Second-Countable T3 Space is T5

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Theorem

Let $T = \struct {S, \tau}$ be a $T_3$ space which is also second-countable.


Then $T$ is a $T_5$ space.


Proof

Let $A, B \subseteq S$ with $A^- \cap B = A \cap B^- = \O$.

For each $x \in A$, since $T$ is $T_3$:

$\exists P, Q \in \tau: x \in P, B^- \subseteq Q, P \cap Q = \O$

Let $\BB$ be a basis for $T$.

Then:

$\exists U \in \BB: x \in U \subseteq P$

Notice that:

\(\ds \) \(\) \(\ds U^- \cap B\)
\(\ds \) \(\subseteq\) \(\ds U^- \cap B^-\) Set Intersection Preserves Subsets; Set is Subset of its Topological Closure
\(\ds \) \(\subseteq\) \(\ds P^- \cap Q\) Set Intersection Preserves Subsets; Topological Closure of Subset is Subset of Topological Closure
\(\ds \) \(=\) \(\ds \O\) Open Set Disjoint from Set is Disjoint from Closure

By Subset of Empty Set, $U^-$ and $B$ are disjoint.


Since $T$ is second-countable, $\BB$ is countable.

Doing the above process for each $x \in A$ yields a subset $\set {U_n}_{n \mathop \in \N}$ of $\BB$.

Doing a similar process for each $y \in B$ yields another subset $\set {V_n}_{n \mathop \in \N}$ of $\BB$.

These sets are open sets by definition.


Define $\ds U'_n = U_n \setminus \bigcup_{i \mathop \le n} V_i^-$ and $\ds V'_n = V_n \setminus \bigcup_{i \mathop \le n} U_i^-$.

Define $\ds U' = \bigcup_{n \mathop \in \N} U'_n$ and $\ds V' = \bigcup_{n \mathop \in \N} V'_n$.

We show that $U'$ and $V'$ are disjoint open sets containing $A$ and $B$ respectively.


For any $n \in \N$, we have that $U_n$ is open.

From Topological Closure is Closed:

$V_i^-$ is closed for each $i \le n$.

From Finite Union of Closed Sets is Closed in Topological Space:

$\ds \bigcup_{i \mathop \le n} V_i^-$ is closed.

By Open Set minus Closed Set is Open:

$\ds U'_n = U_n \setminus \bigcup_{i \mathop \le n} V_i^-$ is open.

By Definition of Topological Space:

$\ds U' = \bigcup_{n \mathop \in \N} U'_n$ is open.

Similarly, $V'$ is open.


Let $y \in B$.

By construction, there is some $k \in \N$ where $y \in V_k$.

From above we see that $U_i^-$ and $B$ are disjoint for all $i \in \N$.

So $y \notin U_i^-$ for every $i \le k$.

Hence $y \in V_k \setminus \bigcup_{i \mathop \le k} U_i^- = V'_k \subseteq V'$.

By Definition of Subset, $B \subseteq V'$.

Similarly, $A \subseteq U'$.


Let $i, j \in \N$.

We show that $U'_i, V'_j$ are disjoint.

Without loss of generality suppose $i \le j$.

Then:

\(\ds U'_i\) \(=\) \(\ds U_i \setminus \bigcup_{k \mathop \le i} V_k^-\)
\(\ds \) \(\subseteq\) \(\ds U_i\) Set Difference is Subset
\(\ds \) \(\subseteq\) \(\ds U_i^-\) Set is Subset of its Topological Closure
\(\ds \) \(\subseteq\) \(\ds \bigcup_{k \mathop \le j} U_k^-\) Set is Subset of Union of Family
\(\ds \leadsto \ \ \) \(\ds U'_i \cap V'_j\) \(=\) \(\ds \O\) Empty Intersection iff Subset of Relative Complement

Now:

\(\ds U' \cap V'\) \(=\) \(\ds \paren {\bigcup_{i \mathop \in \N} U'_i} \cap \paren {\bigcup_{j \mathop \in \N} V'_j}\)
\(\ds \) \(=\) \(\ds \bigcup_{\tuple {i, j} \mathop \in \N \times \N} \paren {U'_i \cap V'_j}\) Intersection Distributes over Union
\(\ds \) \(=\) \(\ds \bigcup_{\tuple {i, j} \mathop \in \N \times \N} \O\)
\(\ds \) \(=\) \(\ds \O\) Union is Empty iff Sets are Empty

Hence $U'$ and $V'$ are disjoint.

Since $A, B$ are arbitrary, $T$ is a $T_5$ space.

$\blacksquare$


Sources

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