Zero Dimensional Space is T3

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Theorem

Let $T = \left({S, \tau}\right)$ be a zero dimensional topological space.

Then $T$ is a $T_3$ space.


Proof

Let $T = \left({S, \tau}\right)$ be a zero dimensional space.

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

Let also $y \notin F$.

Then by definition of closed, $\complement_S \left({F}\right)$ is open in $T$, where $\complement_S \left({F}\right)$ is the complement of $F$ in $S$.


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

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

$\displaystyle \exists \mathcal U \subseteq \mathcal B: \complement_S \left({F}\right) = \bigcup \mathcal U$

that is, $\complement_S \left({F}\right)$ is the union of a subset of elements of $\mathcal B$.

Thus, there is a set $U \in \mathcal U$ such that $y \in U$ and $F \subseteq \complement_S \left({U}\right)$.

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

Thus, by definition, $\complement_S \left({U}\right)$ is also clopen.


So we have that $U$ and $\complement_S \left({U}\right)$ are open sets in $T$ such that:

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

by setting $R = U$ and $W =\complement_S \left({U}\right)$.

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

$\blacksquare$


Sources