nyuuzyou/wikis huggingface dataset:
T1 Space is Weakly Countably Compact iff Countably Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a $T_1$ (Fréchet) topological space.
Then $T$ is weakly countably compact $T$ is countably compact.
== Proof ==
=== Sufficient Condition ===
We have that a Countably Compact Space is Weakly Countably Compact whatever the separation properties.
=== Necessary Condition ===
We need to show that a $T_1$ space which is weakly countably compact is also necessarily countably compact.
‎
So, let $T = \struct {S, \tau}$ be a $T_1$ space which is weakly countably compact.
Let $A \subseteq S$.
Suppose $x$ to be a limit point but not an $\omega$-accumulation point of $A$.
Then there exists some open set $U_x \in \tau$ such that $x \in U_x$ which contains only a finite number of points of $A$, say:
:$U_x = \set {a_1, a_2, \ldots, a_n}$
But this, we have specified, is a $T_1$ space.
So this would imply that $x$ has an open neighborhood which contains no points of $A$.
That is, $x$ is not a limit point of $A$ after all.
Hence if $x$ is a limit point of $A$, it is an $\omega$-accumulation point of $A$.
Thus $S$ is countably compact.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Global Compactness Properties
Category:T1 Spaces
Category:Countably Compact Spaces
Category:Weakly Countably Compact Spaces
T2 Property is Hereditary

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is a $T_2$ (Hausdorff) space.
Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a subspace of $T$.
Then $T_H$ is a $T_2$ (Hausdorff) space.
That is, the property of being a $T_2$ (Hausdorff) space is hereditary.
== Proof ==
Let $T = \struct {S, \tau}$ be a $T_2$ (Hausdorff) space.
Then:
:$\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$ 
That is, for any two distinct elements $x, y \in S$ there exist disjoint open sets $U, V \in \tau$ containing $x$ and $y$ respectively.
We have that the set $\tau_H$ is defined as:
:$\tau_H := \set {U \cap H: U \in \tau}$
Let $x, y \in H$ such that $x \ne y$.
Then as $x, y \in S$ we have that:
:$\exists U, V \in \tau: x \in U, y \in V, U \cap V = \O$
As $x, y \in H$ we have that:
:$x \in U \cap H, y \in V \cap H: \paren {U \cap H} \cap \paren {V \cap H} = \O$
and so the $T_2$ axiom is satisfied in $H$.
== Sources ==
* : $4$: The Hausdorff condition: $4.2$: Separation axioms: Proposition $4.2.4 \ \text{(a)}$
* : $4$: The Hausdorff condition: Exercise $4.3: 2$
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
* : $\S 3$: New Spaces From Old: Subspaces
Category:Hausdorff Spaces
Category:Topological Subspaces
T2 Space is Preserved under Closed Bijection

== Theorem ==
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a closed bijection.
If $T_A$ is a $T_2$ (Hausdorff) space, then so is $T_B$.
== Proof ==
Let $T_A$ be a $T_2$ (Hausdorff) space.
Then:
:$\forall x, y \in S_A, x \ne y: \exists U_A, V_A \in \tau_A: x \in U_A, y \in V_A: U_A \cap V_A = \O$
Suppose that $T_B$ is not Hausdorff.
Then:
:$\exists a, b \in S_B: a \ne b: \forall U_B, V_B \in \tau_B: a \in U_B, b \in V_B \implies U_B \cap V_B \ne \O$
That is, there exists at least one pair of points $a$ and $b$ for which all the open sets containing $a$ and $b$ are not disjoint.
As $\tau_B$ is a topology, it follows that $W_B = U_B \cap V_B$ is also an open set.
Let $U_A = \phi^{-1} \sqbrk {U_B}, V_A = \phi^{-1} \sqbrk {V_B}, W_A = \phi^{-1} \sqbrk {W_B}$.
From Preimage of Intersection under Mapping, we have:
:$\phi^{-1} \sqbrk {U_B \cap V_B} = \phi^{-1} \sqbrk {U_B} \cap \phi^{-1} \sqbrk {V_B}$
that is:
:$U_A \cap V_A = W_A$
By Bijection is Open iff Closed, $\phi$ is an open bijection.
By Bijection is Open iff Inverse is Continuous, it follows that $\phi^{-1}$ is continuous.
As $\phi$ is continuous, all of $U_A, V_A, W_A$ are open in $T_A$.
From the bijective nature of $\phi$, we have that:
:$a \in U_B \implies \map {\phi^{-1} } a \in U_A$
:$b \in V_B \implies \map {\phi^{-1} } b \in V_A$
Let $x = \map {\phi^{-1} } a, y = \map {\phi^{-1} } b$.
Thus we have that:
:$\exists x, y \in S_A: x \ne y: \forall U_A, V_A \in \tau_A: x \in U_A, y \in V_A \implies U_A \cap V_A \ne \O$
contradicting the fact that $T_A$ is a $T_2$ (Hausdorff) space.
Hence $T_B$ must after all be a $T_2$ (Hausdorff) space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:Hausdorff Spaces
T2 Space is Sober Space

== Theorem ==
Let $\struct {S, \tau}$ be a $T_2$ (Hausdorff) space.
Then:
:$\struct {S, \tau}$ is a sober space
== Proof ==
Let $W \in \tau$ be meet-irreducible open proper subset of $S$.
Let $x \in S \setminus W$.
We have $S \setminus W$ is a closed set by definition.
From Closure of Subset of Closed Set of Topological Space is Subset:
:$\set x^- \subseteq S \setminus W$
From Relative Complement inverts Subsets of Relative Complement:
:$W \subseteq S \setminus \set x^-$
Hence:
:$x \notin W$
Let $y \in S \setminus \set x^-$.
Then:
:$y \ne x$
By definition of $T_2$ (Hausdorff) space:
:$\exists W_1, W_2 \in \tau : x \in W_1, y \in W_2 : W_1 \cap W_2 = \O$
Let:
:$U = W \cup W_1$
and
:$V = W \cup W_2$
Hence:
:$x \in U, y \in V$
Since $x \notin W$ and $x \in U$:
:$U \nsubseteq W$
We have:
From Characterization of Meet-Irreducible Open Set:
:$V = W$
Hence:
:$y \in W$
Since $y$ was arbitrary:
:$S \setminus \set x^- \subseteq W$
By definition of set equivalence:
:$W = S \setminus \set x^-$
== Also see ==
* Sober Space is T0 Space
== Sources ==
* : Chapter $1$: Spaces and Lattices of Open Sets, $\S 1$ Sober spaces, Notes $1.2 (2)$
Category:Sober Spaces
Category:Hausdorff Spaces
T2 Space is T1 Space

== Theorem ==
Let $\struct {S, \tau}$ be a $T_2$ (Hausdorff) space.
Then $\struct {S, \tau}$ is also a $T_1$ (Fréchet) space.
== Proof ==
From the definition of $T_2$ (Hausdorff) space:
:$\forall x, y \in S: x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$
As $U \cap V = \O$ it follows from the definition of disjoint sets that:
:$x \in U \implies x \notin V$
:$y \in V \implies y \notin U$
So if $x \in U, y \in V$ then:
:$\exists U \in \tau: x \in U, y \notin U$
:$\exists V \in \tau: y \in V, x \notin V$
which is precisely the definition of a $T_1$ (Fréchet) space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms
Category:T1 Spaces
Category:Hausdorff Spaces
T3 1/2 Property is Hereditary

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is a $T_{3 \frac 1 2}$ space.
Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a subspace of $T$.
Then $T_H$ is a $T_{3 \frac 1 2}$ space.
== Proof ==
Let $T = \struct {S, \tau}$ be a $T_{3 \frac 1 2}$ space.
Then:
: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 $\set y$.
We have that the set $\tau_H$ is defined as:
:$\tau_H := \set {U \cap H: U \in \tau}$
Let $F \subseteq H$ such that $F$ is closed in $H$.
Let $y \in H$ such that $y \notin F$.
From Closed Set in Topological Subspace $F$ is also closed in $T$.
Because $T$ is a $T_{3 \frac 1 2}$ space, we have that there exists an Urysohn function for $F$ and $\set y$:
That is, there exists a continuous mapping $f: S \to \closedint 0 1$, where $\closedint 0 1$ is the closed unit interval, such that:
:$f {\restriction_F} = 0, f {\restriction_{\set y} } = 1$
where $f {\restriction_F}$ denotes the restriction of $f$ to $F$.
That is:
:$\forall a \in F: \map f a = 0$
:$\forall b \in \set y: \map f b = 1$
From Continuity of Composite with Inclusion, as $f$ is continuous on $S$, then $f {\restriction_H}$ is continuous on $H$.
Thus $f {\restriction_H}$ is an Urysohn function for $F$ and $\set y$ in $H$.
So the $T_{3 \frac 1 2}$ axiom is satisfied in $H$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:T3 1/2 Spaces
Category:Topological Subspaces
T3 1/2 Space is Preserved under Homeomorphism

== Theorem ==
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a homeomorphism.
If $T_A$ is a $T_{3 \frac 1 2}$ space, then so is $T_B$.
== Proof ==
Let $F \subseteq S_B$ be closed, and let $y \in S_B$ such that $y \notin F$.
Let $G = \phi^{-1} \sqbrk F$ and let $z = \map {\phi^{-1} } y$.
Since $T_A$ is a $T_{3 \frac 1 2}$ space, there exists an Urysohn function $f: X_A \to \closedint 0 1$ for $G$ and $\set z$.
Define $g: S_B \to \closedint 0 1$ by:
:$\map g x = \map f {\map {\phi^{-1} } x}$
By Composite of Continuous Mappings is Continuous, $g$ is continuous.
Also, for all $x \in F$:
:$\map g x = 0$
as $\map {\phi^{-1} } x \in G$.
Similarly, because $\map {\phi^{-1} } y = z$:
:$\map g y = 1$
Hence $g$ is an Urysohn function for $F$ and $\set y$.
Since $F$ and $y$ were arbitrary, it follows that $T_B$ is a $T_{3 \frac 1 2}$ space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:Homeomorphisms (Topological Spaces)
Category:T3 1/2 Spaces
T3 1/2 Space is T3 Space

== Theorem ==
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.
From the definition of $T_{3 \frac 1 2}$ space:
: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 $\set y$.
Let $F \subseteq S$ be a closed set in $T$ and let $y \in \relcomp S F$.
An Urysohn function for $F$ and $\set y$ is a continuous mapping $f: S \to \closedint 0 1$ where:
:$\forall a \in F: \map f a = 0$
:$\map f y = 1$
Let:
:$U = \map {f^{-1} } 0$
:$V = \map {f^{-1} } 1$
As $f$ is continuous, both $U$ and $V$ are open in $T$.
Suppose $x \in U \cap V$.
Then we would have:
:$x \in U \implies \map f x = 0$
:$x \in V \implies \map f x = 1$
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 $F \subseteq U$, $y \in V$.
which is precisely the definition of a $T_3$ space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Completely Regular Spaces
Category:T3 1/2 Spaces
Category:T3 Spaces
T3 1/2 Space is Uniformizable

== Theorem ==
Let $T$ be a topological space.
Then $T$ is uniformizable $T$ is a $T_{3 \frac 1 2}$ space.
== Proof ==
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Uniformities
Category:Uniformities
Category:T3 1/2 Spaces
T3 1/2 Space is not necessarily T2 Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a be a $T_{3 \frac 1 2}$ space.
Then it is not necessarily the case that $T$ is a $T_2$ (Hausdorff) space.
== Proof ==
Proof by Counterexample:
Let $S$ be a set and let $\PP$ be a partition on $S$ which is specifically not the (trivial) partition of singletons.
Let $T = \struct {S, \tau}$ be the partition space whose basis is $\PP$.
From Partition Topology is $T_{3 \frac 1 2}$, we have that $T$ is a $T_{3 \frac 1 2}$ space.
From Partition Topology is not Hausdorff, $T$ is not a $T_2$ (Hausdorff) space.
The result follows.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Completely Regular Spaces
Category:T3 1/2 Spaces
Category:Hausdorff Spaces
T3 Lindelöf Space is Fully T4 Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a $T_3$ Lindelöf topological space.
Then:
:$T$ is a fully $T_4$ space
== Proof ==
From Lindelöf T3 Space is Paracompact:
:$T$ is a paracompact space
From T3 Space is Fully T4 iff Paracompact:
:$T$ is a fully $T_4$ space
Category:T3 Spaces
Category:Lindelöf Spaces
Category:Fully T4 Spaces
T3 Lindelöf Space is T4 Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a $T_3$ Lindelöf topological space.
Then:
:$T$ is a $T_4$ space.
== Proof 1 ==
== Proof 2 ==
Category:T3 Spaces
Category:Lindelöf Spaces
Category:T4 Spaces
Category:T3 Lindelöf Space is T4 Space
T3 Lindelöf Space is T4 Space/Proof 1

== Theorem ==
== Proof ==
Let $A$ and $B$ be disjoint closed subsets of $T$.
Let $\UU = \set {U \in \tau : U^- \cap B = \O}$.
From Characterization of T3 Space:
:$\forall a \in A : \exists U_a \in \tau: a \in U_a : U_a^- \cap B = \O$
By definition of open cover:
:$\UU$ is an open cover of $A$
From Closed Subspace of Lindelöf Space is Lindelöf Space:
:$\struct {A, \tau_A}$ is a Lindelöf subspace
where $\tau_A$ is the subspace topology on $A$.
By definition of Lindelöf space:
:there exists a countable subcover $\family {U_n}_{n \mathop \in \N}$ of $\UU$ for $A$
Let $\VV = \set {V \in \tau : V^- \cap A = \O}$.
Similar to $A$ above, there exists a countable subcover $\family {V_n}_{n \mathop \in \N}$ of $\VV$ for $B$.
From Countable Open Covers Condition for Separated Sets:
:$A$ and $B$ can be separated in $T$
By definition of separated:
:$\exists U, V \in \tau : A \subseteq U, B \subseteq V, U \cap V = \O$.
Since $A$ and $B$ were arbitrary:
:$\forall A, B \in \map \complement \tau, A \cap B = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$ 
Hence, by definition, $T$ is a $T_4$ space.
== Sources ==
* : Chapter $4$: Embedding and Metrization: Lemma $1$
* : Chapter $4$: Separation and Countability: $\S16$: Countability Properties: Theorem $16.8$
Category:T3 Lindelöf Space is T4 Space
T3 Lindelöf Space is T4 Space/Proof 2

== Theorem ==
== Proof ==
From $T_3$ Lindelöf Space is Fully $T_4$ Space:
:$T$ is a fully $T_4$ space.
From Fully $T_4$ Space is $T_4$ Space:
:$T$ is a $T_4$ space.
Category:T3 Lindelöf Space is T4 Space
T3 Property is Hereditary

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is a $T_3$ space.
Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a subspace of $T$.
Then $T_H$ is a $T_3$ space.
== Proof ==
Let $T = \struct {S, \tau}$ be a $T_3$ space.
Then:
:$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$ 
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 $F \subseteq U$, $y \in V$.
We have that the set $\tau_H$ is defined as:
:$\tau_H := \set {U \cap H: U \in \tau}$
Let $F \subseteq H$ such that $F$ is closed in $H$.
Let $y \in H$ such that $y \notin F$.
From Closed Set in Topological Subspace $F$ is also closed in $T$.
Because $T$ is a $T_3$ space, we have that:
:$\exists U, V \in \tau: F \subseteq U, y \in V, U \cap V = \O$
As $F \subseteq H$ and $y \in H, y \notin F$ we have that:
:$F \subseteq U \cap H, y \in V \cap H: \paren {U \cap H} \cap \paren {V \cap H} = \O$
and so the $T_3$ axiom is satisfied in $H$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:T3 Spaces
Category:Topological Subspaces
T3 Space is Fully T4 iff Paracompact

== Theorem ==
Let $T = \struct {S, \tau}$ be a $T_3$ space.
Then $T$ is fully $T_4$ $T$ is paracompact.
== Proof ==
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Compactness Properties and the $T_i$ Axioms
Category:Paracompact Spaces
Category:T3 Spaces
Category:Fully T4 Spaces
T3 Space is Preserved under Homeomorphism

== Theorem ==
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a homeomorphism.
If $T_A$ is a $T_3$ space, then so is $T_B$.
== Proof ==
Suppose that $T_A$ is a $T_3$ space.
Let $F$ be closed in $T_B$.
Let $y \in S_B$ such that $y \notin F$.
From Preimage of Intersection under Mapping it follows that $\phi^{-1} \sqbrk F$ and $\map {\phi^{-1}} y$ are disjoint.
Also, as $\phi$ is a homeomorphism, it is a fortiori continuous.
Thus by Continuity Defined from Closed Sets, $\phi^{-1} \sqbrk F$ is closed.
Now as $T_A$ is a $T_3$ space, we find disjoint open sets $U_1$ containing $\phi^{-1} \sqbrk F$ and $U_2$ containing $\map {\phi^{-1}} y$.
From Image of Subset is Subset of Image, we have $F = \phi \sqbrk {\phi^{-1} \sqbrk F} \subseteq \phi \sqbrk {U_1}$.
Here, the first equality follows from Subset equals Image of Preimage iff Mapping is Surjection, as $\phi$ is a fortiori surjective, being a homeomorphism.
Then , we deduce also $\set y \subseteq \phi \sqbrk {U_2}$.
From Image of Intersection under Injection it follows that $\phi \sqbrk {U_1}$ and $\phi \sqbrk {U_2}$ are disjoint.
Since $\phi$ is a homeomorphism, they are also both open in $T_B$.
Therewith, we have construed two disjoint open sets in $T_B$, one containing $F$, and the other containing $y$.
Hence $T_B$ is shown to be a $T_3$ space as well.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:Homeomorphisms (Topological Spaces)
Category:T3 Spaces
T3 Space is Semiregular

== Theorem ==
Let $T = \struct {S, \tau}$ be a $T_3$ space.
Then $T$ is a semiregular space.
== Proof ==
Let $\BB = \set {B \subseteq S: B^{- \circ} = B}$.
In other words, $\BB$ is the collection of regular open sets contained in $S$.
It follows immediately from the definition of a basis that our theorem is proved if we can show that:
:$\BB$ is a cover for $S$
:$\forall U, V \in \BB: \forall x \in U \cap V: \exists W \in \BB: x \in W \subseteq U \cap V$
First we show that $\BB$ is a cover for $S$. 
Since $S$ is open:
:$S^\circ = S$
Since $S$ is closed:
:$S^- = S$
Therefore:
:$S^{- \circ} = S$
and so:
:$S \in \BB$
From Set is Subset of Union:
:$S \subseteq \bigcup \BB$
Thus $\BB$ covers $S$.
Next we demonstrate the second condition for $\BB$ to be a basis.
Let $U, V \in \BB$, $x \in U \cap V$.
We will show that for some $W \in \BB$:
:$x \in W \subseteq U \cap V$
By General Intersection Property of Topological Space, $U \cap V \in \tau$. 
Since $T$ is $T_3$, there is a closed neighborhood $N_x$ around $x$ that is contained in $U \cap V$:
:$\exists N_x: \relcomp S {N_x} \in \tau: \exists Q \in \tau: x \in Q \subseteq N_x \subseteq \paren {U \cap V}$
Let $W := {N_x}^{- \circ}$. 
Then by Interior of Closure is Regular Open, $W^{- \circ} = W$ and $W \in \BB$.
On the other hand, since $N_x$ is closed:
:${N_x}^- = N_x$, and thus $W = {N_x}^\circ$
By definition of interior, we have $Q \subseteq W \subseteq N_x$.
Therefore we have $x \in W \subseteq N_x \subseteq U \cap V$, as desired.
It follows that $\BB$ is a basis for $\tau$.
Hence the result by definition of semiregular space.
== Sources ==
* : $\text{II}: \ \S 2$: $(T_3)$ Spaces, Regular and Semiregular Sources
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Additional Separation Properties
Category:T3 Spaces
Category:Semiregular Spaces
T3 Space with Sigma-Locally Finite Basis is Paracompact

== Theorem ==
Let $T = \struct {S, \tau}$ be a $T_3$ topological space.
Let $\BB$ be a $\sigma$-locally finite basis of $T$.
Then:
:$T$ is a paracompact
== Proof ==
Let $\UU$ be an open cover of $T$.
Let $\VV = \set{B \in \BB : \exists U \in \UU : B \subseteq U}$
Hence $\VV \subseteq \BB$.
From Subset of Sigma-Locally Finite Set of Subsets is Sigma-Locally Finite:
:$\VV$ is $\sigma$-locally finite
Let $x \in S$.
By definition of open cover:
:$\exists U \in \UU : x \in U$
By definition of basis:
:$\exists B \in \BB : x \in B \subseteq U$
Hence:
:$B \in \VV$
It follows that $\VV$ is an open cover by definition.
By definition, $\VV$ is an open refinement of $\UU$.
It has been shown that:
:every open cover of $T$ has an open $\sigma$-locally finite refinement
From Characterization of Paracompactness in T3 Space:
:$T$ is paracompact
Category:T3 Spaces
Category:Paracompact Spaces
T3 Space with Sigma-Locally Finite Basis is Perfectly T4 Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a $T_3$ topological space.
Let $\BB$ be a $\sigma$-locally finite basis.
Then:
:$T$ is a perfectly $T_4$ space
== Proof ==
From T3 Space with Sigma-Locally Finite Basis is T4 Space:
:$T$ is a $T_4$ space
It remains to show that every closed set in $T$ is a $G_\delta$ set
Let $\BB = \ds \bigcup_{n \mathop \in \N} \BB_n$ be a $\sigma$-locally finite basis where $\BB_n$ is a locally finite set of subsets for each $n \in \N$.
Let $F$ be closed in $T$.
By definition of closed set:
:$X \setminus F$ is open in $T$.
Let $G = X \setminus F$.
=== Lemma 1 ===
From Complement of F-Sigma Set is G-Delta Set:
:$F$ is a $G_\delta$ set
Since $F$ was arbitrary, we have:
:every closed set in $T$ is a $G_\delta$ set.
It follows that $T$ is a perfectly $T_4$ space by definition.
Category:T3 Spaces
Category:Perfectly T4 Spaces
Category:T3 Space with Sigma-Locally Finite Basis is Perfectly T4 Space
T3 Space with Sigma-Locally Finite Basis is Perfectly T4 Space/Lemma 1

== Theorem ==
Let $T = \struct {S, \tau}$ be a $T_3$ topological space.
Let $\BB$ be a $\sigma$-locally finite basis.
Let $G$ be open in $T$.
Then:
:$G$ is an $F_\sigma$ set
== Proof ==
Let:
:$\CC = \set{B \in \BB : B^- \subseteq G}$
where $B^-$ denotes the closure of $B$ in $T$.
=== Lemma 2 ===
Let $\BB = \ds \bigcup_{n \mathop \in \N} \BB_n$ be a $\sigma$-locally finite basis where $\BB_n$ is a locally finite set of subsets for each $n \in \N$.
For each $n \in \N$, let:
:$\CC_n = \CC \cap \BB_n$
From Subset of Locally Finite Set of Subsets is Locally Finite:
:$\CC_n$ is locally finite
For each $n \in \N$, let:
:$C_n = \cup \set{C^- : C \in \CC_n}$
From Union of Closures of Elements of Locally Finite Set is Closed:
:For each $n \in \N$, $C_n$ is closed in $T$
From Union of Subsets is Subset:
:$\forall n \in \N: C_n \subseteq G$
and
:$\ds \bigcup_{n \mathop \in \N} C_n \subseteq G$
By definition of cover:
:$\forall x \in G : \exists n \in \N : \exists C \in \CC_n : x \in C^-$
From Set is Subset of Union:
:$\forall n \in \N, C \in \CC_n : C^- \subseteq C_n \subseteq \ds \bigcup_{n \mathop \in \N} C_n$
Hence:
:$\forall x \in G : x \in \ds \bigcup_{n \mathop \in \N} C_n$
By definition of subset:
:$G \subseteq \ds \bigcup_{n \mathop \in \N} C_n$
By definition of set equality:
:$G = \ds \bigcup_{n \mathop \in \N} C_n$
Hence $G$ is an $F_\sigma$ set by definition.
Category:T3 Space with Sigma-Locally Finite Basis is Perfectly T4 Space
T3 Space with Sigma-Locally Finite Basis is Perfectly T4 Space/Lemma 2

== Theorem ==
Let $T = \struct {S, \tau}$ be a $T_3$ topological space.
Let $\BB$ be a basis for $T$.
Let $G$ be open in $T$.
Let:
:$\CC = \set{B \in \BB : B^- \subseteq G}$
where $B^-$ denotes the closure of $B$ in $T$.
Then:
:$\CC$ is a is a cover of $G$
== Proof ==
Let $x \in G$.
From Characterization of T3 Space:
:$\exists U \in \tau : x \in U : U^- \subseteq G$
where $U^-$ denotes the closure of $U$ in $T$.
By definition of basis:
:$\exists B \in \BB : x \in B : B \subseteq U$
From Topological Closure of Subset is Subset of Topological Closure:
:$B^- \subseteq U^-$
From Subset Relation is Transitive:
:$B^- \subseteq G$
Hence:
:$B \in \CC$
Since $x$ was arbitrary, we have:
:$\forall x \in G : C \in \CC : x \in C$
Hence $\CC$ is a cover of $G$ by definition.
Category:T3 Space with Sigma-Locally Finite Basis is Perfectly T4 Space
T3 Space with Sigma-Locally Finite Basis is T4 Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a $T_3$ topological space.
Let $\BB$ be a $\sigma$-locally finite basis.
Then:
:$T$ is a $T_4$ space
== Proof 1 ==
== Proof 2 ==
Category:T3 Spaces
Category:T4 Spaces
Category:T3 Space with Sigma-Locally Finite Basis is T4 Space