Zero Dimensional Space is T3

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Theorem

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

Then $T$ is a $T_3$ space.


Proof

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

Let $F \subseteq S$ be closed in $T$.

Let also $y \notin F$.

Then by definition of closed, $\relcomp S F$ is open in $T$, where $\relcomp S F$ is the complement of $F$ in $S$.


As $T$ is zero dimensional, it has a basis $\BB$ which consists entirely of clopen sets.

As $\BB$ is a basis for $T$, it follows that:

$\ds \exists \UU \subseteq \BB: \relcomp S F = \bigcup \UU$

that is, $\relcomp S F$ is the union of a subset of elements of $\BB$.

Thus, there is a set $U \in \UU$ such that $y \in U$ and $F \subseteq \relcomp S U$.

But the elements of $\UU$ are clopen sets, so $U$ is itself clopen.

Thus, by definition, $\relcomp S U$ is also clopen.


So we have that $U$ and $\relcomp S U$ are open sets in $T$ such that:

$\exists W, R \in \tau: F \subseteq W, y \in R: R \cap W = \O$

by setting $R = U$ and $W = \relcomp S U$.

That is, $T$ is a $T_3$ space.

$\blacksquare$


Sources