Equivalence of Definitions of T5 Space

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Theorem

The following definitions of the concept of $T_5$ space are equivalent:


Let $T = \struct {S, \tau}$ be a topological space.

Definition by Open Sets

$\struct {S, \tau}$ is a $T_5$ space if and only if:

$\forall A, B \subseteq S, A^- \cap B = A \cap B^- = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$

That is:

$\struct {S, \tau}$ is a $T_5$ space when for any two separated sets $A, B \subseteq S$ there exist disjoint open sets $U, V \in \tau$ containing $A$ and $B$ respectively.

Definition by Closed Neighborhoods

$\struct {S, \tau}$ is a $T_5$ space if and only if:

$\forall Y, A \subseteq S: (A \subseteq Y^\circ \wedge A^- \subseteq Y) \implies \exists N \subseteq Y: \relcomp S N \in \tau: \exists U \in \tau: A \subseteq U \subseteq N$

That is:

$\struct {S, \tau}$ is a $T_5$ space if and only if every subset $Y \subseteq S$ contains a closed neighborhood of each $A \subseteq Y^\circ$ for which $A^- \subseteq Y$.

In the above, $Y^\circ$ denotes the interior of $Y$ and $A^-$ denotes the closure of $A$.


Proof

Definition by Open Sets implies Definition by Closed Neighborhoods

Suppose that:

$\forall A, B \subseteq S, A^- \cap B = A \cap B^- = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$


Let $Y, A \subseteq S$.

Suppose $A \subseteq Y^\circ$ and $A^- \subseteq Y$.

Then:

\(\ds A^- \cap \relcomp S Y\) \(\subseteq\) \(\ds Y \cap \relcomp S Y\) Set Intersection Preserves Subsets: Corollary: from $A^- \subseteq Y$
\(\ds \) \(=\) \(\ds \O\) Intersection with Complement
\(\ds \leadsto \ \ \) \(\ds A^- \cap \relcomp S Y\) \(=\) \(\ds \O\) Subset of Empty Set

We also have:

\(\ds A \cap \paren {\relcomp S Y}^-\) \(\subseteq\) \(\ds Y^\circ \cap \paren {\relcomp S Y}^-\) Set Intersection Preserves Subsets/Corollary: from $A \subseteq Y^\circ$
\(\ds \) \(=\) \(\ds Y^\circ \cap \relcomp S Y^\circ\) Complement of Interior equals Closure of Complement
\(\ds \) \(=\) \(\ds \O\) Intersection with Complement
\(\ds \leadsto \ \ \) \(\ds A \cap \paren {\relcomp S Y}^-\) \(=\) \(\ds \O\) Subset of Empty Set

We have by hypothesis:

$\exists U, V \in \tau: A \subseteq U, \relcomp S Y \subseteq V, U \cap V = \O$

Then:

\(\ds A\) \(\subseteq\) \(\ds U\) by hypothesis
\(\ds \) \(\subseteq\) \(\ds \relcomp S V\) Empty Intersection iff Subset of Complement: from $U \cap V = \O$
\(\ds \) \(\subseteq\) \(\ds \relcomp S {\relcomp S Y}\) Set Complement inverts Subsets: from $\relcomp S Y \subseteq V$
\(\ds \) \(=\) \(\ds Y\) Relative Complement of Relative Complement

So we have demonstrated that there exists a closed neighborhood $\relcomp S V$ of $A$ contained in $Y$.


As $Y$ and $A$ are arbitrary:

$\forall Y, A \subseteq S: \paren {A \subseteq Y^\circ \wedge A^- \subseteq Y} \implies \exists N \subseteq Y: \relcomp S N \in \tau: \exists U \in \tau: A \subseteq U \subseteq N$

$\Box$


Definition by Closed Neighborhoods implies Definition by Open Sets

Suppose that:

$\forall Y, A \subseteq S: \paren {A \subseteq Y^\circ \wedge A^- \subseteq Y} \implies \exists N \subseteq Y: \relcomp S N \in \tau: \exists U \in \tau: A \subseteq U \subseteq N$


Let $A, B \subseteq S$ be separated sets.

Then $A^- \cap B = A \cap B^- = \O$.

Then:

\(\ds A^-\) \(\subseteq\) \(\ds \relcomp S B\) Empty Intersection iff Subset of Complement: from $A^- \cap B = \O$

We also have:

\(\ds A\) \(\subseteq\) \(\ds \relcomp S {B^-}\) Empty Intersection iff Subset of Complement: from $A \cap B^- = \O$
\(\ds \) \(=\) \(\ds \paren{\relcomp S B}^\circ\) Complement of Interior equals Closure of Complement


We have by hypothesis that there exists a closed neighborhood $N$ of $A$ contained in $\relcomp S B$.

Then:

$\exists U \in \tau: A \subseteq U \subseteq N \subseteq \relcomp S B$

Then we have:

\(\ds B\) \(=\) \(\ds \relcomp S {\relcomp S B}\) Relative Complement of Relative Complement
\(\ds \) \(\subseteq\) \(\ds \relcomp S N\) Set Complement inverts Subsets: from $N \subseteq \relcomp S B$
\(\ds \) \(\in\) \(\ds \tau\) $N$ is closed

Finally:

\(\ds U \cap \relcomp S N\) \(\subseteq\) \(\ds N \cap \relcomp S N\) Set Intersection Preserves Subsets: Corollary: from $U \subseteq N$
\(\ds \) \(=\) \(\ds \O\) Intersection with Complement
\(\ds \leadsto \ \ \) \(\ds U \cap \relcomp S N\) \(=\) \(\ds \O\) Subset of Empty Set


Therefore we have:

\(\ds U, \relcomp S N\) \(\in\) \(\ds \tau\)
\(\ds A\) \(\subseteq\) \(\ds U\)
\(\ds B\) \(\subseteq\) \(\ds \relcomp S N\)
\(\ds U \cap \relcomp S N\) \(=\) \(\ds \O\)


As $A$ and $B$ are arbitrary:

$\forall A, B \subseteq S, A^- \cap B = A \cap B^- = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$

$\blacksquare$


Sources

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