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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 $\omega_1$ are all completely regular spaces. Note: We say that a point $x\in X$ has a neighborhood base at $x$ if there is a collection $\{U_j\}$ of open subsets of $X$ such that $x\in U_j$ for each $j$ and every neighborhood $W$ of $x$ contains some $U_j$. 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.

Tyrone
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Klara
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1 Answers1

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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.

Brian M. Scott
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  • Thank you. I am looking at it, trying to understand and see what you see. – Klara Apr 08 '13 at 23:54
  • Klara, he sees a computer screen and a keyboard. :-) – Asaf Karagila Apr 08 '13 at 23:56
  • @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. – Brian M. Scott Apr 08 '13 at 23:58
  • @brian no I do not know the result you mentioned and we passed disconnected spaces. Im going to check the book again. – Klara Apr 09 '13 at 00:06
  • I found that theorem, i dont remember it, but we might have gone through it a month or 2 ago. – Klara Apr 09 '13 at 00:09
  • @ Brian so are you thinking that H. X\H forms a disconnection in X, since they are nonempty open disjoint subsets of X s. t X=U(H,X\H). – Klara Apr 09 '13 at 00:13
  • @Klara: That’s right. So you can use the older result, or you can prove directly that $f$ is continuous: just check that for any open $U$ in $\Bbb R$, $f^{-1}[U]$ is open in $X$. This is easy, because $f^{-1}[U]$ is always one of four sets: $\varnothing$ (if $0,1\notin U$), $H$, $X\setminus H$, or $X$ itself. – Brian M. Scott Apr 09 '13 at 00:17
  • Brian thank you so much! – Klara Apr 09 '13 at 01:06
  • You’re welcome, @Klara! – Brian M. Scott Apr 09 '13 at 01:09