# Confusion in zero-dimensionality

I saw people wrote that a zero-dimensional space $$X$$ is Tychonoff since the characteristic function of a clopen set is continuous.

The confusion is that to show that X is Tychonoff we need to show that for every closed set $$C$$ and for every point $$d$$ outside $$C$$, there exists a continuous function that separates $$C$$ and $$d$$. Of course if $$C$$ is clopen, then it is very easy. But not every closed set in a zero-dimensional space is clopen.

So how does showing that the characteristic function of a clopen set is continuous prove that $$X$$ is Tychonoff? Thanks.

## 2 Answers

If $$F$$ is closed and $$x \notin F$$ then $$F\complement$$ is open and contains $$x$$.

As the clopen subsets form a base, there is a clopen set $$C$$ such that $$x \in C \subseteq F^\complement$$.

Then $$f=\chi_C$$ is continuous and $$f(x)=1$$ and $$f[F]=0$$.

• Thank you for your reply. I still struggle to see how $X \setminus F$, which is open, can be closed too. Since $X$ has a clopen basis, $X \setminus F$ can be written as a union of clopen sets.But since a union of closed sets is not necessarily closed, how do we ensure that such a union of clopen sets is also closed? – James May 11 '19 at 10:02
• Also how does having a clopen base ensure the existence of $C$? Well, we could take one member from the base that contains $x$ and such member will be clopen but I am not sure how it can be a subset of $X \setminus F$ – James May 11 '19 at 10:10
• @James all we need is one clopen subset containing $x$ and contained in $F^\complement$. And that’s what the base gives us. – Henno Brandsma May 11 '19 at 10:39
• @James $X\setminus F$ is a union of clopen sets. We pick. One that contains $x$. – Henno Brandsma May 11 '19 at 10:40
• Ah, I understand it now, thank you so much for your further explanation. – James May 11 '19 at 10:51

Given a closed set $$F$$ and a point $$x ∉ F$$, it is enough to have a clopen set $$C$$ separating them, i.e. $$x ∈ C$$ and $$C ∩ F = ∅$$. Then the characteristic funtion of $$C$$ also functionally separates $$x$$ and $$F$$.

• Thank you for your explanation. I suppose you meant to write $C \cap F = \emptyset$. My question is that is there such $C$ in a zero-dimensional space $X$? Well $X \setminus F$, which is open, might be a candidate but how do we know it is also closed? – James May 11 '19 at 9:43
• @James $X$ is zero-dimensional iff $X$ has a clopen basis. – YuiTo Cheng May 11 '19 at 9:52
• @YuiToCheng Thank you for your comment. I still struggle to see how $X \setminus F$, which is open, can be closed too. Since $X$ has a clopen basis, $X \setminus F$ can be written as a union of clopen sets.But since a union of closed sets is not necessarily closed, how do we ensure that such a union of clopen sets is also closed? – James May 11 '19 at 10:07