nyuuzyou/wikis huggingface dataset:
Extreme Set in Compact Convex Set contains Extreme Point/Lemma

== Lemma ==
Let $X$ be a vector space over $\R$. 
Let $K$ be a non-empty compact convex subset of $X$.
Let $E$ be an extreme set of $K$. 
Let $P$ be the set of extreme sets in $K$ that are contained in $E$. 
Define an ordering $\preceq$ on $P$ by $A \preceq B$ $B \subseteq A$. 
Then: 
:every non-empty chain in $\struct {P, \preceq}$ has an upper bound.
== Proof ==
Let $C$ be a non-empty chain in $\struct {P, \preceq}$.
Let:
:$\ds D = \bigcap_{E \in C} E$
We first show that: 
:$D \ne \O$
 suppose that $D = \O$.
Then, from de Morgan's laws: 
:$\ds K = \bigcup_{E \in C} \paren {K \setminus E}$
Since each $E \in C$ is closed, we have that: 
:$K \setminus E$ is open in $K$.
So:
:$\set {K \setminus E : E \in C}$ is an open cover for $K$. 
Since $K$ is compact, we have that:
:there exists a finite subcover $\set {K \setminus E_i : i \in \set {1, 2, \ldots, n} }$ of $\set {K \setminus E : E \in C}$ for $K$.
That is: 
:$\ds K = \bigcup_{i \mathop = 1}^n \paren {K \setminus E_i}$
From de Morgan's laws, we then have: 
:$\ds \O = \bigcap_{i \mathop = 1}^n E_i$
However, from Finite Non-Empty Subset of Ordered Set has Maximal and Minimal Elements: Corollary, there exists $i, j$ such that: 
:$E_i \preceq E_k \preceq E_j$
for each $k \in \set {1, 2, \ldots, n}$. 
Then: 
:$\ds E_j \subseteq E_k$
for each $k \in \set {1, 2, \ldots, n}$.
So, from Set Intersection Preserves Subsets, we have: 
:$\ds E_j \subseteq \bigcap_{i \mathop = 1}^n E_i$
Since $E_j$ is non-empty, we have: 
:$\ds \bigcap_{i \mathop = 1}^n E_i \ne \O$
a contradiction. 
So, we have: 
:$D \ne \O$
We now show that $D$ is an extreme set of $K$, and that it is an upper bound for $C$.
From Intersection of Closed Sets is Closed in Normed Vector Space, $D$ is closed.
Let $x, y \in K$ and $t \in \openint 0 1$ be such that $t x + \paren {1 - t} y \in D$.
Then, from the definition of set intersection, we have: 
:$t x + \paren {1 - t} y \in E$
for each $E \in C$.
Since $E$ is an extreme set, we have $x, y \in E$ for each $E \in C$.
So, $x, y \in D$. 
So $D$ is an extreme set of $K$
From Intersection is Subset, we then have: 
:$\ds D = \bigcap_{E \in C} E \subseteq F$
for each $F \in C$. 
So, we have: 
:$F \preceq D$
for each $F \in C$. 
So $D$ is an upper bound for $C$.
Since $C$ was an arbitrary non-empty chain in $\struct {P, \preceq}$, we have: 
:every non-empty chain in $\struct {P, \preceq}$ has an upper bound.
Category:Extreme Set in Compact Convex Set contains Extreme Point
Extreme Value Theorem

== Theorem ==
Let $X$ be a compact metric space and $Y$ a normed vector space.
Let $f: X \to Y$ be a continuous mapping.
Then $f$ is bounded, and there exist $x, y \in X$ such that:
:$\forall z \in X: \norm {\map f x} \le \norm {\map f z} \le \norm {\map f y}$
where $\norm {\map f x}$ denotes the norm of $\map f x$.
Moreover, $\norm f$ attains its minimum and maximum.
=== Extreme Value Theorem for a Real Function ===
=== Extreme Value Theorem for Normed Vector Spaces ===
== Proof ==
By Continuous Image of Compact Space is Compact, $f \sqbrk X \subseteq Y$ is compact.
Therefore, by Compact Subspace of Metric Space is Bounded, $f$ is bounded.
Let $\ds A = \inf_{x \mathop \in X} \norm {\map f x}$.
It follows from the definition of infimum that there exists a sequence $\sequence {y_n}$ in $X$ such that:
:$\ds \lim_{n \mathop \to \infty} \norm {\map f {y_n} } = A$
By Sequence of Implications of Metric Space Compactness Properties, $X$ is sequentially compact.
So there exists a convergent subsequence $\sequence {x_n}$ of $\sequence {y_n}$.
Let $\ds x = \lim_{n \mathop \to \infty} x_n$.
Since $f$ is continuous and a norm is continuous, it follows by Composite of Continuous Mappings at Point is Continuous that:
:$\ds \norm {\map f x} = \norm {\map f {\lim_{n \mathop \to \infty} x_n} } = \norm {\lim_{n \mathop \to \infty} \map f {x_n} } = \lim_{n \mathop \to \infty} \norm {\map f {x_n} } = A$
So $\norm f$ attains its minimum at $x$.
By replacing the infimum with the supremum in the definition of $A$, we also see that $\norm f$ attains its maximum by the same reasoning.
== Historical Note ==
== Sources ==
* 
Category:Metric Spaces
Category:Continuous Functions
Extreme Value Theorem/Historical Note

== Historical Note on Extreme Value Theorem ==
The Extreme Value Theorem in its application to real functions is usually attributed to , as an example of what has been referred to as Weierstrassian rigor.
Hence this result's soubriquet the Weierstrass Extreme Value Theorem.
== Sources ==
* : Chapter $\text {A}.33$: Weierstrass ()
Category:Historical Notes
Extreme Value Theorem/Normed Vector Space

== Theorem ==
Let $X$ be a normed vector space.
Let $K \subseteq X$ be a compact subset.
Suppose $f : X \to \R$ is a continuous mapping at each $x \in K$.
Then:
:$\ds \exists c \in K : \map f c = \sup_{x \mathop \in K} \map f x = \max_{x \mathop \in K} \map f x$
:$\ds \exists d \in K : \map f d = \inf_{x \mathop \in K} \map f x = \min_{x \mathop \in K} \map f x$
== Proof ==
Let $K$ be compact.
By Continuous Mappings preserve Compact Subsets, $\map f K$ is compact.
By Compact Subset of Normed Vector Space is Closed and Bounded, $K$ is bounded.
Hence, $\map f K$ is bounded.
$K$ is nonempty, so $\map f K$ is non-empty.
By Characterizing Property of Supremum of Subset of Real Numbers, non-empty bounded subset of $\R$ has least upper bound.
Let $\ds M := \sup_{x \mathop \in K} \map f x$.
Then $M \in \R$.
Let $n \in \N$.
We have that:
$\ds M - \frac 1 n < M$
However, $M$ is the least upper bound.
Hence, $\ds M - \frac 1 n$ cannot be an upper bound for $\map f K$.
Therefore:
:$\ds \exists x_n \in K : \map f {x_n} > M - \frac 1 n$
Thus we have a sequence $\sequence {x_n}_{n \mathop \in \N}$ contained in $K$.
$K$ is compact.
By definition, there is a convergent subsequence $\sequence {x_{n_k}}_{k \mathop \in \N}$ with a limit $c \in K$.
$f$ is continuous.
By Continuous Mappings preserve Convergent Sequences, $\sequence {\map f {x_{n_k}}}_{k \mathop \in \N}$ converges with a limit $\map f c$.
But from $\ds \map f {x_n} > M - \frac 1 n$ it follows that $\map f c \ge M$.
On the other hand, by definition of $M$, $\map f c \le M$.
Therefore, $\map f c = M$.
Finally, since $c \in K$:
:$\ds \map f c \in \set {\map f x : x \in K}$.
Hence, $M$ is a maximum.
For the second part, let $g : X \to \R$ be such that:
:$\forall x \in X : \map g x := - \map f x$
Then above arguments apply, if infima, lower bounds and minimum are replaced with suprema, upper bounds and maximum.
== Historical Note ==
== Sources ==
* : Chapter $\S 2.2$: Continuous and linear maps. Continuous maps
Category:Continuous Mappings
Category:Suprema
Category:Infima
Category:Max and Min Operations
Extremities of Line Segments containing three Plane Angles any Two of which are Greater than Other form Triangle

== Theorem ==
== Proof ==
:
Let $\angle ABC, \angle DEF, \angle GHK$ be plane angles such that the sum of any two is greater than the remaining one.
That is:
:$\angle ABC + \angle DEF > \angle GHK$
:$\angle DEF + \angle GHK > \angle ABC$
:$\angle GHK + \angle ABC > \angle DEF$
Let the straight lines $AB, BC, DE, EF, GH, GK$ be equal.
Let $AC$, $DF$ and $GK$ be joined.
It is to be demonstrated that it is possible to construct a triangle from straight lines equal to $AC$, $DF$ and $HK$.
If $\angle ABC, \angle DEF, \angle GHK$ are equal to one another, then it is possible to construct an equilateral triangle from $AC$, $DF$ and $HK$.
:
Otherwise, let $\angle ABC, \angle DEF, \angle GHK$ be unequal.
On the straight line $HK$ at the point $H$, let:
:$\angle KHL$ be constructed equal to $\angle ABC$
:$HL$ be constructed equal to one of $AB, BC, DE, EF, GH, GK$
:$KL$ be joined.
We have that:
:$AB$ and $BC$ are equal to $KH$ and $HL$
:$\angle ABC = \angle KHL$
Therefore from :
:$AC = KL$
We have that:
:$\angle GHK + \angle ABC > \angle DEF$
while:
:$\angle ABC = \angle KHL$
Therefore:
:$\angle GHL > \angle DEF$
We have that:
:$GH$ and $HL$ are equal to $DE$ and $EF$
:$\angle GHL > \angle DEF$
Therefore from :
:$GL > DF$
But:
:$GK + KL > GL$
Therefore:
:$GK + KL > DF$
But:
:$KL = AC$
Therefore:
:$AC + GK > DF$
Similarly it can be proved that:
:$AC + DF > GK$
and that:
:$DF + GK > AC$
Hence the result.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Solid Geometry
F-Sigma Set is not necessarily Closed Set

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $X$ be an $F_\sigma$ set of $T$.
Then it is not necessarily the case that $X$ is a closed set of $T$.
== Proof ==
Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.
Let $X \subseteq S$ be an $F_\sigma$ set of $T$.
From $F_\sigma$ and $G_\delta$ Subsets of Uncountable Finite Complement Space:
:$X$ is neither open nor closed in $T$.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction
Category:F-Sigma Sets
Category:Closed Sets
F-Sigma Sets Closed under Intersection

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $F, F'$ be $F_\sigma$ sets of $T$.
Then their intersection $F \cap F'$ is also a $F_\sigma$ set of $T$.
== Proof ==
By definition of $F_\sigma$ set, there exist sequences $\sequence {C_n}_{n \mathop \in \N}$ and $\sequence {C'_n}_{n \mathop \in \N}$ of closed sets of $T$ such that:
:$F = \ds \bigcup_{n \mathop \in \N} C_n$
:$F' = \ds \bigcup_{n \mathop \in \N} C'_n$
Now compute:
By Intersection of Closed Sets is Closed in Topological Space, $C_n \cap C'_m$ is closed, for all $n, m \in \N$.
By Cartesian Product of Countable Sets is Countable, $\N \times \N$ is countable.
Thus $F \cap F'$ is seen to be a countable union of closed sets, hence a $F_\sigma$ set.
== Also see ==
* $F_\sigma$ Sets Closed under Union
* $G_\delta$ Sets Closed under Intersection
Category:F-Sigma Sets
F-Sigma Sets Closed under Union

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $F, F'$ be $F_\sigma$ sets of $T$.
Then their union $F \cup F'$ is also an $F_\sigma$ set of $T$.
== Proof ==
By definition of $F_\sigma$ set, there exist sequences $\sequence {C_n}_{n \mathop \in \N}$ and $\sequence {C'_n}_{n \mathop \in \N}$ of closed sets of $T$ such that:
:$F = \ds \bigcup_{n \mathop \in \N} C_n$
:$F' = \ds \bigcup_{n \mathop \in \N} C'_n$
By General Self-Distributivity of Set Union, we have:
:$F \cup F' = \ds \bigcup_{n \mathop \in \N} \paren {C_n \cup C'_n}$
By Finite Union of Closed Sets is Closed in Topological Space, $C_n \cup C'_n$ is closed, for all $n \in \N$.
Thus $F \cup F'$ is seen to be an $F_\sigma$ set.
== Also see ==
* $F_\sigma$ Sets Closed under Intersection
* $G_\delta$ Sets Closed under Union
Category:F-Sigma Sets
F-Sigma Sets form Lattice

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\FF$ be the collection of all $F_\sigma$ sets of $T$.
Then $\struct {\FF, \subseteq}$ is a lattice, where $\subseteq$ denotes the subset relation.
== Proof ==
From Subset Relation is Ordering, $\subseteq$ is an ordering on $\FF$.
Let $F, F'$ be $F_\sigma$ sets of $T$.
We have $F_\sigma$ Sets Closed under Union, so that $F \cup F' \in \FF$.
From Union is Smallest Superset and Subset of Union, it follows that $F \cup F'$ is the supremum of $F$ and $F'$.
Similarly, we have $F_\sigma$ Sets Closed under Intersection, and so $F \cap F' \in \FF$.
From Intersection is Largest Subset and Intersection is Subset, it follows that $F \cap F'$ is the infimum of $F$ and $F'$.
Thus any two elements of $\FF$ are seen to have both a supremum and an infimum in $\FF$.
Hence $\struct {\FF, \subseteq}$ is a lattice.
== Also see ==
* $G_\delta$ Sets form Lattice
Category:F-Sigma Sets
Category:Lattice Theory
F-Sigma Sets in Indiscrete Topology

== Theorem ==
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $H \subseteq S$.
$H$ is an $F_\sigma$ ($F$-sigma) set of $T$ either $H = S$ or $H = \O$.
== Proof ==
An $F_\sigma$ set is a set which can be written as a countable union of closed sets of $S$.
Hence the only $F_\sigma$ sets of $T$ are made from unions of $T$ and $\O$.
So $T$ and $\O$ are the only $F_\sigma$ sets of $T$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $2$
Category:Indiscrete Topology
Category:F-Sigma Sets
F-Sigma and G-Delta Subsets of Uncountable Finite Complement Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.
Then countably infinite subsets of $S$ are $F_\sigma$ sets and are neither open nor closed sets.
Their relative complements in $S$ are $G_\delta$ sets, and are also neither open nor closed sets.
== Proof ==
Let $U$ be a countably infinite subset of $S$.
Then it can be written as:
:$\ds U = \bigcup_{x \mathop \in U} \set x$
But as $\set x$ is a finite subset of $S$, it is by definition the complement of an open set in $T$.
So by definition $\set x$ is closed in $T$.
So:
:$\ds U = \bigcup_{x \mathop \in U} \set x$
is a countable union of closed sets of $T$.
That is, by definition, $U$ is an $F_\sigma$ set.
As $U$ is countably infinite, its complement is uncountable, not finite, and so $U$ is not open in $T$.
But as $U$ is also not finite, it is not the complement of an open set and so is not closed in $T$ either.
From Complement of $F_\sigma$ Set is $G_\delta$ Set it follows that the complement of $U$ is a $G_\delta$ set, and similarly neither open nor closed.
== Sources ==
* : Part $\text {II}$: Counterexamples: $19$. Finite Complement Topology on an Uncountable Space: $3$
Category:F-Sigma Sets
Category:G-Delta Sets
Category:Finite Complement Topology
Factor Principles/Conjunction on Left/Formulation 1

== Theorem ==
:$p \implies q \vdash \paren {r \land p} \implies \paren {r \land q}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: The Propositional Calculus $1$: $3$ Conjunction and Disjunction: Theorem $18$
Category:Factor Principles
Factor Principles/Conjunction on Left/Formulation 2

== Theorem ==
:$\vdash \paren {p \implies q} \implies \paren {\paren {r \land p} \implies \paren {r \land q} }$
== Proof ==
== Sources ==
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 3$: Theorem $\text{T30}$
Category:Factor Principles
Factor Principles/Conjunction on Right/Formulation 1/Proof by Truth Table

== Theorem ==
== Proof ==
Proof by Truth Table:
$\begin{array}{|ccc||ccccccccccc|}
\hline
p & q & r & (p & \implies & q) & \implies & ((p & \land & r) & \implies & (q & \land & r)) \\
\hline
\T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\
\T & \T & \F & \T & \T & \T & \T & \T & \F & \F & \T & \T & \F & \F \\
\T & \F & \T & \T & \F & \F & \T & \T & \T & \T & \F & \F & \F & \T \\
\T & \F & \F & \T & \F & \F & \T & \T & \F & \F & \T & \F & \F & \F \\
\F & \T & \T & \F & \T & \T & \T & \F & \F & \T & \T & \T & \T & \T \\
\F & \T & \F & \F & \T & \T & \T & \F & \F & \F & \T & \T & \F & \F \\
\F & \F & \T & \F & \T & \F & \T & \F & \F & \T & \T & \F & \F & \T \\
\F & \F & \F & \F & \T & \F & \T & \F & \F & \F & \T & \F & \F & \F \\
\hline
\end{array}$
Category:Factor Principles
Factor Principles/Conjunction on Right/Formulation 2

== Theorem ==
:$\vdash \paren {p \implies q} \implies \paren {\paren {p \land r} \implies \paren {q \land r} }$
== Proof ==
== Sources ==
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 3$: Theorem $\text{T31}$
Category:Factor Principles
Factor Principles/Disjunction on Left/Formulation 2

== Theorem ==
:$\vdash \paren {p \implies q} \implies \paren {\paren {r \lor p} \implies \paren {r \lor q} }$
== Proof ==
== Also see ==
* This is an axiom of the following proof system:
** Definition:Hilbert Proof System/Instance 2
== Sources ==
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Theorem $\text{T56}$
Category:Factor Principles
Factor Principles/Disjunction on Right/Formulation 1/Proof 4

== Theorem ==
== Proof ==
As can be seen for all boolean interpretations by inspection, where the truth value under the main connective on the is $T$, that under the one on the is also $T$:
:$\begin{array}{|ccc||ccc||ccccccc|} \hline
p & q & r & (p & \implies & q) & (p & \lor & r) & \implies & (q & \lor & r) \\
\hline
F & F & F & F & T & F & F & F & F & T & F & F & F \\
F & F & T & F & T & F & F & T & T & T & F & T & T \\ 
F & T & F & F & T & T & F & F & F & T & T & T & F \\
F & T & T & F & T & T & F & T & T & T & T & T & T \\
T & F & F & T & F & F & T & T & F & F & F & F & F \\
T & F & T & T & F & F & T & T & T & T & F & T & T \\
T & T & F & T & T & T & T & T & F & T & T & T & F \\
T & T & T & T & T & T & T & T & T & T & T & T & T \\
\hline
\end{array}$
Category:Factor Principles
Factor Principles/Disjunction on Right/Formulation 2

== Theorem ==
:$\vdash \paren {p \implies q} \implies \paren {\paren {p \lor r} \implies \paren {q \lor r} }$
== Proof ==
== Sources ==
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Theorem $\text{T57}$
Category:Factor Principles
Factor Spaces are T4 if Product Space is T4

== Theorem ==
Let $\mathbb S = \family {\struct{S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of non-empty topological spaces for $\alpha$ in some indexing set $I$.
Let $\ds T = \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\mathbb S$.
Let $T$ be a $T_4$ space.
Then each of $\struct {S_\alpha, \tau_\alpha}$ is a $T_4$ space.
== Proof ==
Since $S_\alpha \ne \O$ we also have $S \ne \O$ by the axiom of choice.
Let $z \in S$.
From Subspace of Product Space is Homeomorphic to Factor Space, every $\struct {S_\alpha, \tau_\alpha}$ is homeomorphic to the subspace $T_\alpha$ of $T$ defined by:
:$T_\alpha = \set {x \in S: \forall \beta \in I \setminus \set \alpha: x_\beta = z_\beta}$
and the homeomorphism $h : S_\alpha \to T_\alpha$ is given by:
:$\forall x_\alpha \in S_\alpha : \map h {x_\alpha} = x : \forall \beta \in I : x_\beta = \begin {cases} x_\alpha & \beta = \alpha \\ z_\beta & \beta \ne \alpha \end {cases}$ 
From Inclusion Mapping is Continuous, the inclusion mapping $i: T_\alpha \to S$ is continuous.
From Composite of Continuous Mappings is Continuous, the composite mapping $i \circ h : S_\alpha \to S$ is continuous.
Let $\pr_\alpha: S \to S_\alpha$ be the projection of $S$ to $S_\alpha$.
For all $x_\alpha \in S_\alpha$:
:$\map {\pr_\alpha} {\map {i \circ h} {x_\alpha}} = x_\alpha$ 
Let $A, B \subseteq S_\alpha$ be an arbitrary pair of disjoint closed sets in $\struct {S_\alpha, \tau_\alpha}$.
Then:
 
From Projection from Product Topology is Open and Continuous:General Result
:$\pr_\alpha: S \to S_\alpha$ is continuous.
From Continuity Defined from Closed Sets:
:$\map {\pr_\alpha^\gets} A, \map {\pr_\alpha^\gets} B$ are disjoint closed sets in $T$
 
By definition of $T_4$ space:
:$\exists U, V \in \tau : \map {\pr_\alpha^\gets} A \subseteq U, \map {\pr_\alpha^\gets} B \subseteq V : U \cap V = \O$
By definition of continuity:
:$\map {\paren {i \circ h}^\gets} U, \map {\paren{i \circ h}^\gets} V$ are open sets in $S_\alpha$.
From Preimage of Intersection under Mapping:
:$\map {\paren {i \circ h}^\gets} U, \map {\paren{i \circ h}^\gets} V$ are disjoint open sets in $S_\alpha$.
From Preimage of Subset is Subset of Preimage:
:$\map {\paren {i \circ h}^\gets} {\map {\pr_\alpha^\gets} A} \subseteq \map {\paren{i \circ h}^\gets} U$
:$\map {\paren {i \circ h}^\gets} {\map {\pr_\alpha^\gets} B} \subseteq \map {\paren{i \circ h}^\gets} V$
Now:
From the definition of set equality:
:$\map {\paren{i \circ h}^\gets} {\map {\pr_\alpha^\gets} A} = A$
Similarly:
:$\map {\paren{i \circ h}^\gets} {\map {\pr_\alpha^\gets} B} = B$
It follows that $A$ and $B$ are contained in the disjoint open sets $\map {\paren {i \circ h}^\gets} U, \map {\paren {i \circ h}^\gets} V$, respectively.
Since $A$ and $B$ were an arbitrary pair of disjoint closed sets in $\struct {S_\alpha, \tau_\alpha}$ then $\struct {S_\alpha, \tau_\alpha}$ is a $T_4$ space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:T4 Spaces
Category:Product Spaces
Factor Spaces are T5 if Product Space is T5

== Theorem ==
Let $\mathbb S = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of topological spaces for $\alpha$ in some indexing set $I$.
Let $\ds T = \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\mathbb S$.
Let $T$ be a $T_5$ space.
Then each of $\struct {S_\alpha, \tau_\alpha}$ is a $T_5$ space.
== Proof ==
Let $T$ be a $T_5$ space.
Let $\struct {S_\alpha, \tau_\alpha}$ be arbitrary.
By Subspace of Product Space is Homeomorphic to Factor Space:
:$\struct {S_\alpha, \tau_\alpha}$
is homeomorphic to a subspace of $T$.
By $T_5$ Property is Hereditary, this subspace is also $T_5$.
Finally, by $T_5$ Space is Preserved under Homeomorphism:
:$\struct {S_\alpha, \tau_\alpha}$
is a $T_5$ space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:T5 Spaces
Category:Product Spaces
Factor Spaces of Hausdorff Product Space are Hausdorff

== Theorem ==
Let $\SS = \family {\struct {S_\alpha, \tau_\alpha} }$ be an indexed family of topological spaces for $\alpha$ in some indexing set $I$.
Let $\ds T = \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\SS$.
Let $T$ be a $T_2$ (Hausdorff) space.
Then:
:for each $\alpha \in I$, $\struct {S_\alpha, \tau_\alpha}$ is a $T_2$ (Hausdorff) space.
== Proof ==
Let $\alpha \in I$.
Let $x, y \in S_\alpha$.
By the Axiom of Choice, there exists $z \in S$.
Define $x', y' \in S$ by:
:$x'_\beta = \begin{cases} z_\beta & \beta \ne \alpha \\ x & \beta = \alpha \end{cases}$
and
:$y'_\beta = \begin{cases} z_\beta & \beta \ne \alpha \\ y & \beta = \alpha \end{cases}$
By definition of a $T_2$ (Hausdorff) space:
:$\exists U, V \in \tau : x' \in U, y' \in V$ and $U \cap V = \O$
Recall that the topology on the product space $T$ is the product topology.
By definition of the product topology the natural basis is a basis for the product topology.
Then:
:$\exists U', V' \in \BB : x' \in U' \subseteq U, y' \in V' \subseteq V$
where $\BB$ is the natural basis for $\tau$.
From Subsets of Disjoint Sets are Disjoint:
:$U' \cap V' = \O$
From Natural Basis of Product Topology:
:$\ds U' = \prod_{\beta \mathop \in I} U_\beta$ where:
::for all $\beta \in I : U_\beta \in \tau_\beta$
::for all but finitely many indices $\beta : U_\beta = X_\beta$
Similarly:
:$\ds V' = \prod_{\beta \mathop \in I} V_\beta$ where:
::for all $\beta \in I : V_\beta \in \tau_\beta$
::for all but finitely many indices $\beta : V_\beta = X_\beta$
From General Case of Cartesian Product of Intersections:
:$\ds \paren {\prod_{\beta \mathop \in I} U_\beta} \bigcap \paren{\prod_{\beta \mathop \in I} V_\beta} = \prod_{\beta \mathop \in I} \paren{U_\beta \cap V_\beta}$
Thus:
:$\ds \prod_{\beta \mathop \in I} \paren{U_\beta \cap V_\beta} = \O$
Now for all $\beta \in I$ such that $\beta \ne \alpha$:
:$z_\beta = x'_\beta \in U_\beta$
and 
:$z_\beta = y'_\beta \in V_\beta$
So:
:$\forall \beta \in I, \beta \ne \alpha : U_\beta \cap V_\beta \ne \O$
From Cartesian Product of Family is Empty iff Factor is Empty it follows that:
:$U_\alpha \cap V_\alpha = \O$
Also we have:
:$x = x'_\alpha \in U_\alpha$
and
:$y = y'_\alpha \in V_\alpha$
Since $U_\alpha$ and $V_\alpha$ are open in $\struct {S_\alpha, \tau_\alpha}$ then $x$ and $y$ are separated by open sets.
Since $x$ and $y$ were arbitrary, then $\struct {S_\alpha, \tau_\alpha}$ is a $T_2$ (Hausdorff) space by definition.
Since $\alpha \in I$ was arbitrary, then for each $\alpha \in I$, $\struct {S_\alpha, \tau_\alpha}$ is a $T_2$ (Hausdorff) space.
Category:Hausdorff Spaces
Category:Product Topology
Factor Theorem

* Polynomial Factor Theorem
* Quotient-Remainder Theorem, also known as the Division Theorem
* A series of theorems in abstract algebra:
** Factor Theorem for Epimorphisms
** Factor Theorem for Group Epimorphisms
** Factor Theorem for Monomorphisms
** Factor Theorem for Sets
** Factor Theorem for Surjections
The terms factor and quotient are in general interchangeable.
Category:Abstract Algebra
Category:Named Theorems