Proposition A countable completely regular space is zero dimensional.
I'm going to need some help showing this is the case. First, suppose $X$ is a completely regular and countable. Then for each closed $E \subset X$ and $a \not\in E$, there is a continuous $f:X \to [0,1]$ so that $f(a) = 0$ and $f(E)=\{1\}$; also $X$ has countably many members.
I'm not sure where to go with this. Any suggestions? I have a couple of ideas.
- Utilize the continuity of $f$ to arrive at a conclusion about openness using the fact that open $V$ in $[0,1]$ implies $f^{-1}(V)$ is open in $X$.
- Assume that $X$ is countable and completely regular and suppose $X$ is not zero dimensional (to arrive at a contradiction)
- Something else regarding the countability of the space
I'm having a hard time seeing how countability ties into this proof. Are any of the suggestions I've made on the right track?
Let $p\in U$ where $U$ is an open subset of $X.$ We show there is an open-and-closed $V$ with $p\in V\subseteq U.$
Let $f:X\to [0,1]$ be continuous with $f(p)=0$ and $f[X\setminus U]\subseteq \{1\}.$ There must exist $r\in (0,1)$ such that $f^{-1}\{r\}=\emptyset.$ Otherwise the set $B=\cup_{r\in (0,1)}f^{-1}\{r\}$ would be the union of an uncountable family of pairwise-disjoint non-empty sets, which would make $B$ an uncountable subset of $X.$
So let $r\in (0,1)$ with $f^{-1}\{r\}=\emptyset.$ Let $V=f^{-1}[0,r).$ Then $V$ is open and $p\in V\subseteq U.$ And $V$ is also closed because $X\setminus V=f^{-1}[r,1]=f^{-1}(r,1]$ is open.