$ be ordered sets. Let $\phi: S \to T$ be an order embedding of $\struct {S, \preceq}$ into $\struct {T, \preceq'}$ Let $p \in S$. Then: :$\map {\phi^{-1} } {\map \phi p^{\succ'} } = p^\succ$ where $\cdot^\succ$ and $\cdot^{\succ'}$ represent strict upper closure with respect to $\preceq$ and $\preceq'$, respectively.


Equivalence of Definitions of Order Embedding/Definition 1 implies Definition 3

Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets. Let $\phi: S \to T$ be a mapping. Let $\phi: S \to T$ be an order embedding by Definition 1: {{:Definition:Order Embedding/Definition 1}} Then $\phi: S \to T$ is an order embedding by Definition 3: {{:Definition:Order Embedding/Definition 3}}


Equivalence of Definitions of Order Embedding/Definition 3 implies Definition 1

Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets. Let $\phi: S \to T$ be a mapping. Let $\phi: S \to T$ be an order embedding by Definition 3: {{:Definition:Order Embedding/Definition 3}} Then $\phi: S \to T$ is an order embedding by Definition 1: {{:Definition:Order Embedding/Definition 1}}


Path as Parameterization of Contour

Let $\left[{a \,.\,.\, b}\right]$ be a closed real interval. Let $\gamma: \left[{a \,.\,.\, b}\right] \to \C$ be a path. Let there exist $n \in \N$ and a subdivision $\left\{{a_0, a_1, \ldots, a_n}\right\}$ of $\left[{a \,.\,.\, b}\right]$ such that: : $\gamma {\restriction_{ \left[{a_{k - 1} \,.\,.\, a_k}\right] } }$ is a smooth path for all $k \in \left\{ {1, \ldots, n}\right\}$ where $\gamma {\restriction_{\left[{a_{k - 1} \,.\,.\, a_k}\right]} }$ denotes the restriction of $\gamma$ to $\left[{a_{k - 1} \,.\,.\, a_k}\right]$. Then there exists a contour $C$ with parameterization $\gamma$ and these properties: :$(1): \quad$ If $\gamma$ is a closed path, then $C$ is a closed contour. :$(2): \quad$ If $\gamma$ is a Jordan arc, then $C$ is a simple contour. :$(3): \quad$ If $\gamma$ is a Jordan curve, then $C$ is a simple closed contour.


Kernel of Induced Homomorphism of Polynomial Forms

Let $R$ and $S$ be commutative rings with unity. Let $\phi: R \to S$ be a ring homomorphism. Let $K = \ker \phi$. Let $R \left[{X}\right]$ and $S \left[{X}\right]$ be the rings of polynomial forms over $R$ and $S$ respectively in the indeterminate $X$. Let $\bar\phi: R \left[{X}\right] \to S \left[{X}\right]$ be the induced morphism of polynomial rings. Then the kernel of $\bar\phi$ is: :$\ker \bar\phi = \left\{{ a_0 + a_1 X + \cdots + a_n X^n \in R \left[{X}\right] : \phi \left({a_i}\right) = 0 \text{ for } i = 0, \ldots, n }\right\}$ Or, more concisely: :$\ker \bar\phi = \left({\ker \phi}\right) \left[{X}\right]$


Boundary of Polygon as Contour

Let $P$ be a polygon embedded in the complex plane $\C$. Denote the boundary of $P$ as $\partial P$. Then there exists a simple closed contour $C$ such that: : $\operatorname{Im} \left({C}\right) = \partial P$ where $\operatorname{Im} \left({C}\right)$ denotes the image of $C$.


Zero Simple Staircase Integral Condition for Primitive

Let $f: D \to \C$ be a continuous complex function, where $D$ is a connected domain. Let $\displaystyle \oint_C \map f z \rd z = 0$ for all simple closed staircase contours $C$ in $D$. Then $f$ has a primitive $F: D \to \C$.


Preordering induces Equivalence Relation

Let $\struct {S, \precsim}$ be a preordered set. Define a relation $\sim$ on $S$ by letting $x \sim y$ {{iff}} $x \precsim y$ and $y \precsim x$. Then $\sim$ is an equivalence relation.


Content of Rational Polynomial is Multiplicative

Let $h \in \Q \sqbrk X$ be a polynomial with rational coefficients. Let $\cont h$ denote the content of $h$. Then for any polynomials $f, g \in \Q \sqbrk X$ with rational coefficients: :$\cont {f g} = \cont f \cont g$


Gauss's Lemma on Irreducible Polynomials

Let $\Z$ be the ring of integers. Let $\Z \sqbrk X$ be the ring of polynomials over $\Z$. Let $h \in \Z \sqbrk X$ be a polynomial. {{TFAE}} :$(1): \quad h$ is irreducible in $\Q \sqbrk X$ and primitive :$(2): \quad h$ is irreducible in $\Z \sqbrk X$.


Antisymmetric Quotient of Preordered Set is Ordered Set

Let $\struct {S, \precsim}$ be a preordered set. Let $\sim$ be the equivalence relation on $S$ induced by $\precsim$. Let $\struct {S / {\sim}, \preceq}$ be the antisymmetric quotient of $\struct {S, \precsim}$. Then: :$\struct {S / {\sim}, \preceq}$ is an ordered set. :$\forall P, Q \in S / {\sim}: \paren {P \preceq Q} \land \paren {p \in P} \land \paren {q \in Q} \implies p \precsim q$ This second statement means that we could just as well have defined $\preceq$ by letting $P \preceq Q$ iff: :$\forall p \in P: \forall q \in Q: p \precsim q$


Ordering on Partition Determines Preordering

Let $S$ be a set. Let $\PP$ be a partition of $S$. Let $\phi: S \to \PP$ be the quotient mapping. Let $\preceq$ be a ordering of $\PP$. Define a relation $\precsim$ on $S$ by letting $p \precsim q$ {{iff}}: :$\map \phi p \preceq \map \phi q$ Then: :$\precsim$ is a preordering on $S$. :$\precsim$ is the only preordering on $S$ that induces the $\preceq$ ordering on $\PP$.


Units of Ring of Polynomial Forms over Commutative Ring

Let $\struct {R, +, \circ}$ be a non-null commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$. Let $R \sqbrk X$ be the ring of polynomial forms in an indeterminate $X$ over $R$. Let $\map P X = a_0 + a_1 X + \cdots + a_n X^n \in R \sqbrk X$. Then: :$\map P X$ is a unit of $R \sqbrk X$ {{iff}}: :$a_0$ is a unit of $R$ Also, for $i = 1, \ldots, n$, $a_i$ is nilpotent in $R$.


Polynomials Closed under Addition/Polynomials over Integral Domain

Let $\left({R, +, \circ}\right)$ be a commutative ring with unity. Let $\left({D, +, \circ}\right)$ be an integral subdomain of $R$. Then $\forall x \in R$, the set $D \left[{x}\right]$ of polynomials in $x$ over $D$ is closed under the operation $+$.


Polynomials Closed under Addition/Polynomials over Ring

Let $\left({R, +, \circ}\right)$ be a ring. Let $\left({S, +, \circ}\right)$ be a subring of $R$. Then $\forall x \in R$, the set $S \left[{x}\right]$ of polynomials in $x$ over $S$ is closed under the operation $+$.


Nilpotent Element is Zero Divisor

Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$. Suppose further that $R$ is not the null ring. Let $x \in R$ be a nilpotent element of $R$. Then $x$ is a zero divisor in $R$.


Units of Ring of Polynomial Forms over Integral Domain

Let $\struct {D, +, \circ}$ be an integral domain. Let $D \sqbrk X$ be the ring of polynomial forms in an indeterminate $X$ over $D$. Then the group of units of $D \sqbrk X$ is precisely the group of elements of $D \sqbrk X$ of degree zero that are units of $D$.


Kernel of Magma Homomorphism is Submagma

Let $\left({S, *}\right)$ and $\left({T, \circ}\right)$ be algebraic structures. Let $\left({T, \circ}\right)$ have an identity $e$. Let $\phi: S \to T$ be a magma homomorphism. Then the kernel of $\phi$ is a submagma of $\left({S, *}\right)$. That is: :$\left({\phi^{-1} \left({e}\right), *}\right)$ is a submagma of $\left({S, *}\right)$ where $\phi^{-1} \left({e}\right)$ denote the preimage of $e$.


Preimage of Zero of Homomorphism is Submagma

Let $\struct {S, *}$ be a magma. Let $\struct {T, \circ}$ be a magma with a zero element $0$. Let $\phi: S \to T$ be a magma homomorphism. Then $\struct {\phi^{-1} \sqbrk 0, *}$ is a submagma of $\struct {S, *}$.


Polynomial over Field is Reducible iff Scalar Multiple is Reducible

Let $K$ be a field. Let $K \left[{X}\right]$ be the ring of polynomial forms over $K$. Let $P \in K \left[{X}\right]$. Let $\lambda \in K \setminus \left\{{0}\right\}$. Then $P$ is irreducible in $K \left[{X}\right]$ iff $\lambda P$ is also irreducible in $K \left[{X}\right]$. {{expand|Investigate whether this result also holds where $K$ is a general ring.}}


Conjunction of Disjunctions Consequence

:$\left({p \lor q}\right) \land \left({r \lor s}\right) \vdash p \lor r \lor \left({q \land s}\right)$


Existence of Ring of Polynomial Forms in Transcendental over Integral Domain

Let $\struct {R, +, \circ}$ be a commutative ring with unity. Let $\struct {D, +, \circ}$ be an integral subdomain of $R$ whose zero is $0_D$. Let $X \in R$ be transcendental over $D$ Then the ring of polynomials $D \sqbrk X$ in $X$ over $D$ exists.


Eisenstein Integers form Integral Domain

The ring of Eisenstein integers $\struct {\Z \sqbrk \omega, +, \times}$ is an integral domain.


Eisenstein Integers form Subring of Complex Numbers

The set of Eisenstein integers $\Z \sqbrk \omega$, under the operations of complex addition and complex multiplication, forms a subring of the set of complex numbers $\C$.


Polynomial Forms over Field form Integral Domain/Formulation 2

Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$. Let $\GF$ be the set of all polynomials over $\struct {F, +, \circ}$ defined as sequences. Let polynomial addition and polynomial multiplication be defined as: :$\forall f = \sequence {a_k} = \tuple {a_0, a_1, a_2, \ldots}, g = \sequence {b_k} = \tuple {b_0, b_1, b_2, \ldots} \in \GF$: ::$f \oplus g := \tuple {a_0 + b_0, a_1 + b_1, a_2 + b_2, \ldots}$ ::$f \otimes g := \tuple {c_0, c_1, c_2, \ldots}$ where $\displaystyle c_i = \sum_{j \mathop + k \mathop = i} a_j \circ b_k$ Then $\struct {\GF, \oplus, \otimes}$ is an integral domain.


Maximal Spectrum of Ring is Nonempty

Let $A$ be a non-trivial commutative ring with unity. Then its maximal spectrum is non-empty: :$\operatorname {Max} \Spec A \ne \O$


Knaster-Tarski Lemma

Let $\left({L, \preceq}\right)$ be a complete lattice. Let $f: L \to L$ be an increasing mapping. Then $f$ has a least fixed point and a greatest fixed point.


Knaster-Tarski Theorem

Let $\left({L, \preceq}\right)$ be a complete lattice. Let $f: L \to L$ be an increasing mapping. Let $F$ be the set (or class) of fixed points of $f$. Then $\left({F, \preceq}\right)$ is a complete lattice.


Degree of Product of Polynomials over Ring/Corollary 2

Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$. Let $D \sqbrk X$ be the ring of polynomials over $D$ in the indeterminate $X$. For $f \in D \sqbrk X$ let $\map \deg f$ denote the degree of $f$. Then: :$\forall f, g \in D \sqbrk X: \map \deg {f g} = \map \deg f + \map \deg g$


Knaster-Tarski Lemma/Power Set

Let $S$ be a set. Let $\powerset S$ be the power set of $S$. Let $f: \powerset S \to \powerset S$ be a $\subseteq$-increasing mapping. That is, suppose that for all $T, U \in \powerset S$: :$T \subseteq U \implies \map f T \subseteq \map f U$ Then $f$ has a greatest fixed point and a least fixed point.


Cantor-Bernstein-Schröder Theorem/Proof 6

Let $A$ and $B$ be sets. Let $f: A \to B$ and $g: B \to A$ be injections. Then there is a bijection $h: A \to B$; so that $A$ and $B$ are equivalent. Furthermore: : For all $x \in A$ and $y \in B$, if $y = h \left({x}\right)$ then either $y = f \left({x}\right)$ or $x = g \left({y}\right)$.


Ring of Polynomial Forms over Integral Domain is Integral Domain

Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$. Let $\struct {D \sqbrk X, \oplus, \odot}$ be the ring of polynomial forms over $D$ in the indeterminate $X$. Then $\struct {D \sqbrk X, \oplus, \odot}$ is an integral domain.


Union of One-to-Many Relations with Disjoint Images is One-to-Many

Let $S_1, S_2, T_1, T_2$ be sets or classes. Let $\mathcal R_1$ be a one-to-many relation on $S_1 \times T_1$. Let $\mathcal R_2$ be a one-to-many relation on $S_2 \times T_2$. Suppose that the images of $\mathcal R_1$ and $\mathcal R_2$ are disjoint. Then $\mathcal R_1 \cup \mathcal R_2$ is a one-to-many relation on $(S_1 \cup S_2) \times (T_1 \cup T_2)$.


Union of Many-to-One Relations with Disjoint Domains is Many-to-One

Let $S_1, S_2, T_1, T_2$ be sets or classes. Let $\RR_1$ be a many-to-one relation on $S_1 \times T_1$. Let $\RR_2$ be a many-to-one relation on $S_2 \times T_2$. Suppose that the domains of $\RR_1$ and $\RR_2$ are disjoint. Then $\RR_1 \cup \RR_2$ is a many-to-one relation on $\paren {S_1 \cup S_2} \times \paren {T_1 \cup T_2}$.


Nth Root of Integer is Integer or Irrational

Let $n$ be a natural number. Let $x$ be an integer. If the $n$th root of $x$ is not an integer, it must be irrational.


Union of Bijections with Disjoint Domains and Codomains is Bijection

Let $A$, $B$, $C$, and $D$ be sets or classes. Let $A \cap B = C \cap D = \varnothing$. Let $f: A \to C$ and $g: B \to D$ be bijections. Then $f \cup g: A \cup B \to C \cup D$ is also a bijection.


Interval in Complete Lattice is Complete Lattice

Let $\left({L, \preceq}\right)$ be a complete lattice. Let $a, b \in L$ with $a \preceq b$. Let $\left[{a \,.\,.\, b}\right]$ be the closed interval between $a$ and $b$. {{explain|Demonstrate that for each $a, b \in L$ that $\left[{a \,.\,.\, b}\right]$ exists and is unique.}} Then $\left[{a \,.\,.\, b}\right]$ is also a complete lattice under $\preceq$.


Set Difference with Subset is Superset of Set Difference

Let $A, B, S$ be sets or classes. Suppose that $A \subseteq B$. Then $S \setminus B \subseteq S \setminus A$, where $\setminus$ represents set difference.


Knaster-Tarski Lemma/Corollary

Let $\struct {L, \preceq}$ be a complete lattice. Let $f: L \to L$ be an increasing mapping. Then $f$ has a fixed point


Knaster-Tarski Lemma/Corollary/Power Set

Let $S$ be a set. Let $\mathcal P \left({S}\right)$ be the power set of $S$. Let $f: \mathcal P \left({S}\right) \to \mathcal P \left({S}\right)$ be a $\subseteq$-increasing mapping. That is, suppose that for all $T, U \in \mathcal P \left({S}\right)$: : $T \subseteq U \implies f \left({T}\right) \subseteq f\left({U}\right)$ Then $f$ has a fixed point.


Alternative Definition of Ordinal in Well-Founded Theory

A set $S$ is an ordinal {{iff}} $S$ is transitive and $\forall x, y \in S: \left({x \in y \lor x = y \lor y \in x}\right)$.


Foundational Relation is Antireflexive/Corollary

Let $\left({S, \preceq}\right)$ be an ordered set. Suppose that $S$ is non-empty. Then $\preceq$ is not a foundational relation.


Reflexive Reduction of Well-Founded Ordering is Foundational Relation

Let $S$ be a set. Let $\preceq$ be a well-founded ordering of $S$. Let $\prec$ be the reflexive reduction of $\preceq$. Then $\prec$ is a foundational relation.


Epsilon Relation is Proper

Let $\mathbb U$ be the universal class. Let $\Epsilon$ be the epsilon relation. Then $\left({\mathbb U, \Epsilon}\right)$ is a proper relational structure.


Relationship between Transitive Closure Definitions

Let $x$ be a set. Let $a$ be the smallest set such that $x \in a$ and $a$ is transitive. Let $b$ be the smallest set such that $x \subseteq b$ and $b$ is transitive. Then $a = b \cup \set x$.


Ordinal is not Element of Itself

Let $x$ be an ordinal. Then $x \notin x$.


Set is Element of Successor

Let $x$ be a set. Let $x^+$ be the successor of $x$. Then $x \in x^+$.


Element of Ordinal is Ordinal

Let $n$ be an ordinal. Let $m \in n$. Then $m$ is also an ordinal.


Modulus of Exponential of Imaginary Number is One

Let $\cmod z$ denote the modulus of a complex number $z$. Let $e^z$ be the complex exponential of $z$. Let $x$ be wholly real. Then: :$\cmod {e^{i x} } = 1$


Absolute Value of Power

Let $x$, $y$ be real numbers. Let $x^y$, $x$ to the power of $y$, be real. Then: :$\size {x^y} = \size x^y$


Count of Rows of Truth Table

Let $P$ be a WFF of propositional logic. Suppose $\mathcal P$ is of finite size such that it contains $n$ different letters. Then a truth table constructed to express $P$ will contain $2^n$ rows.


Denumerable Class is Set

Let $A$ be a class. Let $\N$ be the natural numbers. Suppose that $F: \N \to A$ is a bijection. Then $A$ is a set.


Relative Complement inverts Subsets

Let $S$ be a set. Let $A \subseteq S, B \subseteq S$ be subsets of $S$. Then: :$A \subseteq B \iff \relcomp S B \subseteq \relcomp S A$ where $\complement_S$ denotes the complement relative to $S$.


Union of Subsets is Subset/Family of Sets

Let $\family {S_i}_{i \mathop \in I}$ be a family of sets indexed by $I$. Then for all sets $X$: :$\displaystyle \paren {\forall i \in I: S_i \subseteq X} \implies \bigcup_{i \mathop \in I} S_i \subseteq X$ where $\displaystyle \bigcup_{i \mathop \in I} S_i$ is the union of $\family {S_i}$.


Union Distributes over Union/Sets of Sets

Let $A$ and $B$ denote sets of sets. Then: :$\displaystyle \bigcup \left({A \cup B}\right) = \left({\bigcup A}\right) \cup \left({\bigcup B}\right)$ where $\displaystyle \bigcup A$ denotes the union of $A$.


Union Distributes over Union/Families of Sets

Let $I$ be an indexing set. Let $\family {A_\alpha}_{\alpha \mathop \in I}$ and $\family {B_\alpha}_{\alpha \mathop \in I}$ be indexed families of subsets of a set $S$. Then: :$\displaystyle \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cup B_\alpha} = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cup \paren {\bigcup_{\alpha \mathop \in I} B_\alpha}$ where $\displaystyle \bigcup_{\alpha \mathop \in I} A_\alpha$ denotes the union of $\family {A_\alpha}_{\alpha \mathop \in I}$.


Intersection Distributes over Intersection/Sets of Sets

Let $A$ and $B$ denote sets of sets. Then: :$\displaystyle \bigcap \paren {A \cap B} = \paren {\bigcap A} \cap \paren {\bigcap B}$ where $\displaystyle \bigcap A$ denotes the intersection of $A$.


Intersection Distributes over Intersection/Families of Sets

Let $I$ be an indexing set. Let $\family {A_\alpha}_{\alpha \mathop \in I}$ and $\family {B_\alpha}_{\alpha \mathop \in I}$ be indexed families of subsets of a set $S$. Then: :$\displaystyle \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cap B_\alpha} = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cap \paren {\bigcap_{\alpha \mathop \in I} B_\alpha}$ where $\displaystyle \bigcap_{\alpha \mathop \in I} A_i$ denotes the intersection of $\family {A_\alpha}$.


Intersection Distributes over Intersection/General Result

Let $\left\langle{\mathbb S_i}\right\rangle_{i \in I}$ be an $I$-indexed family of sets of sets. Then: :$\displaystyle \bigcap_{i \mathop \in I} \bigcap \mathbb S_i = \bigcap \bigcap_{i \mathop \in I} \mathbb S_i$


Intersection Distributes over Union/Family of Sets

Let $I$ be an indexing set. Let $\family {A_\alpha}_{\alpha \mathop \in I}$ be a indexed family of subsets of a set $S$. Let $B \subseteq S$. Then: :$\displaystyle \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cap B} = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cap B$ where $\displaystyle \bigcup_{\alpha \mathop \in I} A_\alpha$ denotes the union of $\family {A_\alpha}_{\alpha \mathop \in I}$.


Union Distributes over Intersection/Family of Sets

Let $I$ be an indexing set. Let $\family {A_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of a set $S$. Let $B \subseteq S$. Then: :$\displaystyle \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cup B} = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cup B$ where $\displaystyle \bigcap_{\alpha \mathop \in I} A_\alpha$ denotes the intersection of $\family {A_\alpha}_{\alpha \mathop \in I}$.


Equivalence of Definitions of Transitive Closure (Relation Theory)/Finite Chain is Smallest

Let $S$ be a set or class. Let $\RR$ be a relation on $S$. Let $\RR^+$ be the transitive closure of $\RR$ by the finite chain definition. That is, for $x, y \in S$ let $x \mathrel {\RR^+} y$ {{iff}} for some natural number $n > 0$ there exist $s_0, s_1, \dots, s_n \in S$ such that $s_0 = x$, $s_n = y$, and: :$\forall k \in \N_n: s_k \mathrel \RR s_{k+1}$ Then $\RR^+$ is transitive and if $\QQ$ is a transitive relation on $S$ such that $\RR \subseteq \QQ$ then $\RR \subseteq \QQ$.


Order-Preserving Bijection on Wosets is Order Isomorphism

Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be well-ordered sets. Let $\phi: S \to T$ be a bijection such that $\phi: S \to T$ is order-preserving: :$\forall x, y \in S: x \preceq_1 y \implies \map \phi x \preceq_2 \map \phi y$ Then: :$\forall x, y \in S: \map \phi x \preceq_2 \map \phi y \implies x \preceq_1 y$ That is, $\phi: S \to T$ is an order isomorphism.


Inverse Image of Set under Set-Like Relation is Set

Let $A$ be a class. Let $\RR$ be a set-like endorelation on $A$. Let $B \subseteq A$ be a set. Then $\map {\RR^{-1} } B$, the inverse image of $B$ under $\RR$, is also a set.


Reciprocal of Holomorphic Function

Let $f: \C \to \C$ be a complex function. Let $U \subseteq \C$ be an open set such that $f$ has no zeros in $U$. Suppose further that $f$ is holomorphic in $U$. Then the complex function :$\dfrac 1 {f_{\restriction U} } : U \to \C$ is holomorphic.


Transitive Closure of Set-Like Relation is Set-Like

Let $A$ be a class. Let $\RR$ be a set-like endorelation on $A$. Let $\RR^+$ be the transitive closure of $\RR$. Then $\RR^+$ is also a set-like relation.


Minimal WRT Restriction

Let $A$ be a set or class. Let $\mathcal R$ be a relation on $A$. Let $B$ be a subset or subclass of $A$. Let $\mathcal R'$ be the restriction of $\mathcal R$ to $B$. Let $m \in B$. Then: :$m$ is $\mathcal R$-minimal in $B$ {{iff}}: :$m$ is $\mathcal R'$-minimal in $B$.


Intersection of Ordinals is Ordinal

Let $A$ be a non-empty class of ordinals. Then $\bigcap A$ is an ordinal.


Meet with Complement is Bottom

Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra, defined as in Definition 2. Then: :$\exists \bot \in S: \forall a \in S: a \wedge \neg a = \bot$ where $\wedge$ denotes the meet operation in $S$. This element $\bot$ is unique for any given $S$, and is named '''bottom'''.


Join with Complement is Top

Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra, defined as in Definition 2. Then: :$\exists \top \in S: \forall a \in S: a \vee \neg a = \top$ where $\wedge$ denotes the meet operation in $S$. This element $\top$ is unique for any given $S$, and is named '''top'''.


Ordering is Equivalent to Subset Relation/Lemma

Let $\struct {S, \preceq}$ be an ordered set. Then: : $\forall a_1, a_2 \in S: \paren ({a_1 \preceq a_2 \implies {a_1}^\preceq \subseteq {a_2}^\preceq}$ where ${a_1}^\preceq$ denotes the lower closure of $a_1$.


Smallest Element WRT Restricted Ordering

Let $S$ be a set or class. Let $\preceq$ be an ordering on $S$. Let $T$ be a subset or subclass of $S$. Let $\preceq'$ be the restriction of $\preceq$ to $T$. Let $m \in T$. Then $m$ is the $\preceq$-smallest element of $T$ iff $m$ is the $\preceq'$-smallest element of $T$.


Restriction of Well-Founded Ordering

Let $S$ be a set or class. Let $T$ be a subset or subclass of $S$. Let $\preceq$ be a well-founded ordering of $A$. Let $\preceq'$ be the restriction of $\preceq$ to $T$. Then $\preceq'$ is a well-founded ordering of $T$.


Restriction of Well-Ordering is Well-Ordering

Let $S$ be a set or class. Let $\preceq$ be a well-ordering of $S$. Let $T$ be a subset or subclass of $S$. Let $\preceq'$ be the restriction of $\preceq$ to $T$. Then $\preceq'$ well-orders $T$.


Ordering is Equivalent to Subset Relation/Proof 2

{{:Ordering is Equivalent to Subset Relation}} Specifically: Let :$\mathbb S := \set {a^\preceq: a \in S}$ where $a^\preceq$ is the lower closure of $a$. That is: :$a^\preceq := \set {b \in S: b \preceq a}$ Let the mapping $\phi: S \to \mathbb S$ be defined as: :$\map \phi a = a^\preceq$ Then $\phi$ is an order isomorphism from $\struct {S, \preceq}$ to $\struct {\mathbb S, \subseteq}$.


Event Space contains Sample Space

:$\Omega \in \Sigma$


Event Space contains Empty Set

:$\O \in \Sigma$


Power Set of Sample Space is Event Space

Let $\EE$ be an experiment whose sample space is $\Omega$. Let $\powerset \Omega$ be the power set of $\Omega$. Then $\powerset \Omega$ is an event space of $\EE$.


Event Space from Single Subset of Sample Space

Let $\EE$ be an experiment whose sample space is $\Omega$. Let $\O \subsetneqq A \subsetneqq \Omega$. Then $\Sigma := \set {\O, A, \Omega \setminus A, \Omega}$ is an event space of $\EE$.


Intersection of Events is Event

:$A, B \in \Sigma \implies A \cap B \in \Sigma$


Set Difference of Events is Event

:$A, B \in \Sigma \implies A \setminus B \in \Sigma$


Symmetric Difference of Events is Event

:$A, B \in \Sigma \implies A \ast B \in \Sigma$


Characterization of Affine Transformations

Let $\mathcal E$ and $\mathcal F$ be affine spaces over a field $k$. Let $\mathcal L: \mathcal E \to \mathcal F$ be a mapping. Then $\mathcal L$ is an affine transformation {{iff}} for all points $p, q \in \mathcal E$ and all $\lambda \in k$: :$\mathcal L \left({\lambda p + \left({1 - \lambda}\right) q}\right) = \lambda \mathcal L \left({p}\right) + \left({1 - \lambda}\right) \mathcal L \left({q}\right)$ where $\lambda p + \left({1 - \lambda}\right) q$ and $\lambda \mathcal L \left({p}\right) + \left({1 - \lambda}\right) \mathcal L \left({q}\right)$ denote barycenters.


Probability of Empty Event is Zero

:$\map \Pr \O = 0$


Ordinal Membership is Asymmetric

Let $m$ and $n$ be ordinals. Then it is not the case that $m \in n$ and $n \in m$.


Intersection of Ordinals is Smallest

Let $A$ be a non-empty set or class of ordinals. Let $m = \bigcap A$ be the intersection of all the elements of $A$. Then $m$ is the smallest element of $A$.


Ordinal Class is Strongly Well-Ordered by Subset

Let $\On$ be the class of all ordinals. Then the restriction of the subset relation, $\subseteq$, to $\On$ is a strong well-ordering. That is: :$\subseteq$ is an ordering on $\On$. :If $A$ is a non-empty subclass of $\On$, then $A$ has a $\subseteq$-smallest element.


Subset is Compatible with Ordinal Successor

Let $x$ and $y$ be ordinals and let $x^+$ denote the successor set of $x$. Let $x \in y$. Then: :$x^+ \in y^+$


Successor is Less than Successor/Sufficient Condition/Proof 1

Let $x$ and $y$ be ordinals and let $x^+$ denote the successor set of $x$. Let $x^+ \in y^+$. Then: : $x \in y$


Successor is Less than Successor/Sufficient Condition/Proof 2


== Proof 1 == {{:Successor is Less than Successor/Sufficient Condition/Proof 1}}


Probability of Event not Occurring

:$\forall A \in \Sigma: \map \Pr {\Omega \setminus A} = 1 - \map \Pr A$


Delta-Algebra is Sigma-Algebra

Every $\delta$-algebra is a $\sigma$-algebra.


Countable Intersection of Events is Event

:$\quad A_1, A_2, \ldots \in \Sigma \implies \ds \bigcap_{i \mathop = 1}^\infty A_i \in \Sigma$


Subset Relation is Ordering/General Result

Let $\mathbb S$ be a set of sets or class. Then $\subseteq$ is an ordering on $\mathbb S$. In other words, let $\left({\mathbb S, \subseteq}\right)$ be the relational structure defined on $\mathbb S$ by the relation $\subseteq$. Then $\left({\mathbb S, \subseteq}\right)$ is an ordered set.


Noetherian Domain is Factorization Domain

Let $R$ be a noetherian integral domain. Then $R$ is a factorization domain.


Probability Measure is Monotone

Let $A, B \in \Sigma$ such that $A \subseteq B$. Then: :$\map \Pr A \le \map \Pr B$


Transfinite Induction/Principle 1/Proof 2

Let $\On$ denote the class of all ordinals. Let $A$ denote a class. Suppose that: :For each element $x$ of $\On$, if $\forall y \in \On: \paren {y < x \implies y \in A}$ then $x$ is an element of $A$. Then $\On \subseteq A$.


Immediate Successor is Unique in Toset

Let $(S, \preceq)$ be a totally ordered set. Let $x, y \in S$. Suppose that $y$ is an immediate successor of $x$. Then $y$ is the ''only'' immediate successor of $x$.


Mapping from Set to Ordinal Class is Bounded Above

Let $x$ be a set. Let $\operatorname{On}$ be the class of all ordinals. Let $f: x \to \operatorname{On}$ be a mapping. Then $f$ has an upper bound.


Equivalence of Definitions of Independent Events

Let $\EE$ be an experiment with probability space $\struct {\Omega, \Sigma, \Pr}$. Let $A, B \in \Sigma$ be events of $\EE$ such that $\map \Pr A > 0$ and $\map \Pr B > 0$. {{TFAE|def = Independent Events}}


Union of Subset of Ordinals is Ordinal/Corollary

Let $y$ be a set. Let $\On$ be the class of all ordinals. Let $F: y \to \On$ be a mapping. Then: :$\displaystyle \bigcup \map F y \in \On$ where $\map F y$ is the image of $y$ under $F$.


Independent Events are Independent of Complement/General Result

Let $A_1, A_2, \ldots, A_m$ be events in a probability space $\struct {\Omega, \Sigma, \Pr}$. Then $A_1, A_2, \ldots, A_m$ are independent {{iff}} $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_m$ are also independent.


Probability of Independent Events Not Happening/Corollary

Let $A$ be an event in an event space of an experiment $\EE$ whose probability space is $\struct {\Omega, \Sigma, \Pr}$. Let $\map \Pr A = p$. Suppose that the nature of $\EE$ is that its outcome is independent of previous trials of $\EE$. Then the probability that $A$ does not occur during the course of $m$ trials of $\EE$ is $\paren {1 - p}^m$.


Probability of Limit of Sequence of Events/Increasing

Let $\left \langle{A_n}\right \rangle_{n \mathop \in \N}$ be an increasing sequence of events. Let $\displaystyle A = \bigcup_{i \mathop \in \N} A_i$ be the limit of $\left \langle{A_n}\right \rangle_{n \mathop \in \N}$. Then: :$\displaystyle \Pr \left({A}\right) = \lim_{n \to \infty} \Pr \left({A_n}\right)$


Probability of Limit of Sequence of Events/Decreasing

Let $\left \langle{B_n}\right \rangle_{n \mathop \in \N}$ be a decreasing sequence of events. Let $\displaystyle B = \bigcap_{i \mathop \in \N} B_i$ be the limit of $\left \langle{B_n}\right \rangle_{n \mathop \in \N}$. Then: :$\displaystyle \Pr \left({B}\right) = \lim_{n \to \infty} \Pr \left({B_n}\right)$


Sum of Discrete Random Variables

Let $U: \Omega \to \R$ be defined as: :$\forall \omega \in \Omega: \map U \omega = \map X \omega + \map Y \omega$ Then $U$ is also a discrete random variable on $\struct {\Omega, \Sigma, \Pr}$.


Product of Discrete Random Variables

Let $V: \Omega \to \R$ be defined as: :$\forall \omega \in \Omega: V \left({\omega}\right) = X \left({\omega}\right) Y \left({\omega}\right)$ Then $V$ is also a discrete random variable on $\left({\Omega, \Sigma, \Pr}\right)$.


Bernoulli Process as Geometric Distribution/Shifted

Let $\left \langle{Y_i}\right \rangle$ be a Bernoulli process with parameter $p$. Let $\mathcal E$ be the experiment which consists of performing the Bernoulli trial $Y_i$ as many times as it takes to achieve a success, and then stop. Let $k$ be the number of Bernoulli trials to achieve a success. Then $k$ is modelled by a shifted geometric distribution with parameter $p$.


Geometric Distribution Gives Rise to Probability Mass Function/Shifted

Let $Y$ be a discrete random variable on a probability space $\left({\Omega, \Sigma, \Pr}\right)$. Let $Y$ have the shifted geometric distribution with parameter $p$ (where $0 < p < 1$). Then $Y$ gives rise to a probability mass function.


Negative Binomial Distribution Gives Rise to Probability Mass Function

Let $X$ be a discrete random variable on a probability space $\struct {\Omega, \Sigma, \Pr}$.


Negative Binomial Distribution Gives Rise to Probability Mass Function/First Form

Let $X$ have the negative binomial distribution (first form) with parameters $n$ and $p$ ($0 < p < 1$). Then $X$ gives rise to a probability mass function.


Bernoulli Process as Negative Binomial Distribution/First Form

Let $\EE$ be the experiment which consists of performing the Bernoulli trial $X_i$ until a total of $n$ failures have been encountered. Let $X$ be the discrete random variable defining the number of successes before $n$ failures have been encountered. Then $X$ is modeled by a negative binomial distribution of the first form.


Bernoulli Process as Negative Binomial Distribution/Second Form

Let $\EE$ be the experiment which consists of performing the Bernoulli trial $X_i$ as many times as it takes to achieve a total of $n$ successes, and then stops. Let $Y$ be the discrete random variable defining the number of trials before $n$ successes have been achieved. Then $X$ is modeled by a negative binomial distribution of the second form.


Negative Binomial Distribution as Generalized Geometric Distribution/Second Form

The second form of the negative binomial distribution is a generalization of the shifted geometric distribution: Let $\sequence {Y_i}$ be a Bernoulli process with parameter $p$. Let $\FF$ be the experiment which consists of: :Perform the Bernoulli trial $Y_i$ as many times as it takes to achieve $n$ successes, and then stop. Let $k$ be the number of Bernoulli trials that need to be taken in order to achieve up to (and including) the $n$th success. Let $\FF'$ be the experiment which consists of: :Perform the Bernoulli trial $Y_i$ until '''one''' success is achieved, and then stop. Then $k$ is modelled by the experiment: :Perform experiment $\FF'$ until $n$ failures occur, and then stop.


Derivative of Geometric Sequence/Corollary

:$\displaystyle \sum_{n \mathop \ge 1} n \paren {n + 1} x^{n - 1} = \frac 2 {\paren {1 - x}^3}$


Equivalence of Definitions of Countable Set

Let $S$ be a set. {{TFAE|def = Countable Set}}


Infinite Set of Natural Numbers is Countably Infinite

Let $\N$ be the set of natural numbers. Let $S$ be an infinite subset of $\N$. Then $S$ is countably infinite. That is, there is a bijection $f: \N \to S$.


Generating Function for Natural Numbers

Let $\sequence {a_n}$ be the sequence defined as: : $\forall n \in \N_{> 0}: a_n = n - 1$ That is: :$\sequence {a_n} = 0, 1, 2, 3, 4, \ldots$ Then the generating function for $\sequence {a_n}$ is given as: :$G \paren z = \dfrac 1 {\paren {1 - z}^2}$


Injection has Surjective Left Inverse Mapping

Let $S$ and $T$ be sets such that $S \ne \O$. Let $f: S \to T$ be a injection. Then there exists a surjection $g: T \to S$ such that: :$g \circ f = I_S$


Many-to-One Relation Extends to Mapping

Let $S$ and $T$ be sets. Let $T$ be non-empty. Let $\mathcal R \subset S \times T$ be a many-to-one relation. Then there exists a mapping $f: S \to T$ such that $\mathcal R \subseteq f$.


Probability Generating Function as Expectation

Let $X$ be a discrete random variable whose codomain, $\Omega_X$, is a subset of the natural numbers $\N$. Let $p_X$ be the probability mass function for $X$. Let $\Pi_X \left({s}\right)$ be the probability generating function for $X$. Then: :$\Pi_X \left({s}\right) = E \left({s^X}\right)$ where $E \left({s^X}\right)$ denotes the expectation of $s^X$.


Probability Generating Function of Negative Binomial Distribution/First Form

Let $X$ be a discrete random variable with the negative binomial distribution (first form) with parameters $n$ and $p$. Then the p.g.f. of $X$ is: :$\Pi_X \left({s}\right) = \left({\dfrac q {1 - ps}}\right)^n$ where $q = 1 - p$.


Probability Generating Function of Negative Binomial Distribution/Second Form

Let $X$ be a discrete random variable with the negative binomial distribution (second form) with parameters $n$ and $p$. Then the p.g.f. of $X$ is: :$\displaystyle \Pi_X \left({s}\right) = \left({\frac {ps} {1 - qs}}\right)^n$ where $q = 1 - p$.


Probability Generating Function of Scalar Multiple of Random Variable

Let $X$ be a discrete random variable whose probability generating function is $\Pi_X \left({s}\right)$. Let $k \in \Z_{\ge 0}$ be a positive integer. Let $Y$ be a discrete random variable such that $Y = m X$. Then :$\Pi_Y \left({s}\right) = \Pi_X \left({s^m}\right)$. where $\Pi_Y \left({s}\right)$ is the probability generating function of $Y$.


Derivatives of Probability Generating Function at One

Let $X$ be a discrete random variable whose probability generating function is $\Pi_X \left({s}\right)$. Then the $n$th derivative of $\Pi_X \left({s}\right)$ at $s = 1$ is given by: :$\dfrac {\mathrm d^n} {\mathrm d s^n} \Pi_X \left({1}\right) = E \left({X \left({X - 1}\right) \cdots \left({X - n + 1}\right)}\right)$ for $n = 1, 2, \ldots$


Differentiation of Power Series/Corollary

The value of $\displaystyle \frac {\mathrm d^n}{\mathrm d x^n} \sum_{m \mathop \ge 0} a_m \left({x - \xi}\right)^m$ at $x = \xi$ is: :$\displaystyle \left.{\frac {\mathrm d^n}{\mathrm d x^n} \sum_{m \mathop \ge 0} a_m \left({x - \xi}\right)^m}\right|_{x = \xi} = a_n n!$


Non-Equivalence of Proposition and Negation/Formulation 1

:$p \implies \neg p, \neg p \implies p \vdash \bot$


Non-Equivalence of Proposition and Negation/Formulation 2

:$\vdash \neg \left({p \iff \neg p}\right)$


No Injection from Power Set to Set/Lemma

Let $S$ be a set. Let $\powerset S$ be the power set of $S$. Then there does not exist a set $B$ such that there is an injection from $B$ into $S$ and a surjection from $B$ onto $\powerset S$.


No Injection from Power Set to Set

Let $S$ be a set. Let $\powerset S$ be the power set of $S$. Then there is no injection from $\powerset S$ into $S$.


Inverse Image Mapping of Injection is Surjection

Let $S$ and $T$ be sets. Let $f: S \to T$ be a injection. Let $f^\gets: \powerset T \to \powerset S$ be the inverse image mapping of $f$. Then $f^\gets$ is a surjection.


Expectation of Negative Binomial Distribution/Second Form

Let $X$ be a discrete random variable with the negative binomial distribution (second form) with parameters $n$ and $p$. Then the expectation of $X$ is given by: :$E \left({X}\right) = \dfrac n p$


Derivatives of PGF of Negative Binomial Distribution/Second Form

Let $X$ be a discrete random variable with the negative binomial distribution (second form) with parameters $n$ and $p$. Then the derivatives of the PGF of $X$ {{WRT|Differentiation}} $s$ are: :$\dfrac {\mathrm d^k} {\mathrm d s^k} \Pi_X \left({s}\right) = ...$


First Derivative of PGF of Negative Binomial Distribution/Second Form

Let $X$ be a discrete random variable with the negative binomial distribution (second form) with parameters $n$ and $p$. Then the first derivative of the PGF of $X$ {{WRT|Differentiation}} $s$ is: :$\dfrac \d {\d s} \map {\Pi_X} s = n p \paren {\dfrac {\paren {p s}^{n - 1} } {\paren {1 - q s}^{n + 1} } }$


Nth Derivative of Reciprocal of Mth Power/Corollary

The $n$th derivative of $\dfrac 1 x$ {{WRT|Differentiation}} $x$ is: :$\dfrac {\d^n} {\d x^n} \dfrac 1 x = \dfrac {\paren {-1}^n n!} {x^{n + 1} }$ where $n!$ denotes $n$ factorial.


First Derivative of PGF of Negative Binomial Distribution/First Form

Let $X$ be a discrete random variable with the negative binomial distribution (first form) with parameters $n$ and $p$. Then the first derivative of the PGF of $X$ w.r.t. $s$ is: :$\dfrac \d {\d s} \map {\Pi_X} s = \dfrac {n p} q \paren {\dfrac q {1 - p s} }^{n + 1}$


Expectation of Negative Binomial Distribution/First Form

Let $X$ be a discrete random variable with the negative binomial distribution (first form) with parameters $n$ and $p$. Then the expectation of $X$ is given by: :$E \left({X}\right) = \dfrac {n p} q$


Closure Condition for Hausdorff Space

Let $\struct {X, \tau}$ be a topological space. Then $\struct {X, \tau}$ is a Hausdorff space {{iff}}: :For all $x, y \in X$ such that $x \ne y$, there exists an open set $U$ such that $x \in U$ and $y \notin U^-$, where $U^-$ is the closure of $U$.


Second Derivative of PGF of Negative Binomial Distribution/Second Form

Let $X$ be a discrete random variable with the negative binomial distribution (second form) with parameters $n$ and $p$. Then the second derivative of the PGF of $X$ w.r.t. $s$ is: :$\dfrac {\mathrm d^2} {\mathrm d s^2} \Pi_X \left({s}\right) = \left({\dfrac {ps} {1 - qs} }\right)^{n+2} \left({\dfrac {n \left({n-1}\right) + 2 n q s} {\left({p s^2}\right)^2} }\right)$


Derivatives of PGF of Negative Binomial Distribution/First Form

Let $X$ be a discrete random variable with the negative binomial distribution (second form) with parameters $n$ and $p$. Then the derivatives of the PGF of $X$ {{WRT|Differentiation}} $s$ are: :$\dfrac {\mathrm d^k} {\mathrm d s^k} \Pi_X \left({s}\right) = \dfrac {n^{\overline k} p^k} {q^k} \left({\dfrac q {1 - p s} }\right)^{n + k}$ where: : $n^{\overline k}$ is the rising factorial: $n^{\overline k} = n \left({n + 1}\right) \left({n + 2}\right) \cdots \left({n + k - 1}\right)$ : $q = 1 - p$


Variance of Negative Binomial Distribution/Second Form

Let $X$ be a discrete random variable with the negative binomial distribution (second form) with parameters $n$ and $p$. Then the variance of $X$ is given by: :$\displaystyle \operatorname{var} \left({X}\right) = \frac {n q} {p^2}$


PGF of Sum of Independent Discrete Random Variables/General Result

Let: :$Z = X_1 + X_2 + \cdots + X_n$ where each of $X_1, X_2, \ldots, X_n$ are independent discrete random variables with PGFs $\map {\Pi_{X_1} } s, \map {\Pi_{X_2} } s, \ldots, \map {\Pi_{X_n} } s$. Then: :$\displaystyle \map {\Pi_Z} s = \prod_{j \mathop = 1}^n \map {\Pi_{X_j} } s$


Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods/Lemma

Let $\left({S, \tau}\right)$ be a Hausdorff space. Let $C$ be a compact subspace of $S$. Let $x \in S \setminus C$. Then there exist open sets $U$ and $V$ such that $x \in U$, $C \subseteq V$, and $U \cap V = \varnothing$.


Sum of Independent Binomial Random Variables

Let $X$ and $Y$ be discrete random variables with a binomial distribution: :$X \sim \operatorname{B} \left({m, p}\right)$ and :$Y \sim \operatorname{B} \left({n, p}\right)$ Let $X$ and $Y$ be independent. Then their sum $Z = X + Y$ is distributed as: :$Z \sim \operatorname{B} \left({\left({m + n}\right), p}\right)$


Field has Algebraic Closure

Every field has an algebraic closure.


Point Finite Set of Open Sets in Separable Space is Countable

Let $\struct {X, \tau}$ be a separable space. Let $\mathcal F$ be a point finite set of open sets of $X$. Then $\mathcal F$ is countable.


Countable Union of Finite Sets is Countable

The following statements are equivalent in $\mathrm{ZF}^-$: {{explain|$\mathrm{ZF}^-$}} :The Axiom of Countable Choice for Finite Sets holds. :The union of any countable set of finite sets is countable.


Reverse Triangle Inequality/Normed Vector Space

Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space. Then: :$\forall x, y \in X: \norm {x - y} \ge \size {\norm x - \norm y}$


Equivalence of Forms of Axiom of Countable Choice

The following forms of the Axiom of Countable Choice are equivalent in $\mathrm{ZF}^-$: {{explain|$\mathrm{ZF}^-$}}


GCD from Generator of Ideal

Let $m, n \in \Z$, with either $m \ne 0$ or $n \ne 0$. Let $I = \ideal {m, n}$ be the ideal generated by $m$ and $n$. Let $d$ be a non-negative generator for the principal ideal $I$. Then: :$\gcd \set {m, n} =  d$ where $\gcd \set {m, n}$ denotes the greatest common divisor of $m$ and $n$.


Finite Union of Finite Sets is Finite

Let $S$ be a finite set of finite sets. Then the union of $S$ is finite.


Open Set Characterization of Denseness

Let $\left({X, \tau}\right)$ be a topological space. Let $S \subseteq X$. Then $S$ is (everywhere) dense in $X$ {{iff}} every non-empty ($\tau$-)open set contains an element of $S$.


Union of Set of Sets when a Set Intersects All

Let $F$ be a set of sets. Let $S$ be a set or class. Suppose that: :$\forall A \in F: A \cap S \ne \O$ Then: :$\ds F = \bigcup_{x \mathop \in S} \set {A \in F: x \in A}$


Order of Group of Units of Integers Modulo m

Let $n \in \Z_{\ge 0}$ be an integer. Let $\struct {\Z / n \Z, +, \cdot}$ be the ring of integers modulo $n$. Let $U = \struct {\paren {\Z / n \Z}^\times, \cdot}$ denote the group of units of this ring. Then: :$\order U = \map \phi n$ where $\phi$ denotes the Euler $\phi$-function.


Cyclicity Condition for Units of Ring of Integers Modulo m

Let $n \in \Z_{\ge 0}$ be a positive integer. Let $\struct {\Z / n \Z, +, \times}$ be the ring of integers modulo $n$. Let $U = \struct {\paren {\Z / n \Z}^\times, \times}$ denote the group of units of $\struct {\Z / n \Z, +, \times}$. Then $U$ is cyclic {{iff}} either: :$n = p^\alpha$ or: :$n = 2 p^\alpha$ where $p \ge 3$ is prime and $\alpha \ge 0$.


Chinese Remainder Theorem/Corollary

Let $n_1, n_2, \ldots, n_r$ be positive integers such that $n_i \perp n_j$ for all $i \ne j$ (that is, $\gcd \left\{{n_i, n_j}\right\} = 1$). Let $N = n_1 \cdots n_r$. For an integer $k$, let $\Z / k \Z$ denote the ring of integers modulo $k$. Then we have a ring isomorphism: :$\Z / N \Z \simeq \Z / n_1 \Z \times \cdots \times \Z / n_r \Z$


Minkowski's Theorem

Let $L$ be a lattice in $\R^n$. Let $d$ be the covolume of $L$. Let $\mu$ be a translation invariant measure on $\R^n$ Let $S$ be a convex subset of $\R^n$ that is symmetric about the origin, i.e. such that: :$\forall p \in S : -p \in S$  Let the volume of $S$ be greater than $2^n d$. Then $S$ contains a non-zero point of $L$.


Component of Finite Union in Ultrafilter

Let $S$ be a non-empty set. Let $\mathcal U$ be a ultrafilter on $S$. Let $\left\{{Y_1, \dots, Y_n}\right\}$ be a pairwise disjoint set of subsets of $S$ such that $S = Y_1 \cup \cdots \cup Y_n$. Then there is a unique $k \in \left\{{1, \dots, n}\right\}$ such that $Y_k \in \mathcal U$.


Restricted Tukey-Teichmüller Theorem/Weak Form

Let $X$ be a set. Let $\mathcal A$ be a non-empty set of subsets of $X$. Let $'$ be a unary operation on $X$. Let $\mathcal A$ have finite character. For all $A \in \mathcal A$ and all $x \in X$, let either: :$A \cup \set x \in \mathcal A$ or: :$A \cup \set {x'} \in \mathcal A$ Then there exists a $B \in \mathcal A$ such that for all $x \in X$, either $x \in B$ or $x' \in B$.


Restricted Tukey-Teichmüller Theorem/Strong Form

Let $X$ be a set. Let $\mathcal A$ be a non-empty set of subsets of $X$. Let $'$ be a unary operation on $X$. Let $\mathcal A$ have finite character. For all $A \in \mathcal A$ and all $x \in X$, let either: : $A \cup \left\{ {x}\right\} \in \mathcal A$ or: :$A \cup \left\{ {x'}\right\} \in \mathcal A$ Then for each $A \in \mathcal A$ there exists a $C \in \mathcal A$ such that: :$A \subseteq C$ and: :for all $x \in X$, either $x \in C$ or $x' \in C$.


Boolean Prime Ideal Theorem/Extension Lemma

Let $\struct {B, \vee, \wedge, \neg, \bot, \top}$ be a Boolean algebra. Let $J \subseteq B$ have the finite join property. Let $z \in B$. Then either $J \vee z$ or $J \vee \neg z$ also has the finite join property.


Finite Character for Sets of Mappings

Let $S$ and $T$ be sets. Let $\mathcal F$ be a set of mappings from subsets of $S$ to $T$. That is, let $\mathcal F$ be a set of partial mappings from $S$ to $T$. Then the following are equivalent: {{begin-axiom}} {{axiom | n = 1         | t = $\mathcal F$ has finite character in the sense of Definition:Finite Character/Mappings. }} {{axiom | n = 2         | t = $\mathcal F$ has finite character as a set of subsets of $S \times T$ in the sense of Definition:Finite Character. }} {{end-axiom}}


Cowen-Engeler Lemma

Let $X$ and $Y$ be sets. Let $M$ be a set of mappings from subsets of $X$ to $Y$. Thus each element of $M$ is a partial mapping from $X$ to $Y$. Define a mapping $\Phi: X \to \powerset Y$ thus: :$\map \Phi x = \set {\map f x: f \in M \text{ and } x \in \Dom f}$ That is: :$\map \Phi x = \set {y \in Y: \exists f \in M: \tuple {x, y} \in f}$ Suppose that the following hold: {{begin-axiom}} {{axiom | n = 1         | t = $\map \Phi x$ is finite for each $x \in X$. }} {{axiom | n = 2         | t = For each finite subset $F$ of $X$, there exists an element $f \in M$ such that the domain of $f$ is $F$. }} {{axiom | n = 3         | t = $M$ has finite character. }} {{axiom | t = That is, a mapping $f$ from some subset of $X$ to $Y$ is an element of $M$ {{iff}} its restriction to each finite subset of $\Dom f$ is in $M$. }} {{end-axiom}} Then there exists an element of $M$ with domain $X$.


Ultrafilter Lemma/Corollary

Let $S$ be a non-empty set. Let $\mathcal A$ be a set of subsets of $S$. Suppose that $\mathcal A$ has the finite intersection property. Then there is an ultrafilter $\mathcal U$ on $S$ such that $\mathcal A \subseteq \mathcal U$.


Order-Extension Principle/Strict

Let $S$ be a set. Let $\prec$ be a strict ordering on $S$. Then there exists a strict total ordering $<$ on $S$ such that: :$\forall a, b \in S: a \prec b \implies a < b$


Order-Extension Principle/Strict/Finite Set

Let $T$ be a finite set. Let $\prec$ be a strict ordering on $T$. Then there exists a strict total ordering $<$ on $T$ such that: :$\forall a, b \in T: \left({a \prec b \implies a < b}\right)$


Finitely Satisfiable Theory has Maximal Finitely Satisfiable Extension/Lemma

Let $T$ be a finitely satisfiable $\mathcal L$-theory. Let $\phi$ be an $\mathcal L$-sentence. Then either $T \cup \left\{ {\phi}\right\}$ or $T \cup \left\{ {\neg \phi}\right\}$ is finitely satisfiable.