avewright/proofwiki-math huggingface dataset:
Theorem: Product Space is T0 iff Factor Spaces are T0/Sufficient Condition

Let $\SS = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of non-empty topological spaces where $I$ is an arbitrary index set.
Let $\ds T = \struct {S, \tau} = \prod \struct {S_\alpha, \tau_\alpha}$ be the product space of $\SS$.
Let $T$ be a $T_0$ space.
Then each of $\struct {S_\alpha, \tau_\alpha}$ is a $T_0$ space.

Proof:
Let $T$ be a $T_0$ space.
{{AimForCont}} $\exists \beta: \struct {S_\beta, \tau_\beta}$ is not a $T_0$ space.
Then $\exists a, b \in S_\beta$ such that $\forall U_\beta \in \tau_\beta$, either $a, b \in U_\beta$ or $a, b \notin U_\beta$.
Consider the elements  $y, z \in S$ defined as:
:<nowiki>$y = \family {x_\alpha} : x_\alpha = \begin {cases}
s_\alpha & : \alpha \ne \beta \\
a & : \alpha = \beta \end {cases}$</nowiki>
:<nowiki>$z = \family {x_\alpha} : x_\alpha = \begin {cases}
s_\alpha & : \alpha \ne \beta \\
b & : \alpha = \beta \end {cases}$</nowiki>
That is, $y$ and $z$ match (arbitrarily) on all coordinates except that for $\beta$.
Let $H \subseteq S: y \in H$.
Then $z \in H$ as $\forall U_\beta \in \map {\pr_\beta} H : b \in U_\beta$
Similarly, let $K \subseteq S: z \in K$.
Then $y \in K$ as $\forall U_\beta \in \map {\pr_\beta} H : a \in U_\beta$
So $T$ is not a $T_0$ space.
Hence by Proof by Contradiction it follows that each of $\struct {S_\alpha, \tau_\alpha}$ is a $T_0$ space.
Theorem: Product Space is T1 iff Factor Spaces are T1/Sufficient Condition

Let $\SS = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of non-empty topological spaces where $I$ is an arbitrary index set.
Let $\ds T = \struct {S, \tau} = \prod \struct {S_\alpha, \tau_\alpha}$ be the product space of $\SS$.
Let $T$ be a $T_1$ space.
Then each of $\struct {S_\alpha, \tau_\alpha}$ is a $T_1$ space.

Proof:
Let $T$ be a $T_1$ space.
As $S_\alpha \ne \O$ we also have $S \ne \O$ by the axiom of choice.
From Subspace of Product Space is Homeomorphic to Factor Space, $\struct {S_\alpha, \tau_\alpha}$ is homeomorphic to a subspace $T_\alpha$ of $T$.
From $T_1$ property is hereditary, $T_\alpha$ is a $T_1$ space.
From $T_1$ Property is Preserved under Homeomorphism, $\struct {S_\alpha, \tau_\alpha}$ is a $T_1$ space.
As $\alpha \in I$ was arbitrary, it follows that each of $\struct {S_\alpha, \tau_\alpha}$ is a $T_1$ space.
Theorem: Product Space is T1 iff Factor Spaces are T1/Necessary Condition

Let $\SS = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of non-empty topological spaces where $I$ is an arbitrary index set.
Let $\ds T = \struct {S, \tau} = \prod \struct {S_\alpha, \tau_\alpha}$ be the product space of $\SS$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be a $T_1$ space.
Then $T$ is a $T_1$ space.

Proof:
Let each of $\struct {S_\alpha, \tau_\alpha}$ be a $T_1$ space.
{{Recall|T1 Space|$T_1$ space|1}}
{{:Definition:T1 Space/Definition 1}}
{{AimForCont}} $T$ is not a $T_1$ space.
Then $\exists a, b \in S, a \ne b$ such that for all $U \in \tau$, $a \in U \implies b \in U$.
Then $a$ and $b$ are different in at least one coordinate.
Suppose, $a_\alpha = p, b_\alpha = q$ for some coordinate $\alpha$.
Then for each $U_\alpha \in \tau_\alpha$:
:$\ds a \in U_\alpha \times \prod_{\beta \mathop \ne \alpha} S_\beta \implies b \in U_\alpha \times \prod_{\beta \mathop \ne \alpha} S_\beta$
But:
:$\ds a \in U_\alpha \times \prod_{\beta \mathop \ne \alpha} S_\beta$ {{iff}} $p \in U_\alpha$
:$\ds b \in U_\alpha \times \prod_{\beta \mathop \ne \alpha} S_\beta$ {{iff}} $q \in U_\alpha$
from which we infer, for each $U_\alpha \in \tau_\alpha$:
:$p \in U_\alpha \implies q \in U_\alpha$
It follows that $\struct {S_\alpha, \tau_\alpha}$ is not a $T_1$ space.
Hence by Proof by Contradiction it follows that $T$ is a $T_0$ space.
Theorem: Product Space is T2 iff Factor Spaces are T2/Sufficient Condition

Let $\SS = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of non-empty topological spaces where $I$ is an arbitrary index set.
Let $\ds T = \struct {S, \tau} = \prod \struct {S_\alpha, \tau_\alpha}$ be the product space of $\SS$.
Let $T$ be a $T_2$ space.
Then each of $\struct {S_\alpha, \tau_\alpha}$ is a $T_2$ space.

Proof:
Let $T$ be a $T_2$ space.
As $S_\alpha \ne \O$ we also have $S \ne \O$ by the axiom of choice.
From Subspace of Product Space is Homeomorphic to Factor Space, $\struct {S_\alpha, \tau_\alpha}$ is homeomorphic to a subspace $T_\alpha$ of $T$.
From $T_2$ property is hereditary, $T_\alpha$ is a $T_2$ space.
From $T_2$ Property is Preserved under Homeomorphism, $\struct {S_\alpha, \tau_\alpha}$ is a $T_2$ space.
As $\alpha \in I$ was arbitrary, it follows that each of $\struct {S_\alpha, \tau_\alpha}$ is a $T_2$ space.
Theorem: Product Space is T2 iff Factor Spaces are T2/Sufficient Condition

Let $\SS = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of non-empty topological spaces where $I$ is an arbitrary index set.
Let $\ds T = \struct {S, \tau} = \prod \struct {S_\alpha, \tau_\alpha}$ be the product space of $\SS$.
Let $T$ be a $T_2$ space.
Then each of $\struct {S_\alpha, \tau_\alpha}$ is a $T_2$ 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.
{{qed}}
Theorem: Product Space is Regular iff Factor Spaces are Regular

:$T$ is a regular space {{iff}} each of $\struct {S_\alpha, \tau_\alpha}$ is a regular space.

Proof:
{{Recall|Regular Space|index = 1}}
{{:Definition:Regular Space/Definition 1}}
The result follows from:
:Product Space is $T_0$ iff Factor Spaces are $T_0$
:Product Space is $T_3$ iff Factor Spaces are $T_3$
{{qed}}
Theorem: Product Space is Completely Regular iff Factor Spaces are Completely Regular

:$T$ is a completely regular space {{iff}} each of $\struct {S_\alpha, \tau_\alpha}$ is a completely regular space.

Proof:
{{Recall|Completely Regular Space|completely regular space|index = 1}}
{{:Definition:Completely Regular Space/Definition 1}}
The result follows from:
:Product Space is $T_0$ iff Factor Spaces are $T_0$
:Product Space is $T_{3 \frac 1 2}$ iff Factor Spaces are $T_{3 \frac 1 2}$
{{qed}}
Theorem: Factor Spaces are Normal if Product Space is Normal

Let $\SS = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of non-empty topological spaces where $I$ is an arbitrary index set.
Let $\ds T = \struct {S, \tau} = \prod \struct {S_\alpha, \tau_\alpha}$ be the product space of $\SS$.
Let $T$ be a normal space.
Then each of $\struct {S_\alpha, \tau_\alpha}$ is a normal space.

Proof:
{{Recall|Normal Space|normal space|index = 1}}
{{:Definition:Normal Space/Definition 1}}
The result follows from:
:Product Space is $T_1$ iff Factor Spaces are $T_1$
:Factor Spaces are $T_4$ if Product Space is $T_4$
{{qed}}
Theorem: Factor Spaces are Completely Normal if Product Space is Completely Normal

Let $\SS = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of non-empty topological spaces where $I$ is an arbitrary index set.
Let $\ds T = \struct {S, \tau} = \prod \struct {S_\alpha, \tau_\alpha}$ be the product space of $\SS$.
Let $T$ be a completely normal space.
Then each of $\struct {S_\alpha, \tau_\alpha}$ is a completely normal space.

Proof:
{{Recall|Completely Normal Space|completely normal space|index = 1}}
{{:Definition:Completely Normal Space/Definition 1}}
The result follows from:
{{begin-itemize}}
{{item||Product Space is $T_1$ iff Factor Spaces are $T_1$}}
{{item||Factor Spaces are $T_5$ if Product Space is $T_5$.}}
{{end-itemize}}
{{qed}}
Theorem: Product of Normal Factor Spaces is not necessarily Normal

Let $\SS = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of non-empty topological spaces where $I$ is an arbitrary index set.
Let $\ds T = \struct {S, \tau} = \prod \struct {S_\alpha, \tau_\alpha}$ be the product space of $\SS$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be a normal space.
Then it is not necessarily the case that $T$ is also a normal space.

Proof:
From Existence of Completely Normal Space whose Product Space is Not Normal, there exist completely normal topological spaces $T_A$ and $T_B$ such that their product space $T_A \times T_B$ is not a normal space.
From Completely Normal Space is Normal, $T_A$ and $T_B$ are both normal spaces.
Hence there exist normal spaces $T_A$ and $T_B$ whose product space is not a normal space.
{{qed}}
Theorem: Product of Completely Normal Factor Spaces is not necessarily Completely Normal

Let $\SS = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of non-empty topological spaces where $I$ is an arbitrary index set.
Let $\ds T = \struct {S, \tau} = \prod \struct {S_\alpha, \tau_\alpha}$ be the product space of $\SS$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be a completely normal space.
Then it is not necessarily the case that $T$ is also a completely normal space.

Proof:
From Existence of Completely Normal Space whose Product Space is Not Normal, there exist completely normal topological spaces $T_A$ and $T_B$ such that their product space $T = T_A \times T_B$ is not a normal space.
As $T$ is not a normal space, it follows from Completely Normal Space is Normal that $T$ is not a completely normal space either.
Hence there exist completely normal spaces $T_A$ and $T_B$ whose product space is not a completely normal space.
{{qed}}
Theorem: Product of T4 Factor Spaces is not necessarily T4

Let $\SS = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of non-empty topological spaces where $I$ is an arbitrary index set.
Let $\ds T = \struct {S, \tau} = \prod \struct {S_\alpha, \tau_\alpha}$ be the product space of $\SS$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be a $T_4$ space.
Then it is not necessarily the case that $T$ is also a $T_4$ space.

Proof:
From Product of Normal Factor Spaces is not necessarily Normal, there exist normal topological spaces $T_A$ and $T_B$ such that their product space $T = T_A \times T_B$ is not a normal space.
{{Recall|Normal Space|normal space|index = 1}}
{{:Definition:Normal Space/Definition 1}}
As $T_A$ and $T_B$ are both normal spaces, it follows that they are both $T_1$ spaces.
Hence, from Product Space is $T_1$ iff Factor Spaces are $T_1$, their product space is also a $T_1$ space.
But as $T$ is not a normal space, it must therefore be the case that $T$ is not a $T_4$ space.
Hence there exist $T_4$ spaces $T_A$ and $T_B$ whose product space is not a $T_4$ space.
{{qed}}
Category:T4 Spaces
Category:Separation Properties Preserved under Topological Product
Category:Product Spaces
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Theorem: Product of T5 Factor Spaces is not necessarily T5

Let $\SS = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of non-empty topological spaces where $I$ is an arbitrary index set.
Let $\ds T = \struct {S, \tau} = \prod \struct {S_\alpha, \tau_\alpha}$ be the product space of $\SS$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be a $T_5$ space.
Then it is not necessarily the case that $T$ is also a $T_5$ space.

Proof:
From Product of Completely Normal Factor Spaces is not necessarily Completely Normal, there exist normal topological spaces $T_A$ and $T_B$ such that their product space $T = T_A \times T_B$ is not a completely normal space.
Let $T = T_A \times T_B$ be a product space such that:
{{begin-itemize}}
{{item||$T_A$ and $T_B$ are normal spaces}}
{{item||$T {{=}} T_A \times T_B$ is not a completely normal space.}}
{{end-itemize}}
{{Recall|Completely Normal Space|completely normal space|index = 1}}
{{:Definition:Completely Normal Space/Definition 1}}
As $T_A$ and $T_B$ are both completely normal spaces, it follows that they are both $T_1$ spaces.
Hence, from Product Space is $T_1$ iff Factor Spaces are $T_1$, their product space is also a $T_1$ space.
But as $T$ is not a completely normal space, it must therefore be the case that $T$ is not a $T_5$ space.
Hence there exist $T_5$ spaces $T_A$ and $T_B$ whose product space is not a $T_5$ space.
{{qed}}
Category:T5 Spaces
Category:Separation Properties Preserved under Topological Product
Category:Product Spaces
2dek7qrrqfwal35dl4zuala0b67nxxl
Theorem: T0 Property is Preserved under Expansion

If $T_A$ is a $T_0$ space, then so is $T_B$.

Proof:
Let $T_A = \struct {S, \tau_1}$ be a $T_0$ space.
Let $T_B = \struct {S, \tau_2}$ be such that $\tau_2$ is an expansion of $\tau_1$.
From Identity Mapping to Expansion is Closed, we have that $I_S$ is closed.
We also have Identity Mapping is Bijection.
Hence we can directly apply:
:$T_0$ Property is Preserved under Closed Bijection
and it follows that as $T_A$ is a $T_0$ space, then $T_B$ is also a $T_0$ space.
{{qed}}
Theorem: T1 Property is Preserved under Expansion

If $T_A$ is a $T_1$ space, then so is $T_B$.

Proof:
Let $T_A = \struct {S, \tau_1}$ be a $T_1$ space.
Let $T_B = \struct {S, \tau_2}$ be such that $\tau_2$ is an expansion of $\tau_1$.
From Identity Mapping to Expansion is Closed, we have that $I_S$ is closed.
We also have Identity Mapping is Bijection.
Hence we can directly apply:
:$T_1$ Property is Preserved under Closed Bijection
and it follows that as $T_A$ is a $T_1$ space, then $T_B$ is also a $T_1$ space.
{{qed}}
Theorem: T2 Property is Preserved under Expansion

If $T_A$ is a $T_2$ (Hausdorff) space, then so is $T_B$.

Proof:
Let $T_A = \struct {S, \tau_1}$ be a $T_2$ space.
Let $T_B = \struct {S, \tau_2}$ be such that $\tau_2$ is an expansion of $\tau_1$.
From Identity Mapping to Expansion is Closed, we have that $I_S$ is closed.
We also have Identity Mapping is Bijection.
Hence we can directly apply:
:$T_2$ Property is Preserved under Closed Bijection
and it follows that as $T_A$ is a $T_2$ space, then $T_B$ is also a $T_2$ space.
{{qed}}
Theorem: T2.5 Property is Preserved under Expansion

If $T_A$ is a $T_{2 \frac 1 2}$ space, then so is $T_B$.

Proof:
Let $T_A = \struct {S, \tau_1}$ be a $T_{2 \frac 1 2}$ space.
Let $T_B = \struct {S, \tau_2}$ be such that $\tau_2$ is an expansion of $\tau_1$.
From Identity Mapping to Expansion is Closed, we have that $I_S$ is closed.
We also have Identity Mapping is Bijection.
Hence we can directly apply:
:$T_{2 \frac 1 2}$ Property is Preserved under Closed Bijection
and it follows that as $T_A$ is a $T_{2 \frac 1 2}$ space, then $T_B$ is also a $T_{2 \frac 1 2}$ space.
{{qed}}
Theorem: T3 Property is Not Preserved under Expansion

If $T_A$ is a $T_3$ space, then it is not necessarily the case that so is $T_B$.

Proof:
;Proof by Counterexample
Let $\struct {\R, \tau_1}$ be the set of real numbers under the usual (Euclidean) topology.
Let $\struct {\R, \tau_2}$ be the indiscrete rational extension of $\struct {\R, \tau_1}$.
By definition, $\struct {\R, \tau_2}$ is an expansion of $\struct {\R, \tau_1}$.
From Metric Space is $T_3$:
:$\struct {\R, \tau_1}$ is a $T_3$ space.
But we have:
:Indiscrete Extension of Reals is not $T_3$
Hence the result.
{{qed}}
Category:T3 Spaces
Category:Separation Properties Not Preserved under Expansion
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Theorem: Metric Space is T1

A metric space $M = \struct {A, d}$ is a $T_1$ space.

Proof:
From Metric Space is $T_2$, $M$ is a $T_1$ space.
From $T_2$ Space is $T_1$ it follows that $M$ is a $T_1$ space.
{{qed}}
Category:Metric Space fulfils all Separation Axioms
Category:Examples of T1 Spaces
8f1b54yf4ivjqtdvmirz6w81bhf83me
Theorem: Metric Space is T0

A metric space $M = \struct {A, d}$ is a $T_0$ space.

Proof:
From Metric Space is $T_1$, $M$ is a $T_1$ space.
From $T_1$ Space is $T_0$ it follows that $M$ is a $T_0$ space.
{{qed}}
Category:Metric Space fulfils all Separation Axioms
Category:Examples of T0 Spaces
nckw1120qksd27ur9etfquhok49r3te
Theorem: Metric Space is Normal

A metric space $M = \struct {A, d}$ is a normal space.

Proof:
{{Recall|Normal Space|normal space|index = 1}}
{{:Definition:Normal Space/Definition 1}}
We have that:
:a Metric Space is $T_4$
:a Metric Space is $T_1$.
Hence the result.
{{qed}}
Category:Metric Space fulfils all Separation Axioms
Category:Examples of Normal Spaces
9twedg5d85es49rml0o0qkuj9ezzz39
Theorem: Metric Space is Completely Regular

A metric space $M = \struct {A, d}$ is a completely regular space.

Proof:
We have:
* Metric Space is Normal
* Normal Space is Completely Regular
Hence the result.
{{qed}}
Category:Metric Space fulfils all Separation Axioms
Category:Examples of Completely Regular Spaces
amdzzwkx14h4m77l8xn7qfrvg6u9rku
Theorem: Metric Space is T3.5

A metric space $M = \struct {A, d}$ is a $T_{3 \frac 1 2}$ space.

Proof:
{{Recall|Completely Regular Space}}
{{:Definition:Completely Regular Space}}
We have:
:Metric Space is Completely Regular
Hence the result.
{{qed}}
Category:Metric Space fulfils all Separation Axioms
Category:Examples of T3.5 Spaces
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Theorem: Metric Space is Urysohn

A metric space $M = \struct {A, d}$ is an Urysohn space.

Proof:
We have:
:Metric Space is Completely Regular
:Completely Regular Space is Urysohn.
Hence the result.
{{qed}}
Category:Metric Space fulfils all Separation Axioms
Category:Examples of Urysohn Spaces
ikofrwbmmmto628fmz8r65wrwfr0h1d
Theorem: Metric Space is Regular

A metric space $M = \struct {A, d}$ is a regular space.

Proof:
We have:
:Metric Space is Completely Regular
:Completely Regular Space is Regular.
Hence the result.
{{qed}}
Theorem: Metric Space is Regular

A metric space $M = \struct {A, d}$ is a regular space.

Proof:
We have:
:Metric Space is Fully Normal
:Fully Normal Space is Normal
:Normal Space is Regular.
Hence the result.
{{qed}}
Theorem: Completely Regular Space is T1

Let $\struct {S, \tau}$ be a completely regular space.
Then $\struct {S, \tau}$ is also a $T_1$ space.

Proof:
We have:
:Completely Regular Space is $T_2$
:$T_2$ Space is $T_1$
Hence the result.
{{qed}}
Theorem: Completely Regular Space is T2

Let $\struct {S, \tau}$ be a completely regular space.
Then $\struct {S, \tau}$ is also a $T_2$ (Hausdorff) space.

Proof:
{{Recall|Regular Space|regular space}}
{{:Definition:Regular Space/Definition 3}}
We have:
:Completely Regular Space is Regular
Hence the result.
{{qed}}
Theorem: Metric Space is T3

A metric space $M = \struct {A, d}$ is a $T_3$ space.

Proof:
{{Recall|Regular Space|index = 1}}
{{:Definition:Regular Space/Definition 1}}
We have:
:Metric Space is Regular Space
Hence the result.
{{qed}}
Category:Metric Space fulfils all Separation Axioms
Category:Examples of T3 Spaces
o73eog6uxvibvpuwms8bvxr43fo15h0
Theorem: Metric Space is T2.5

A metric space $M = \struct {A, d}$ is a $T_{2 \frac 1 2}$ space.

Proof:
We have:
:Metric Space is Regular
:Regular Space is $T_{2 \frac 1 2}$ Space
Hence the result.
{{qed}}
Category:Metric Space fulfils all Separation Axioms
Category:Examples of T2.5 Spaces
si4kug551qywib65obvw4um48t4vter
Theorem: Metric Space is Semiregular

A metric space $M = \struct {A, d}$ is a semiregular space.

Proof:
We have:
:Metric Space is Regular
:Regular Space is Semiregular.
Hence the result.
{{qed}}
Category:Metric Space fulfils all Separation Axioms
Category:Examples of Semiregular Spaces
d4dkcb119jsvxld73mt69idx08lc16g
Theorem: T4 Property is Not Preserved under Expansion

If $T_A$ is a $T_4$ space, then it is not necessarily the case that so is $T_B$.

Proof:
;Proof by Counterexample
Let $\struct {\R, \tau_1}$ be the set of real numbers under the usual (Euclidean) topology.
Let $\struct {\R, \tau_2}$ be the indiscrete rational extension of $\struct {\R, \tau_1}$.
By definition, $\struct {\R, \tau_2}$ is an expansion of $\struct {\R, \tau_1}$.
From Metric Space is $T_4$:
:$\struct {\R, \tau_1}$ is a $T_4$ space.
But we have:
:Indiscrete Extension of Reals is not $T_4$
Hence the result.
{{qed}}
Category:T4 Spaces
Category:Separation Properties Not Preserved under Expansion
k2ty1sqhwnfu3li8mp28qzfsyq7k3ky
Theorem: T5 Property is Not Preserved under Expansion

If $T_A$ is a $T_5$ space, then it is not necessarily the case that so is $T_B$.

Proof:
;Proof by Counterexample
Let $\struct {\R, \tau_1}$ be the set of real numbers under the usual (Euclidean) topology.
Let $\struct {\R, \tau_2}$ be the indiscrete rational extension of $\struct {\R, \tau_1}$.
By definition, $\struct {\R, \tau_2}$ is an expansion of $\struct {\R, \tau_1}$.
From Metric Space is $T_5$:
:$\struct {\R, \tau_1}$ is a $T_5$ space.
But we have:
:Indiscrete Extension of Reals is not $T_5$
Hence the result.
{{qed}}
Category:T5 Spaces
Category:Separation Properties Not Preserved under Expansion
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Theorem: Regular Property is Not Preserved under Expansion

If $T_A$ is a regular space, then it is not necessarily the case that so is $T_B$.

Proof:
Let $T_A$ be a regular space.
{{Recall|Regular Space|index = 1}}
{{:Definition:Regular Space/Definition 1}}
Thus $T_A$ is a $T_3$ space.
From $T_0$ Property is Preserved under Expansion we have that $T_B$ is a $T_0$ space.
But from $T_3$ Property is Not Preserved under Expansion, it is not necessarily the case that $T_B$ is also a $T_3$ space.
Hence it is not necessarily the case that $T_B$ is also a regular space.
Hence the result.
{{qed}}
Category:Regular Spaces
Category:Separation Properties Not Preserved under Expansion
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Theorem: T3.5 Property is Not Preserved under Expansion

If $T_A$ is a $T_{3 \frac 1 2}$ space, then it is not necessarily the case that so is $T_B$.

Proof:
Let $T_A$ be a $T_{3 \frac 1 2}$ space.
From $T_{3 \frac 1 2}$ Space is $T_3$, it follows that $T_A$ is a $T_3$ space.
But from $T_3$ Property is Not Preserved under Expansion, it is not necessarily the case that $T_B$ is also a $T_3$ space.
But again, we have:
:$T_{3 \frac 1 2}$ Space is $T_3$
Hence, by the Rule of Transposition, it is not necessarily the case that $T_B$ is a $T_{3 \frac 1 2}$ space.
Hence the result.
{{qed}}
Category:T3.5 Spaces
Category:Separation Properties Not Preserved under Expansion
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