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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.

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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 $f[F]=0$.

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  • $\begingroup$ 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? $\endgroup$ – James May 11 '19 at 10:02
  • $\begingroup$ 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 $X \setminus F$ $\endgroup$ – James May 11 '19 at 10:10
  • $\begingroup$ @James all we need is one clopen subset containing $x$ and contained in $F^\complement$. And that’s what the base gives us. $\endgroup$ – Henno Brandsma May 11 '19 at 10:39
  • $\begingroup$ @James $X\setminus F$ is a union of clopen sets. We pick. One that contains $x$. $\endgroup$ – Henno Brandsma May 11 '19 at 10:40
  • $\begingroup$ Ah, I understand it now, thank you so much for your further explanation. $\endgroup$ – James May 11 '19 at 10:51
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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 $F$.

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    $\begingroup$ 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? $\endgroup$ – James May 11 '19 at 9:43
  • $\begingroup$ @James $X$ is zero-dimensional iff $X$ has a clopen basis. $\endgroup$ – YuiTo Cheng May 11 '19 at 9:52
  • $\begingroup$ @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? $\endgroup$ – James May 11 '19 at 10:07

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