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I need to show that: All zero dimensional spaces are completely regular.

Here are my definitions: Recall that a space is called zero dimensional if each point has a neighborhood base consisting of sets which are both open and closed. In particular the Michael line M, the Sorgenfrey line S, and the countable ordinals w_1 are all completely regular spaces. Note: We say that a point x ∈X has a neighborhood base at x if there is a collection {Uxj: j=1-> ∞} of open subsets of X such that every neighborhood W of x contains some Uxj. Definition of completely regular space A completely regular space is a topological space in which, for every point and a closed set not containing the point, there is a continuous function that has value 0 at the given point and value 1 at each point in the closed set.

I was thinking to prove it by contradiction. So here I go: Suppose X is not completely regular then there exists one point and every closed set not containing the point,s.t. there is not a continuous function that has value 0 at the given point and value 1 at each point in the closed set. then what?!

Please help.

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HINT: Don’t make it too hard; a direct proof is easy. If $H$ is a clopen subset of a space $X$, then the map

$$f:X\to\Bbb R:x\mapsto\begin{cases}0,&\text{if }x\in H\\1,&\text{if }x\in X\setminus H\end{cases}$$

is continuous. Prove this, and you’re almost done.

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  • $\begingroup$ Thank you. I am looking at it, trying to understand and see what you see. $\endgroup$ – Klara Apr 8 '13 at 23:54
  • $\begingroup$ Klara, he sees a computer screen and a keyboard. :-) $\endgroup$ – Asaf Karagila Apr 8 '13 at 23:56
  • $\begingroup$ @Klara: Do you know the result that a space $X$ is disconnected if and only if there is a continuous function from $X$ onto the two-point discrete space $\{0,1\}$? If you do, thinking about that might help you to see what I see (besides my screen and keyboard :-)). If not, ignore this comment. $\endgroup$ – Brian M. Scott Apr 8 '13 at 23:58
  • $\begingroup$ @brian no I do not know the result you mentioned and we passed disconnected spaces. Im going to check the book again. $\endgroup$ – Klara Apr 9 '13 at 0:06
  • $\begingroup$ I found that theorem, i dont remember it, but we might have gone through it a month or 2 ago. $\endgroup$ – Klara Apr 9 '13 at 0:09

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