nyuuzyou/wikis huggingface dataset:
Characterization of Prime Element in Meet Semilattice

== Theorem ==
Let $L = \struct {S, \wedge, \preceq}$ be a meet semilattice.
Let $p \in S$,
Then:
:$p$ is prime element
:
:for all non-empty finite subsets $A$ of $S$:
::if $\inf A \preceq p$, then there exists element $x$ of $A$ such that $x \preceq p$.
== Proof ==
=== Sufficient Condition ===
Let $p$ be prime element.
Let $A$ be non-empty finite subsets of $S$.
Define
:$\map P X : \equiv X \ne \O \land \inf X \preceq p \implies \exists x \in X: x \preceq p$
where $X \subseteq S$.
We will prove that
:$\forall x \in A, B \subseteq A: \map P B \implies \map P {B \cup \set x}$
Let $x \in A, B \subseteq A$ such that
:$\map P B$
and
:$B \cup \set x \ne \O$ and $\map \inf {B \cup \set x} \preceq p$
Case $B = \O$.
By Union with Empty Set:
:$B \cup \set x = \set x$
By Infimum of Singleton:
:$\inf \set x = x$
By definition of singleton:
:$x \in \set x$
Thus
:$\exists z \in B \cup \set x: z \preceq p$
Case $B \ne \O$.
By Subset of Finite Set is Finite:
:$B$ is finite.
By Existence of Non-Empty Finite Infima in Meet Semilattice:
:$B, \set x$ admit infima.
By Infimum of Infima:
:$\map \inf {B \cup \set x} = \inf B \wedge \inf \set x$
By definition of prime element:
:$\inf B \preceq p$ or $x \preceq p$
Case $\inf B \preceq p$.
By assumption:
:$\exists z \in B: z \preceq p$
By definition of union:
:$z \in B \cup \set x$
Thus
:$\exists z \in B \cup \set x: z \preceq p$
Case $x \preceq p$.
By definitions of union and singleton:
:$x \in B \cup \set x$
Thus
:$\exists z \in B \cup \set x: z \preceq p$
By definition of empty set:
:$\map P \O$
By Induction of Finite Set:
:$\map P A$
Thus
:if $\inf A \preceq p$, then there exists element $x$ of $A$ such that $x \preceq p$.
=== Necessary Condition ===
Suppose that
:for all non-empty finite subsets $A$ of $S$:
::if $\inf A \preceq p$, then there exists element $x$ of $A$ such that $x \preceq p$.
Let $x, y \in S$ such that
:$x \wedge y \preceq p$
Define $A := \set {x, y}$
By definitions of non-empty set and finite set:
:$A$ is non-empty finite subset of $S$.
By definition of meet:
:$\inf A = x \wedge y$
By assumption:
:$\exists z \in A: z \preceq p$
Thus by definition of unordered tuple:
:$x \preceq p$ or $y \preceq p$
== Sources ==
* 
* 
Category:Prime Elements
Characterization of Prime Filter by Finite Suprema

== Theorem ==
Let $L = \struct {S, \vee, \preceq}$ be a join semilattice.
Let $F$ be a filter in $L$.
Then
:$F$ is a prime filter
:
:for all non-empty finite subset $A$ of $S: \paren {\sup A \in F \implies \exists a \in A: a \in F}$
== Proof ==
=== Sufficient Condition ===
Let $F$ be a prime ideal.
Define $\map \PP X: \equiv X \ne \O \land \sup X \in F \implies \exists x \in X: x \in F$
where $X$ is subset of $S$.
Let $A$ be a non-empty finite subset of $S$.
By definition of empty set:
:$\map \PP \O$
We will prove that:
:$\forall x \in A, B \subseteq A: \map \PP B \implies \map \PP {B \cup \set x}$
Let $x \in A, B \subseteq A$ such that:
:$\map \PP B$ (Induction Hypothesis)
Assume that:
:$B \cup \set x \ne \O$ and $\map \sup {B \cup \set x} \in F$
Case $B = \O$:
By Union with Empty Set:
:$B \cup \set x = \set x$
By Supremum of Singleton:
:$\sup \set x = x$
By definition of singleton:
:$x \in \set x$
Thus
:$\exists a \in B \cup \set x: a \in F$
Case $B \ne \O$:
By Subset of Finite Set is Finite:
:$B$ is finite.
By Existence of Non-Empty Finite Suprema in Join Semilattice:
:$B$ admits a supremum.
By Supremum of Singleton:
:$\set x$ admits a supremum.
By definition of prime filter:
:$\sup B \in F$ or $x \in F$
Case $\sup B \in F$:
By Induction Hypothesis:
:$\exists a \in B: a \in F$
By definition of union:
:$a \in B \cup \set x$
Thus:
:$\exists a \in B \cup \set x: a \in F$
Case $x \in F$:
By definition of union:
:$x \in B \cup \set x$
Thus:
:$\exists a \in B \cup \set x: a \in F$
By Induction of Finite Set:
:$\map \PP A$
Thus the result.
=== Necessary Condition ===
Suppose that
:for all non-empty finite subset $A$ of $S: \paren {\sup A \in F \implies \exists a \in A: a \in F}$
Let $x, y \in S$ such that
:$x \vee y \in F$
By Unordered Pair is Finite:
:$\set {x, y}$ is a finite set.
By definition of unordered tuple:
:$x \in \set {x, y}$
By definition of non-empty set:
:$\set {x, y}$ is a non-empty set.
By definition of join:
:$\sup \set {x, y} = x \vee y$
By assumption:
:$\exists a \in \set {x, y}: a \in F$
Thus by definition of unordered tuple:
:$x \in F$ or $y \in F$
Hence $I$ is prime filter.
== Sources ==
* 
* 
Category:Prime Ideals (Order Theory)
Characterization of Prime Ideal

== Theorem ==
Let $L = \struct {S, \wedge, \preceq}$ be a meet semilattice.
Let $I$ be an proper ideal in $L$.
Then
:$I$ is a prime ideal
:
:$\forall x, y \in S: \paren {x \wedge y \in I \implies x \in I \lor y \in I}$
== Proof ==
=== Sufficient Condition ===
Assume that
:$I$ is a prime ideal.
Let $x, y \in S$ such that
:$x \wedge y \in I$
By definition of relative complement:
:$x \wedge y \notin \relcomp S I$
By definition of prime ideal:
:$\relcomp S I$ is filter in $L$.
By Filtered in Meet Semilattice:
:$x \notin \relcomp S I$ or $y \notin \relcomp S I$
Thus by definition of relative complement:
:$x \in I$ or $y \in I$
=== Necessary Condition ===
Assume that
:$\forall x, y \in S: \paren {x \wedge y \in I \implies x \in I \lor y \in I}$
Define $F := \relcomp S I$.
By definition of proper subset:
:$\exists x_0 \in S: x_0 \notin I$
By definition of relative complement:
:$x_0 \in F$
Thus by definition
:$F$ is a non-empty set.
We will prove that
:$F$ is filtered.
Let $x, y \in F$.
By definition of relative complement:
:$x \notin I$ and $y \notin I$
By assumption:
:$x \wedge y \notin I$
By definition of relative complement:
:$x \wedge y \in F$
By Meet Precedes Operands:
:$x \wedge y \preceq x$ and $x \wedge y \preceq y$
Thus
:$\exists z \in F: z \preceq x \land z \preceq y$
We will prove that:
:$F$ is an upper section.
Let $x \in F, y \in S$ such that
:$x \preceq y$
By definition of lower section:
:$y \in I \implies x \in I$
By definition of relative complement:
:$x \notin I$
Then
:$y \notin I$
Thus by definition of relative complement:
:$y \in F$
Thus by definition:
:$F$ is a filter in $L$.
== Sources ==
* 
* 
Category:Prime Ideals (Order Theory)
Characterization of Prime Ideal by Finite Infima

== Theorem ==
Let $L = \struct {S, \wedge, \preceq}$ be a meet semilattice.
Let $I$ be an ideal in $L$.
Then
:$I$ is a prime ideal
:for all non-empty finite subset $A$ of $S: \paren {\inf A \in I \implies \exists a \in A: a \in I}$
== Proof ==
=== Sufficient Condition ===
Let $I$ be a prime ideal.
Define:
:$\map P X: \equiv X \ne \O \land \inf X \in I \implies \exists x \in X: x \in I$
where $X$ is subset of $S$.
Let $A$ be a non-empty finite subset of $S$.
By definition of empty set:
:$\map P \O$
We will prove that:
:$\forall x \in A, B \subseteq A: \map P B \implies \map P {B \cup \set x}$
Let $x \in A, B \subseteq A$ such that:
:$\map P B$
This will be used as an induction hypothesis.
Assume that:
:$B \cup \set x \ne \O$ and $\map \inf {B \cup \set x} \in I$
;Case $B = \O$
By Union with Empty Set:
:$B \cup \set x = \set x$
By Infimum of Singleton:
:$\inf \set x = x$
By definition of singleton:
:$x \in \set x$
Thus
:$\exists a \in B \cup \set x: a \in I$
;Case $B \ne \O$
By Subset of Finite Set is Finite:
:$B$ is finite.
By Existence of Non-Empty Finite Infima in Meet Semilattice:
:$B$ admits an infimum.
By Infimum of Singleton:
:$\set x$ admits an infimum.
By Characterization of Prime Ideal:
:$\inf B \in I$ or $x \in I$
;Case $\inf B \in I$
By the induction hypothesis:
:$\exists a \in B: a \in I$
By definition of union:
:$a \in B \cup \set x$
Thus:
:$\exists a \in B \cup \set x: a \in I$
;Case $x \in I$:
By definition of union:
:$x \in B \cup \set x$
Thus
:$\exists a \in B \cup \set x: a \in I$
By Induction of Finite Set:
:$\map P A$
Thus the result.
=== Necessary Condition ===
Suppose that:
:for all non-empty finite subset $A$ of $S: \paren {\inf A \in I \implies \exists a \in A: a \in I}$
Let $x, y \in S$ such that:
:$x \wedge y \in I$
By Unordered Pair is Finite:
:$\set {x, y}$ is a finite set.
By definition of unordered tuple:
:$x \in \set {x, y}$
By definition of non-empty set:
:$\set {x, y}$ is a non-empty set.
By definition of meet:
:$\inf \set {x, y} = x \wedge y$
By assumption:
:$\exists a \in \set {x, y}: a \in I$
Thus by definition of unordered tuple:
:$x \in I$ or $y \in I$
Hence by Characterization of Prime Ideal:
:$I$ is prime ideal.
== Sources ==
* 
* 
Category:Prime Ideals (Order Theory)
Characterization of Probability Density Function

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X: \Omega \to \R$ be a continuous random variable on $\struct {\Omega, \Sigma, \Pr}$.
Let $\Omega_X = \Img X$, the image of $X$.
Let the probability density function of $X$ is the mapping $f_X: \R \to \closedint 0 1$ be defined as:
:$\forall x \in \R: \map {f_X} x = \begin {cases}
\ds \lim_{\epsilon \mathop \to 0^+} \frac {\map \Pr {x - \frac \epsilon 2 \le X \le x + \frac \epsilon 2} } \epsilon & : x \in \Omega_X \\
0 & : x \notin \Omega_X
\end {cases}$
Suppose that the cumulative distribution function of $X$ defines a continuously differentiable real function $F_X: x \mapsto \map \Pr {X \le x}$.
Then the probability density function of $X$ is the mapping $f_X: \R \to \closedint 0 1$ satisfies:
:$\dfrac \d {\d x} \map {F_X} x = \map {f_X} x$.
== Proof ==
== Sources ==
* : $\S 4.1$
Category:Probability
Characterization of Pseudoprime Element by Finite Infima

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a continuous lattice.
Let $p \in S$ be a pseudoprime element.
Let $A$ be a non-empty finite subset of $S$ such that
:$\inf A \ll p$
where $\ll$ denotes the way below relation.
Then $\exists a \in A: a \preceq p$
== Proof ==
By definition of pseudoprime element:
:there exists a prime ideal $P$ in $L$: $p = \sup P$
By definition of way below closure:
:$\inf A \in p^\ll$
By definition of reflexivity:
:$p \preceq \sup P$
By Continuous iff Way Below Closure is Ideal and Element Precedes Supremum:
:$p^\ll \subseteq P$
By definition of subset:
:$\inf A \in P$
Thus by Characterization of Prime Ideal by Finite Infima
:$\exists a \in A: a \in P$
By definition of up-complete:
:$P$ admits a supremum.
By definition of supremum:
:$p$ is upper bound for $P$.
Thus by definition of upper bound:
:$a \preceq p$
== Sources ==
* 
* 
Category:Prime Elements
Characterization of Pseudoprime Element when Way Below Relation is Multiplicative

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below continuous lattice such that
:$\ll$ is multiplicative relation
where $\ll$ denotes the way below relation of $L$.
Let $p \in S$.
Then $p$ is pseudoprime element 
:$\forall a, b \in S: a \wedge b \ll p \implies a \preceq p \lor b \preceq p$
== Proof ==
=== Sufficient Condition ===
Let $p$ be pseudoprime element.
Let $a, b \in S$ such that
:$a \wedge b \ll p$
By definition of meet:
:$\inf \left\{ {a, b}\right\} \ll p$
By Characterization of Pseudoprime Element by Finite Infima:
:$\exists c \in \left\{ {a, b}\right\}: c \preceq p$
Thus
:$a \preceq p$ or $b \preceq p$
=== Necessary Condition ===
Suppose
:$\forall a, b \in S: a \wedge b \ll p \implies a \preceq p \lor b \preceq p$
:
:$p$ is not a pseudoprime element.
By Prime is Pseudoprime (Order Theory):
:$p$ is not a prime element.
By definition of prime element:
:$\exists x, y \in S: x \wedge y \preceq p$ and $x \npreceq p$ and $y \npreceq p$
By definition of continuous:
:$\forall z \in S: z^\ll$ is directed.
and
:$L$ satisfies axiom of approximation.
By Axiom of Approximation in Up-Complete Semilattice:
:$\exists u \in S: u \ll x \land u \npreceq p$
and
:$\exists v \in S: v \ll y \land v \npreceq p$
By Way Below Relation is Auxiliary Relation:
:$\ll$ is auxiliary relation.
By Multiplicative Auxiliary Relation iff Congruent:
:$u \wedge v \ll x \wedge y$
By definition of transitivity:
:$u \wedge v \ll p$
By assumption:
:$u \preceq p$ or $v \preceq p$
This contradicts $u \npreceq p$ and $v \npreceq p$.
== Sources ==
* 
* 
Category:Prime Elements
Category:Way Below Relation
Characterization of Rational P-adic Integer

== Theorem ==
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.
Let $\Z_p$ be the $p$-adic integers for some prime $p$.
Let $\Q$ be the rational numbers.
Then:
:$\Z_p \cap \Q = \set{\dfrac a b \in \Q : p \nmid b}$
== Proof ==
Let $\norm{\,\cdot\,}^\Q _p$ denote the $p$-adic norm on the rational numbers.
We have:
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$: Lemma $3.3.4 \ \text{(i)}$
* : $\S 1.6$ The $p$-adic expansion of rational numbers: Exercise $32 \ (1)$
Category:P-adic Integers
Characterization of Rational P-adic Unit

== Theorem ==
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.
Let $\Z^\times_p$ be the $p$-adic units.
Let $\Q$ be the rational numbers.
Then:
:$\Z^\times_p \cap \Q = \set{\dfrac a b \in \Q : p \nmid ab}$
== Proof ==
Let $\norm{\,\cdot\,}^\Q _p$ denote the $p$-adic norm on the rational numbers.
We have:
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$
* : $\S 1.6$ The $p$-adic expansion of rational numbers: Exercise $32 \ (2)$
Category:P-adic Integers
Characterization of Reducing Subspaces

== Theorem ==
Let $\HH$ be a Hilbert space.
Let $A \in \map B \HH$ be a bounded linear operator.
Let $M$ be a closed linear subspace of $\HH$.
Let $P$ denote the orthogonal projection on $M$.
Let $\begin{pmatrix}
 W & X \\
 Y & Z
\end{pmatrix}$ be the matrix notation for $A$ with respect to $M$.
Then the following four statements are equivalent:
:$(1): \quad M$ is a reducing subspace for $A$
:$(2): \quad P A = A P$
:$(3): \quad X = Y = 0$
:$(4): \quad M$ is an invariant subspace for both $A$ and its adjoint $A^*$
== Proof ==
== Sources ==
* $\text {II}.3.7$
Category:Hilbert Spaces
Category:Linear Transformations on Hilbert Spaces
Characterization of Set Equals Union of Sets

== Theorem ==
Let $A$ be a set.
Let $\BB$ be a set of sets.
Then $A = \ds \bigcup \BB$ :
:$\forall a \in A : \exists B \in \BB : a \in B$
:$\forall B \in \BB : B \subseteq A$
== Proof ==
=== Necessary Condition ===
Let $A = \ds \bigcup \BB$.
By definition of set union:
:$\forall a \in A = \ds \bigcup \BB : \exists B \in \BB : a \in B$
From Set is Subset of Union:
:$\forall B \in \BB : B \subseteq \ds \bigcup \BB = A$
=== Sufficient Condition ===
Let:
:$\forall a \in A : \exists B \in \BB : a \in B$
:$\forall B \in \BB : B \subseteq A$
From set union
:$\forall a \in A : a \in \bigcup \BB$
By definition of subset:
:$A \subseteq \bigcup \BB$
From Union of Subsets is Subset:
:$\bigcup \BB \subseteq A$
By definition of set equality:
:$A = \bigcup \BB$
Category:Set Union
Characterization of Sigma-Algebra Generated by Collection of Mappings

== Theorem ==
Let $\struct {X_i, \Sigma_i}$ be measurable spaces, with $i \in I$ for some index set $I$.
Let $X$ be a set, and let, for $i \in I$, $f_i: X \to X_i$ be a mapping.
Then:
:$\map \sigma {f_i: i \in I} = \map \sigma {\ds \bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} }$
where:
:$\map \sigma {f_i: i \in I}$ is the $\sigma$-algebra generated by $\family {f_i}_{i \mathop \in I}$
:$\map \sigma {\ds \bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} }$ is the $\sigma$-algebra generated by $\ds \bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i}$
:$\map {f_i^{-1} } {\Sigma_i}$ denotes the pre-image $\sigma$-algebra on $X$ by $f$
== Proof ==
For each $i \in I$, one has by definition of generated $\sigma$-algebra:
:$\ds \map {f_i^{-1} } {\Sigma_i} \subseteq \bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} \subseteq \map \sigma {\bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} }$
which shows that each of the $f_i$ is measurable.
Next, suppose that $\Sigma$ is a $\sigma$-algebra such that each of the $f_i$ is $\Sigma \,/\, \Sigma_i$-measurable.
Then for all $i \in I$, one has:
:$\map {f_i^{-1} } {\Sigma_i} \subseteq \Sigma$
and hence by Union is Smallest Superset: Family of Sets:
:$\ds \bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} \subseteq \Sigma$
Finally, by Generated Sigma-Algebra Preserves Subset, it follows that:
:$\ds \map \sigma {\bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} } \subseteq \Sigma$
Thus:
:$\ds \map \sigma {\bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} } = \map \sigma {f_i : i \in I}$
by definition of the latter.
== Sources ==
* : $7.5$
* : $\S 7$: Problem $3 \ \text{(ii)}$
Category:Sigma-Algebras Generated by Collection of Mappings
Characterization of Spectrum of Bounded Linear Operator in Hilbert Space in terms of Approximate Eigenvalues/Corollary

== Corollary ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $T : \HH \to \HH$ be a self-adjoint bounded linear operator.
Let $\map \sigma T$ be the spectrum of $T$.
Then $\lambda \in \map \sigma T$ $\lambda$ is an approximate eigenvalue of $T$.
== Proof ==
From Adjoint is Involutive, we have $T^{\ast \ast} = T$.
Hence from Characterization of Spectrum of Bounded Linear Operator in Hilbert Space in terms of Approximate Eigenvalues, we have that $\lambda \in \map \sigma T$ $\lambda$ is an approximate eigenvalue of $T$ or $\overline \lambda$ is an eigenvalue of $T$.
From Spectrum of Self-Adjoint Bounded Linear Operator is Real, we have:
:$\map \sigma T \subseteq \R$ 
Hence we have that $\lambda \in \map \sigma T$ only if $\lambda \in \R$. 
So we obtain that $\lambda \in \map \sigma T$ $\lambda$ is an approximate eigenvalue of $T$ or $\lambda$ is an eigenvalue of $T$.
From Eigenvalue of Bounded Linear Operator is Approximate Eigenvalue, an eigenvalue is an approximate eigenvalue, so we conclude that $\lambda \in \map \sigma T$ $\lambda$ is an approximate eigenvalue of $T$.
Category:Characterization of Spectrum of Bounded Linear Operator in Hilbert Space in terms of Approximate Eigenvalues
Characterization of Stationary Gaussian Process

== Theorem ==
Let $S$ be a Gaussian stochastic process giving rise to a time series $T$.
Let the the mean of $S$ be fixed.
Let the autocovariance matrix of $S$ also be fixed.
Then $S$ is stationary.
== Proof ==
From Characterization of Multivariate Normal Distribution, the normal distribution is completely characterized by its expectation and its variance.
The result follows.
== Sources ==
* :
::Part $\text {I}$: Stochastic Models and their Forecasting:
:::$2$: Autocorrelation Function and Spectrum of Stationary Processes:
::::$2.1$ Autocorrelation Properties of Stationary Models:
:::::$2.1.3$ Positive Definiteness and the Autocovariance Matrix: Gaussian processes
Category:Stochastic Processes
Characterization of Strictly Increasing Mapping on Woset

== Theorem ==
Let $J$ and $E$ be well-ordered sets.
Let $h: J \to E$ be a mapping.
Let $S_\alpha$ denote an initial segment determined by $\alpha$.
:$(1): \quad h$ is strictly increasing and its image is either all of $E$ or an initial segment of $E$
:$(2): \quad \forall \alpha \in J: \map h \alpha = \map \min {E \setminus h \sqbrk {S_\alpha} }$, and $h \sqbrk {S_\alpha} = S_{\map h \alpha}$
where:
:$h \sqbrk {S_\alpha}$ denotes the image of $S_\alpha$ under $h$
:$\min$ denotes the smallest element of the set.
== Proof ==
=== $(1)$ implies $(2)$ ===
Suppose $h$ satisfies:
:$h$ is strictly increasing and its image is either all of $E$ or an initial segment of $E$
Then for any $x,y \in J$:
=== $(2)$ implies $(1)$ ===
Suppose $h$ satisfies:
:$\map h \alpha = \map \min {E \setminus h \sqbrk {S_\alpha} }$
By the Principle of Recursive Definition for Well-Ordered Sets, $h$ is thus uniquely determined.
Then:
Thus for every $x \in S_y$, we have that:
:$\map h x \in S_{\map h y}$.
Therefore $h$ is an strictly increasing mapping.
Furthermore, the image set of $h$ is the union of initial segments in $E$.
By Union of Initial Segments is Initial Segment or All of Woset:
:$h \sqbrk J$ 
is an initial segment of $E$ or all of $E$.
== Sources ==
* : Supplementary Exercise $1.2$
Category:Well-Orderings
Characterization of Supremum Precedes Element

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$ admit a supremum $a$
Let $b \in S$.
Then:
:$a \preceq b$
:
:$\forall t \in T : t \preceq b$
== Proof ==
=== Necessary Condition ===
Let $a \preceq b$.
By definition of supremum:
:$a$ is an upper bound for $T$
By definition of upper bound:
:$\forall t \in T : t \preceq a$
From :
:$\forall t \in T : t \preceq b$
=== Sufficient Condition ===
Let:
:$\forall t \in T : t \preceq b$
By definition of upper bound:
:$b$ is an upper bound for $T$
By definition of supremum:
:$a \preceq b$
Category:Suprema
Characterization of T0 Space by Closed Sets

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a topological space.
Then
:$T$ is a $T_0$ space 
::for every points $x, y \in S$ if $x \ne y$ then
:::there exists a closed subset $F$ of $S$ such that $x \in F$ and $y \notin F$
::or
:::there exists a closed subset $F$ of $S$ such that $x \notin F$ and $y \in F$
== Proof ==
=== Sufficient Condition ===
Let $T$ be a $T_0$ space.
Let $x, y \in S$ such that
:$x \ne y$
By definition of $T_0$ space:
:$\left({\exists U \in \tau: x \in U \land y \notin U}\right) \lor \left({\exists U \in \tau: x \notin U \land y \in U}\right)$
WLOG: Suppose:
:$\exists U \in \tau: x \in U \land y \notin U$
By definition:
:$\complement_S\left({U}\right)$ is closed
where $\complement_S\left({U}\right)$ denotes the relative complement of $U$ in $S$.
By definition of relative complement:
:$x \notin \complement_S\left({U}\right) \land y \in \complement_S\left({U}\right)$
Thus:
:$\exists F \subseteq S: F$ is closed $\land\, x \notin F \land y \in F$
=== Necessary Condition ===
This statement follows mutatis mutandis.
== Sources ==
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Category:T0 Spaces
Characterization of T0 Space by Closures of Singletons

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Then
:$T$ is a $T_0$ space :
::$\forall x, y \in S: x \ne y \implies x \notin \set y^- \lor y \notin \set x^-$
where $\set y^-$ denotes the closure of $\set y$.
== Proof ==
=== Sufficient Condition ===
Let $T$ be a $T_0$ space.
Let $x, y \in S$ such that
:$x \ne y$
:$x \in \set y^- \land y \in \set x^-$
Then:
By Characterization of $T_0$ Space by Distinct Closures of Singletons:
:$\set x^- \ne \set y^-$
This contradicts the equality.
Thus the result by Proof by Contradiction
=== Necessary Condition ===
Assume that:
:$(1): \quad \forall x, y \in S: x \ne y \implies x \notin \set y^- \lor y \notin \set x^-$
By Characterization of $T_0$ Space by Distinct Closures of Singletons it suffices to prove
:$\forall x, y \in S: x \ne y \implies \set x^- \ne \set y^-$
Let $x, y \in S$ such that $x \ne y$.
:$(2): \quad \set x^- = \set y^-$
Then:
This contradicts the assumption $(1)$.
Thus the result by Proof by Contradiction.
== Sources ==
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Category:T0 Spaces