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Let $M$ be a metric space that is locally compact. Let $O_i \subset M$. Let $C$ be an open cover of $O_i$, and let $C'\subset C$. Define $U \subset O_i$ to be an open neighborhood of some $x\in O_i$ such that there exists $\epsilon$ with the property that if $y$ with $d(x,y)<\epsilon$, then $y \in U$. Define $B_r(x)$ and $B_r(y)$ for some $r>0$ to be the open balls about $x$ and $y$ respectively, and let $B_r(x),B_r(y)\subset U$. Now, $C' = \bigcup B_r$ is countable, and covers $O_i$. Can we then have $C=\bigcup U$ to cover $O_i$, and hence have $O_i$ be compact?

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    $\begingroup$ Is your metric space countable? You write $x_i$, and in your union it is implicit that the $U_i$ are countable. $\endgroup$ – Adam Saltz Apr 2 '13 at 17:09
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    $\begingroup$ $M$ is a neighbourhood of each $x \in M$. (There is some ugliness with the empty metric space, of course). As phrased your problem is trivially solved by taking $U=M$ for each $x$ in $M$. $\endgroup$ – kahen Apr 2 '13 at 17:13
  • $\begingroup$ @kahen I modified the answer as to define $U \subset M$, to avoid triviality. $\endgroup$ – Othmar Reinhard Apr 2 '13 at 17:26
  • $\begingroup$ In response to your latest edit: I see you can make open covers of $M$, but I see nothing in your question that hints at how you want to extract finite subcovers. Might I suggest you watch Professor Su's presentation of compactness on Youtube. Lectures 11, 12 and 13. If I recall correctly, he's teaching from "Baby Rudin". $\endgroup$ – kahen Apr 2 '13 at 17:33
  • $\begingroup$ Thanks for your concern. Misunderstanding arises from the fact that I did not make myself clear. I tried to fix that be writing the question from the beginning. $\endgroup$ – Othmar Reinhard Apr 2 '13 at 18:01
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for all x in M the open ball B(x,1) is an open neighborhood

B(x,1) contains B(x,r) for all r<1

and the union of B(x,1) cover M

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  • $\begingroup$ Wouldn't the unions of $B(x,r)$ also cover $M$? $\endgroup$ – Othmar Reinhard Apr 2 '13 at 17:21
  • $\begingroup$ yes of course it cover M $\endgroup$ – masmoudihoussem Apr 2 '13 at 17:23
  • $\begingroup$ But, $B(x,r) \subset B(x,1)$, and so $M$ is compact. $\endgroup$ – Othmar Reinhard Apr 2 '13 at 17:25
  • $\begingroup$ @OthmarReinhard why does $B(x,r) \subset B(x,1)$ make you think that $M$ should be compact? That makes no sense. $\endgroup$ – kahen Apr 2 '13 at 17:27
  • $\begingroup$ NO, M must be covered by finite number of neighborhoods $\endgroup$ – masmoudihoussem Apr 2 '13 at 17:28
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(Original answer no longer relevant after OP's edit.)

I'm no longer sure what you're asking, now. Let me go through and point out some issues that need to be dealt with, step by step.

  1. You've fixed some subset $O_i$ of $M$. I'm not sure why you bothered to index it, since you've chosen only one subset of $M$. Is this part of some larger problem? If so, then providing the full context would probably clear up a lot of confusion.
  2. You've then chosen an open cover $C$ of $O_i$ (apparently so that you can show that $O_i$ is compact), followed by fixing an arbitrary subcollection $C'$ of $C$. We'll address this again, later.
  3. Now, you've taken a subset $U$ of $O_i$ that is an open neighborhood of some $x\in O_i$. Is $U$ open in $M$, or just open in $O_i$?
  4. You've further required that there is some $\epsilon>0$ such that $B_\epsilon(x)$ is contained in $U$. This requirement is redundant if $U$ is open in $M$, so are you specifying it because $U$ is only open in $O_i$ and not in $M$?
  5. Next, you've apparently taken some arbitrary $y\in U$ (I can't tell why), and found $r>0$ such that both $B_r(x),B_r(y)$ are contained in $U$.
  6. At this point, you seem to be saying that $C'$ is a union of some unspecified collection of open balls of radius $r$, which would mean that $C'$ is some open subset of $M$. But you already declared earlier that $C'$ is a collection of open subsets of $M$. Which is it? Why do you then assert that it is countable? I can't see any reason for it, regardless of whether it is an arbitrary subcollection of $C$ or whether it is an open subset of $M$.
  7. Now, if $C'$ is a subset of $M$, then we say it contains $O_i$ if and only if $O_i\subset C'$. If $C'$ is a collection of subsets of $M$, then we say it covers $O_i$ if and only if $O_i\subset\bigcup C'$. If it is an arbitrary subcollection of $C$, then there's no reason to suspect that it should cover $O_i$--after all, $C'$ could be empty. If $C'$ is meant to be a subset of $M$, then it might well contain $O_i$, but it isn't really made clear by your arguments so far that it does.

If you will edit your question to fix these issues (especially the lack of context issue), I would be glad to take another crack at answering your question.

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  • $\begingroup$ I want the open neighborhoods to contain the open balls $r>0$ about $x$, so that the metric space is compact. $\endgroup$ – Othmar Reinhard Apr 2 '13 at 17:17
  • $\begingroup$ @OthmarReinhard I don't see what compactness has to do with your question as written other than you can produce an open cover. But as I commented, that's trivial: $\{M\}$ covers $M$ and that tells us nothing about compactness. Consider editing it again and be more specific. $\endgroup$ – kahen Apr 2 '13 at 17:21
  • $\begingroup$ @Othmar: In that case, you can take each $U_i$ to be a ball about $x_i$ of radius at least $r$, and you're fine. Alternately, kahen's suggestion is simpler yet. What does compactness have to do with your post? $\endgroup$ – Cameron Buie Apr 2 '13 at 17:26

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