0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $f[F]=0$.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $F$.

$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.