# Base of clopen and $T_{0}$ implies $X$ is Tychonoff

Let $X$ be a topological space and assume $X$ has a base $\mathcal{B}$ of clopen sets. Show $X$ is completely regular and a $T_{0}$ space.

My try:

First it is not hard to show that if $B \subset X$ then $\chi_{B}$, the characteristic function of $B$ is cts iff $B$ is clopen.

So let $F \subset X$ be a closed set and let $x \in X \setminus F$. Then since $X \setminus$ is open we can find $B \in \mathcal{B}$ such that $x \in B \subseteq X \setminus F$. Now define $\phi: X \rightarrow [0,1]$ by $\phi(x)= \chi_{B}(x)$ then since $B$ is clopen $\phi$ is a continuous map, $\phi(F)=\{0\}$ and $\phi(x)=1$, therefore $X$ is completely regular.

EDIT:

Sorry, Brian Scott is right, I'm trying to prove the following, if $X$ is $T_{0}$ and has a base of clopen sets then $X$ is completely regular and $T_{1}$. So I think the above proof is correct (i.e showing it is completely regular), how to show it is $T_{1}$?

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This is false. A trivial space always has a basis of clopens, but (if it has more than one point) is not $T_0$. – Chris Eagle Jul 5 '11 at 19:36
What about $X = \{a,b\}$ with the trivial topology? It has the base $\mathcal B = \{X, \varnothing\}$ consisting of clopen sets, but it is not $T_0$. – Sam Jul 5 '11 at 19:37
What is true is that if $X$ is $T_0$ and has a base of clopen sets, then it's $T_1$ and completely regular, i.e., Tikhonov. – Brian M. Scott Jul 5 '11 at 19:44
@Brian M. Scott: you're right, my mistake. Can you please have a look? – user10 Jul 6 '11 at 0:19

Suppose $x$ and $y$ are distinct points in $X$. $X$ is $T_0$, so WLOG there is an open set $U$ containing $x$ and not $y$. To show $X$ is $T_1$, we need to find an open set containing $y$ and not $x$. But since $X$ has a basis of clopen sets, there's a clopen $V$ containing $x$ and contained in $U$. The complement of $V$ is then the open set we want.

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