avewright/proofwiki-math huggingface dataset:
Theorem: Exterior of Intersection contains Union of Exteriors

Let $T = \struct {S, \tau}$ be a topological space.
Let $\mathbb H$ be a set of subsets of $TS$.
That is, let $\mathbb H \subseteq \powerset S$ where $\powerset S$ is the power set of $S$.
Then:
:$\ds \bigcup_{H \mathop \in \mathbb H} H^e \subseteq \paren {\bigcap_{H \mathop \in \mathbb H} H}^e$
where $H^e$ denotes the exterior of $H$.

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \bigcup_{H \mathop \in \mathbb H} H^e
      | r = \bigcup_{H \mathop \in \mathbb H} \paren {S \setminus H^-})
      | c = {{Defof|Exterior (Topology)|Exterior}}
}}
{{eqn | r = S \setminus \bigcap_{H \mathop \in \mathbb H} H^-
      | c = De Morgan's Laws: Difference with Intersection
}}
{{end-eqn}}
From Closure of Intersection is Subset of Intersection of Closures:
:$\ds \paren {\bigcap_{H \mathop \in \mathbb H} H^-} \subseteq \bigcap_{H \mathop \in \mathbb H} H^-$
From Set Complement inverts Subsets:
:$\ds S \setminus \paren {\bigcap_{H \mathop \in \mathbb H} H^-} \supseteq S \setminus \bigcap_{H \mathop \in \mathbb H} H^-$
From the definition of exterior:
:$\ds S \setminus \paren {\bigcup_{H \mathop \in \mathbb H} H}^- = \paren {\bigcup_{H \mathop \in \mathbb H} H}^e$
Putting this together:
:$\ds \bigcup_{H \mathop \in \mathbb H} H^e \subseteq \paren {\bigcap_{H \mathop \in \mathbb H} H}^e$
{{qed}}
Theorem: Separated Sets are Disjoint

Let $T = \struct {S, \tau}$ be a topological space.
Let $A, B \subseteq S$ such that $A$ and $B$ are separated in $T$.
Then $A$ and $B$ are disjoint:
:$A \cap B  = \O$

Proof:
Let $A$ and $B$ be separated in $T$.
Then:
{{begin-eqn}}
{{eqn | l = A^- \cap B
      | r = \O
      | c = {{Defof|Separated Sets}}: $A^-$ is the closure of $A$
}}
{{eqn | ll= \leadsto
      | l = \paren {A \cup A'} \cap B
      | r = \O
      | c = {{Defof|Closure (Topology)|Set Closure}}: $A'$ is the derived set of $A$
}}
{{eqn | ll= \leadsto
      | l = \paren {A \cap B} \cup \paren {A' \cap B}
      | r = \O
      | c = Intersection Distributes over Union
}}
{{eqn | ll= \leadsto
      | l = A \cap B
      | r = \O
      | c = Union is Empty iff Sets are Empty
}}
{{end-eqn}}
{{qed}}
Theorem: Second-Countable Space is First-Countable

Let $T = \struct {S, \tau}$ be a topological space which is a second-countable space.
Then $T$ is also a first-countable space.

Proof:
{{Recall|First-Countable Space|first-countable space}}
{{:Definition:First-Countable Space}}
{{Recall|Second-Countable Space|second-countable space}}
{{:Definition:Second-Countable Space}}
Consider the underlying set $S$ as an open set.
From Set is Open iff Neighborhood of all its Points, $S$ has that property.
As $T$ has a countable basis, then (trivially) every point in $T$ has a countable local basis.
So a second-countable space is trivially first-countable.
{{qed}}
Theorem: Second-Countability is Hereditary

Let $T = \struct {S, \tau}$ be a topological space which is second-countable.
Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a subspace of $T$.
Then $T_H$ is second-countable.
That is, the property of second-countability is hereditary.

Proof:
From the definition of second-countable, $\struct {S, \tau}$ has a countable basis.
That is, $\exists \BB \subseteq \tau$ such that:
:for all $U \in \tau$, $U$ is a union of sets from $\BB$
:$\BB$ is countable.
As $H \subseteq S$ it follows that a $H$ itself is a union of sets from $\BB$.
The result follows from Basis for Topological Subspace.
{{qed}}
Theorem: First-Countability is Hereditary

Let $T = \struct {S, \tau}$ be a topological space which is first-countable.
Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a subspace of $T$.
Then $T_H$ is first-countable.
That is, the property of first-countability is hereditary.

Proof:
From the definition of first-countable, every point in $S$ has a countable local basis in $T$.
The intersection of $H$ with the countable local basis of $S$ provides a countable local basis for $H$.
As every point in $H$ is also a point in $S$, the result follows from Basis for Topological Subspace.
{{qed}}
Theorem: Basis for Topological Subspace

Let $T = \struct {A, \tau}$ be a topological space.
Let $\O \subseteq H \subseteq A$ and so let $T_H = \struct {H, \tau_H}$ be a subspace of $T$.
Let $\BB$ be a (synthetic) basis for $T$.
Let $\BB_H$ be defined as:
:$\BB_H = \set {U \cap H: U \in \BB}$
Then $\BB_H$ is a (synthetic) basis for $H$.

Proof:
$\BB \subseteq \powerset A$ is a synthetic basis for $T$ {{iff}}:
:$(\text B 1): \quad$ $A$ is a union of sets from $\BB$
:$(\text B 2): \quad$ If $B_1, B_2 \in B$, then $B_1 \cap B_2$ is a union of sets from $\BB$.
Let $A = \mathbb S$ be a union of sets from $\BB$.
Then:
:$\ds A = \bigcup_{S \mathop \in \mathbb S} S$
for some $\mathbb S \subseteq \BB$.
Hence:
{{begin-eqn}}
{{eqn | l = H
      | r = A \cap H
      | c = Intersection with Subset is Subset
}}
{{eqn | r = \paren {\bigcup_{S \mathop \in \mathbb S} S} \cap H
      | c = 
}}
{{eqn | r = \bigcup_{S \mathop \in \mathbb S} \paren {S \cap H}
      | c = Intersection Distributes over Union
}}
{{end-eqn}}
But if $S \in \BB$, then $S \cap H \in \BB_H$
So $H$ is a union of sets from $\BB_H$.
In the same way we investigate $U_1$ and $U_2$.
Let $U_1, U_2 \in \BB_H$.
Then $U_1 = B_1 \cap H, U_2 = B_2 \cap H$ for some $B_1, B_2 \in \BB$.
Then $U_1 \cap U_2 = B_1 \cap H \cap B_2 \cap H = \paren {B_1 \cap B_2} \cap H$.
As $\BB$ is a (synthetic) basis for $A$, we have that:
:$\ds B_1 \cap B_2 = \bigcup_{S \mathop \in \mathbb S} S$
for some $\mathbb S \subseteq \BB$.
Hence:
{{begin-eqn}}
{{eqn | l = U_1 \cap U_2
      | r = \paren {B_1 \cap B_2} \cap H
      | c = 
}}
{{eqn | r = \paren {\bigcup_{S \mathop \in \mathbb S} S} \cap H
      | c = 
}}
{{eqn | r = \bigcup_{S \mathop \in \mathbb S} \paren {S \cap H}
      | c = Intersection Distributes over Union
}}
{{end-eqn}}
But if $S \in \BB$, then $S \cap H \in \BB_H$.
So $U_1 \cap U_2$ is a union of sets from $\BB_H$.
So $\BB_H$ fulfils the conditions to be a synthetic basis for $H$.
{{qed}}
Theorem: Characterisation of Local Rings

Let $R$ be a ring.
Let $J \lhd R$ be a maximal ideal.
:$(1): \quad$ If the set $R \setminus J$ is precisely the group of units $R^\times$ of $R$, then $\tuple {R, J}$ is a local ring.
:$(2): \quad$ If $1 + x$ is a unit in $R$ for all $x \in J$ then $\tuple {R, J}$ is local.
{{explain|The specific interpretation of the notation $\tuple {R, J}$ is unclear. The usual notation for a ring is something like $\struct {R, +, \times}$.}}

Proof:
$(1): \quad$ Suppose that $J$ is the set of non-units of $R$.
Then by Ideal of Unit is Whole Ring, every ideal not equal to $R$ is contained in $J$.
Therefore $J$ is the unique maximal ideal of $R$, so $\tuple {R, J}$ is local.
$(2): \quad$ Let $x \in R \setminus J$.
Then:
:$J \subsetneq \map I {J \cup \set x} \subseteq R$
where $\map I S$ is the ideal generated by $S$.
Since $J$ is maximal, by definition, $x$ and $J$ generate all of $R$.
Therefore $t x + m = 1$ for some $m \in J$, $t \in R$.
Thus $t x = 1 - m \in 1 + J$ is a unit by hypothesis.
Therefore $x$ is a unit.
Now use part $(1)$.
{{qed}}
Category:Local Rings
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Theorem: Continuity Defined by Closure

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: T_1 \to T_2$ be a mapping.
Then $f$ is continuous {{iff}}:
:$\forall H \subseteq S_1: f \sqbrk {H^-} \subseteq \paren {f \sqbrk H}^-$
where $H^-$ denotes the closure of $H$ in $T_1$.
That is, {{iff}} the image of the closure is a subset of the closure of the image.

Proof:
=== Necessary Condition ===
{{:Continuity Defined by Closure/Necessary Condition/Proof 1}}{{qed|lemma}}
Theorem: Continuity Defined by Closure

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: T_1 \to T_2$ be a mapping.
Then $f$ is continuous {{iff}}:
:$\forall H \subseteq S_1: f \sqbrk {H^-} \subseteq \paren {f \sqbrk H}^-$
where $H^-$ denotes the closure of $H$ in $T_1$.
That is, {{iff}} the image of the closure is a subset of the closure of the image.

Proof:
Let $f$ be continuous.
Let $y \in f \sqbrk {\map \cl H}$.
Then:
:$\exists x \in \map \cl H: y = \map f x$
Let $U$ be an open set of $T_2$ such that $y \in U$.
Then by definition of continuous mapping:
:$f^{-1} \sqbrk U$ is an open set of $T_1$ such that:
::$x \in f^{-1} \sqbrk U$
Hence:
:$f^{-1} \sqbrk U \cap H \ne \O$
as $x \in \map \cl H$.
Hence:
{{begin-eqn}}
{{eqn | l = U \cap f \sqbrk H
      | o = \supseteq
      | r = f \sqbrk {f^{-1} \sqbrk U \cap H}
      | c = 
}}
{{eqn | o = \ne
      | r = \O
      | c = 
}}
{{end-eqn}}
and:
:$y \in \map \cl {f \sqbrk H}$
That is:
:$f \sqbrk {\map \cl H} \subseteq \map \cl {f \sqbrk H}$
Theorem: Valuation Ring is Local

Let $R$ be a valuation ring.
Then $R$ is a local ring.

Proof:
By definition of local ring:
:$R$ is a '''local ring''' {{iff}}:
::$R$ is non-trivial
::the sum of two arbitrary non-units is a non-unit.
First we recall that as a valuation ring is an integral domain, then {{afortiori}}:
:$R$ is non-trivial
:$R$ has no (proper) zero divisors
:$R$ is a commutative and unitary ring.
Let $0$ and $1$ be the zero and unity of $R$ respectively.
Let $x, y \in R$ be non-units.
We shall show that $x + y$ is non-unit.
{{WLOG}}, let $x = 0$.
Then:
:$x + y = y$
and so $x + y$ is a non-unit {{hypothesis}}.
Similarly for $y = 0$.
Let $x \ne 0$ and $y \ne 0$.
By definition of non-unit, neither $x$ nor $y$ has an inverse in $R$.
Let $K$ be the field of quotients of $R$.
In particular, $x^{-1} \in K$ and $y^{-1} \in K$.
First note that:
{{begin-eqn}}
{{eqn | l = \paren {x y^{-1} }^{-1}
      | r = y x^{-1}
      | c = Inverse of Group Product
}}
{{eqn | r = x^{-1} y
      | c = 
}}
{{end-eqn}}
By definition of valuation ring, either:
:$x y^{-1} \in R$
or:
{{begin-eqn}}
{{eqn | l = \paren {x y^{-1} }^{-1}
      | r = y x^{-1}
      | c = Inverse of Group Product
}}
{{eqn | r = x^{-1} y
      | c = {{Ring-axiom|C}}
}}
{{eqn | o = \in
      | r = R
      | c = 
}}
{{end-eqn}}
{{WLOG}}, let $x y^{-1} \in R$.
{{AimForCont}} there exists a $u \in R$ such that:
:$u \paren {x + y} = 1$
Then:
{{begin-eqn}}
{{eqn | l = y^{-1}
      | r = u \paren {x + y} y^{-1} 
      | c = {{Defof|Unity of Ring}}
}}
{{eqn | r = u \paren {x y^{-1} + y y^{-1} }
      | c = {{Ring-axiom|D}}
}}
{{eqn | r = u \paren {x y^{-1} + 1}
      | c = {{Defof|Inverse Element}}
}}
{{eqn | o = \in
      | r = R
      | c = {{Ring-axiom|A0}} and {{Ring-axiom|M0}}
}}
{{end-eqn}}
That is, the inverse of $y$ is in $R$.
This contradicts the fact that $y$ is a non-unit.
Therefore $x + y$ is a non-unit.
{{qed}}
Category:Valuation Rings
Category:Local Rings
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Theorem: Bijection is Open iff Inverse is Continuous

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: T_1 \to T_2$ be a bijection.
Then $f$ is open {{iff}} $f^{-1}$ is continuous.

Proof:
Let $f$ be a bijection.
Let $g := f^{-1}$.
By Bijection iff Inverse is Bijection we have that $g$ is a bijection and that $g^{-1} = f$.
Let $f$ be open.
Then by definition of open mapping:
:$\forall H \in \tau_1: f \sqbrk H \in \tau_2$
taking $H \in \tau_1$ by definition of open in $T_1$.
But $f = g^{-1}$ and so:
:$\forall H \in \tau_1: g^{-1} \sqbrk H \in \tau_2$
which is exactly the definition for $g$ to be continuous.
The argument works the other way.
Let $g$ be continuous.
Then by definition of continuous mapping:
:$\forall H \in \tau_1: g^{-1} \sqbrk H \in \tau_2$
But $g^{-1} = f$ and so:
:$\forall H \in \tau_1: f \sqbrk H \in \tau_2$
which is exactly the definition for $f$ to be open.
{{qed}}
Theorem: Bijection is Open iff Closed

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: T_1 \to T_2$ be a bijection.
Then $f$ is open {{iff}} $f$ is closed.

Proof:
Let $f$ be a bijection.
Suppose $f$ is an open mapping.
From the definition of open mapping:
:$\forall H \in \tau_1: f \sqbrk H \in \tau_2$
As $f$ is a bijection:
:$f \sqbrk {S_1 \setminus H} = f \sqbrk {S_1} \setminus f \sqbrk H = S_2 \setminus f \sqbrk H$
By definition of closed set:
:$S_1 \setminus H$ is closed in $T_1$
and as $f$ is an open mapping:
:$f \sqbrk {S_1 \setminus H} = S_2 \setminus f \sqbrk H$
is closed in $T_2$.
Hence by definition $f$ is a closed mapping.
A similar argument demonstrates that if $f$ is closed then it is open.
{{qed}}
Theorem: T2 Space is T1

Let $T = \struct {S, \tau}$ be a $T_2$ (Hausdorff) space.
Then $T$ is also a $T_1$ space.

Proof:
{{Recall|T1 Space|$T_1$ space}}
{{:Definition:T1 Space/Definition 1}}
Let $T$ be a $T_2$ space.
{{Recall|T2 Space|$T_2$ space}}
{{:Definition:T2 Space/Definition 1}}
As $U \cap V = \O$ it follows from the definition of disjoint sets that:
{{begin-itemize}}
{{item||$x \in U \implies x \notin V$}}
{{item||$y \in V \implies y \notin U$}}
{{end-itemize}}
So if $x \in U, y \in V$ then:
:$\exists U \in \tau: x \in U, y \notin U$
As both $x$ and $y$ are arbitrary, tt follows that $T$ is a $T_1$ space.
{{qed}}
Theorem: T1 Space is T0

Let $\struct {S, \tau}$ be a $T_1$ space.
Then $\struct {S, \tau}$ is also a $T_0$ space.

Proof:
{{Recall|T0 Space|$T_0$ space|index = 1}}
{{:Definition:T0 Space/Definition 1}}
Let $\struct {S, \tau}$ be a $T_1$ space.
Let $x, y \in S: x \ne y$.
{{Recall|T1 Space|$T_1$ space|index = 1}}
{{:Definition:T1 Space/Definition 1}}
From the Rule of Addition:
:'''Either'''
::$\exists U \in \tau: x \in U, y \notin U$
:'''or''':
::$\exists V \in \tau: y \in V, x \notin V$
which is precisely the definition of a $T_0$ space.
{{qed}}
Theorem: T5 Space is T4

Let $\struct {S, \tau}$ be a $T_5$ space.
Then $\struct {S, \tau}$ is also a $T_4$ space.

Proof:
Let $\struct {S, \tau}$ be a $T_5$ space.
{{Recall|T5 Space|$T_5$ space}}
{{:Definition:T5 Space/Definition 1}}
Let $C, D \subseteq S$ be disjoint sets which are closed in $T$.
Thus $C, D \in \map \complement \tau$ from the definition of closed set.
From Topological Closure is Closed:
:$C^- = C, D^- = D$
and so from $C \cap D = \O$:
:$C^- \cap D = C \cap D^- = \O$
Thus from the definition of $T_5$ space:
:$\forall C, D \in \map \complement \tau, C \cap D = \O: \exists U, V \in \tau: C \subseteq U, D \subseteq V$ 
which is precisely the definition of a $T_4$ space.
{{qed}}
Theorem: Completely Normal Space is Normal

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

Proof:
Let $\struct {S, \tau}$ be a completely normal space.
{{Recall|Completely Normal Space|completely normal space|index = 1}}
{{:Definition:Completely Normal Space/Definition 1}}
We have that a $T_5$ space is $T_4$.
{{Recall|Normal Space|normal space|index = 1}}
{{:Definition:Normal Space/Definition 1}}
Hence the result.
{{qed}}
Theorem: Space which is T3 and T0 is also T2

Let $\struct {S, \tau}$ be a topological space which is both a $T_3$ space and a $T_0$ space.
Then $\struct {S, \tau}$ is also a $T_2$ (Hausdorff) space.

Proof:
{{Recall|T2 Space|$T_2$ Space|index = 1}}
{{:Definition:T2 Space/Definition 1}}
Let $T = \struct {S, \tau}$ be both a $T_3$ space and a $T_0$ space.
{{Recall|T0 Space|$T_0$ Space|index = 1}}
{{:Definition:T0 Space/Definition 1}}
Let $x, y \in S$.
{{WLOG}}, suppose that $\exists V \in \tau: y \in V, x \notin V$.
Then by definition of relative complement:
:$x \in \relcomp S V$
Let $F := \relcomp S V$.
Then $V := \relcomp S F$.
As $V$ is open, by definition of closed set we have that $F = \relcomp S V$ is closed.
That is:
:$V = \relcomp S F \in \tau$
As $y \in V$ it follows that $y \notin F$, that is:
:$y \in \relcomp S F = V$
{{Recall|T3 Space|$T_3$ Space|index = 1}}
{{:Definition:T3 Space/Definition 1}}
So we have that:
:$x \in F \subseteq U \implies x \in U$
:$y \notin F, y \in V$
such that $U \cap V = \O$.
So:
:$\forall x, y \in S: x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$
which is precisely the definition of a $T_2$ (Hausdorff) space.
{{qed}}
Theorem: Normal Space is T3

Let $\struct {S, \tau}$ be a normal space.
Then $\struct {S, \tau}$ is also a $T_3$ space.

Proof:
{{Recall|T3 Space|$T_3$ space|index = 1}}
{{:Definition:T3 Space/Definition 1}}
{{Recall|Normal Space|normal space|index = 1}}
{{:Definition:Normal Space/Definition 1}}
{{Recall|T1 Space|$T_1$ space|index = 3}}
{{:Definition:T1 Space/Definition 3}}
Let $F$ be an arbitrary closed set in $T$.
Let $y \in \relcomp S F$, that is, $y \in S$ such that $y \notin F$.
As $T$ is a $T_1$ space it follows that $\set y$ is closed.
{{Recall|T4 Space|$T_4$ space|index = 1}}
{{:Definition:T4 Space/Definition 1}}
We note that $F$ and $\set y$ are disjoint closed sets for which those disjoint open sets $U$ and $V$ exist.
But $F$ and $\set y$ are also such that $y \in \relcomp S F$.
As $F$ is arbitrary:
:$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$ 
which is precisely the definition of a $T_3$ space.
{{qed}}
Theorem: T2.5 Space is T2

Let $\struct {S, \tau}$ be a $T_{2 \frac 1 2}$ space.
Then $\struct {S, \tau}$ is also a $T_2$ (Hausdorff) space.

Proof:
{{Recall|T2 Space|$T_2$ space|index = 1}}
{{:Definition:T2 Space/Definition 1}}
Let $T = \struct {S, \tau}$ be a $T_{2 \frac 1 2}$ space.
{{Recall|T2.5 Space|$T_{2 \frac 1 2}$ space|index = 1}}
{{:Definition:T2.5 Space/Definition 1}}
We have that Set is Subset of its Topological Closure and so $U \subseteq U^-$ and $V \subseteq V^-$.
This leads to:
:$\forall x, y \in S: x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$
which is precisely the definition of a $T_2$ space.
{{qed}}
Theorem: Regular Space is T2.5

Let $\struct {S, \tau}$ be a regular space.
Then $\struct {S, \tau}$ is also a $T_{2 \frac 1 2}$ space.

Proof:
{{Recall|T2.5 Space|$T_{2 \frac 1 2}$ space|index = 2}}
{{:Definition:T2.5 Space/Definition 2}}
Let $T = \struct {S, \tau}$ be a regular space.
{{Recall|Regular Space|index = 3}}
{{:Definition:Regular Space/Definition 3}}
Let $x, y \in S$ with $x \ne y$.
{{Recall|T2 Space|$T_2$ space|index = 1}}
{{:Definition:T2 Space/Definition 1}}
Thus there exist disjoint open sets containing $x$ and $y$ respectively:
:$\exists U_x, U_y \in \tau: x \in U_x, y \in U_y, U_x \cap U_y = \O$
{{Recall|T3 Space|$T_3$ space|index = 2}}
{{:Definition:T3 Space/Definition 2}}
Specifically for $x$ and $y$:
:$\exists N_x: \relcomp S {N_x} \in \tau: \exists V_x \in \tau: x \in V_x \subseteq N_x \subseteq U_x$
:$\exists N_y: \relcomp S {N_y} \in \tau: \exists V_y \in \tau: y \in V_y \subseteq N_y \subseteq U_y$
From Subsets of Disjoint Sets are Disjoint:
:$N_x \cap N_y = \O$
Hence $x$ and $y$ are separated by closed neighborhoods.
The result follows.
{{qed}}
Theorem: Equivalence of Definitions of T0 Space

{{TFAE|def = T0 Space|view = $T_0$ (Kolmogorov) space}}
Let $T = \struct {S, \tau}$ be a topological space.

Proof:
=== Definition by Open Sets implies Definition by Limit Points ===
Let $T = \struct {S, \tau}$ be a topological space for which:
:$\forall x, y \in S$, either:
::$(1): \quad \exists U \in \tau: x \in U, y \notin U$
::$(2): \quad \exists U \in \tau: y \in U, x \notin U$
Let $x, y \in S$ such that $x \ne y$.
By the definition of limit point of a point:
:$(3): \quad x$ is a limit point of $y$ if every open set $U \in \tau$ such that $x \in U$ contains $y$.
:$(4): \quad y$ is a limit point of $x$ if every open set $U \in \tau$ such that $y \in U$ contains $x$.
Thus:
:If $(3)$ holds then $(1)$ is false.
:If $(4)$ holds then $(2)$ is false.
So both $(3)$ and $(4)$ can not both hold at once.
Hence $T = \struct {S, \tau}$ is a topological space for which no two points can be limit points of each other.
{{qed|lemma}}
Theorem: Equivalence of Definitions of T1 Space

{{TFAE|def = T1 Space|view = $T_1$ space}}
Let $T = \struct {S, \tau}$ be a topological space.

Proof:
=== Definition by Open Sets implies Definition by Closed Points ===
Let $T = \struct {S, \tau}$ be a topological space for which:
:$\forall x \in S: \forall y \in S, x \ne y: \exists U \in \tau: x \in U, y \notin U$
Let $x, y \in S$.
By the definition of limit point of a point, the above condition means:
:$x$ is a limit point of $y$ if every open set $U \in \tau$ such that $x \in U$ contains $y$.
Hence we have that $y$ is ''not'' a limit point of $x$.
As $x$ and $y$ are any two points in $S$, it follows that $x$ has no limit points.
Thus it holds vacuously that all the limit point of $x$ are in $\set x$.
By Closed Set iff Contains all its Limit Points, we have that $\set x$ is a closed set in $T$.
This holds for all $x \in S$.
Hence $T = \struct {S, \tau}$ is a topological space for which all points are closed.
{{qed|lemma}}
Theorem: T3.5 Space is T3

Let $T$ be a $T_{3 \frac 1 2}$ space.
Then $T$ is also a $T_3$ space.

Proof:
Let $T = \struct {S, \tau}$ be a $T_{3 \frac 1 2}$ space.
{{Recall|T3.5 Space|$T_{3 \frac 1 2}$ space}}
{{:Definition:T3.5 Space/Definition 1}}
{{Recall|Urysohn Function|Urysohn function}}
{{:Definition:Urysohn Function}}
Let $F \subseteq S$ be a closed set in $T$.
Let $y \in \relcomp S F$.
Let:
{{begin-eqn}}
{{eqn | l = U
      | r = \map {f^{-1} } 0
}}
{{eqn | l = V
      | r = \map {f^{-1} } 1
}}
{{end-eqn}}
As $f$ is continuous, both $U$ and $V$ are open in $T$.
Suppose $x \in U \cap V$.
Then we would have:
{{begin-eqn}}
{{eqn | l = x \in U
      | o = \implies
      | r = \map f x = 0
}}
{{eqn | l = x \in V
      | o = \implies
      | r = \map f x = 1
}}
{{end-eqn}}
which contradicts the many-to-one nature of a mapping.
So $U \cap V = \O$.
Thus we have:
:$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$ 
That is, for any closed set $F \subseteq S$ and any point $y \in S$ such that $y \notin F$ there exist disjoint open sets $U, V \in \tau$ such that $F \subseteq U$, $y \in V$.
This is precisely the definition of a $T_3$ space.
{{qed}}
Theorem: Completely Regular Space is Regular

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

Proof:
Let $T = \struct {S, \tau}$ be a completely regular space.
{{Recall|Completely Regular Space|completely regular space|index = 1}}
{{:Definition:Completely Regular Space/Definition 1}}
We have:
:$T_{3 \frac 1 2}$ Space is $T_3$.
{{Recall|Regular Space|regular space|index = 1}}
{{:Definition:Regular Space/Definition 1}}
Hence the result.
{{qed}}
Theorem: Normal Space is Completely Regular

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

Proof:
{{Recall|Completely Regular Space|completely regular space|index = 2}}
{{:Definition:Completely Regular Space/Definition 2}}
{{Recall|Normal Space|normal space|index = 1}}
{{:Definition:Normal Space/Definition 1}}
From Normal Space is $T_{3 \frac 1 2}$, we have that $T$ is a $T_{3 \frac 1 2}$ space.
So, by definition, $T$ is a completely regular space.
{{qed}}
Theorem: T4 and T3 Space is T3.5

Let a topological space $T = \struct {S, \tau}$ be:
:a $T_4$ space
and also:
:a $T_3$ space.
Then $T$ is also a $T_{3 \frac 1 2}$ space.

Proof:
Let $T = \struct {S, \tau}$ be a $T_4$ space which is also a $T_3$ space.
{{Recall|T3 Space|$T_3$ Space|index = 1}}
{{:Definition:T3 Space/Definition 1}}
Consider this $U \in \tau$, which is disjoint from $\set y$.
Then $\relcomp S U$ is a closed set which is disjoint from $F$ but such that $\set y \subseteq \relcomp S U$.
As $T$ is a $T_4$ space, we have that from Urysohn's Lemma there exists an Urysohn function $f$ for $F$ and $\relcomp S U$.
As $\set y \subseteq \relcomp S U$, this function $f$ is a Urysohn function for $F$ and $\set y$ as well.
So:
:For any closed set $F \subseteq S$ and any point $y \in S$ such that $y \notin F$, there exists an Urysohn function for $F$ and $\set y$
which is precisely the definition of a $T_{3 \frac 1 2}$ space.
{{qed}}
Theorem: T0 Property is Preserved under Homeomorphism

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

Proof:
By definition of homeomorphism, $\phi$ is a closed continuous bijection.
The result follows from $T_0$ Property is Preserved under Closed Bijection.
{{qed}}