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What I wanna ask is what is the use of $V=$ interception of all neighborhood of some point $p$? It make me hard to understand the proofs.

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Note that more or less the same proof works in any Hausdorff topological space. –  M Turgeon Sep 19 '12 at 14:59
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Since by construction $V_{q_i}$ and $W_{q_i}$ are disjoint, it follows that the intersection of all $V_{q_i}$ is disjoint to the union of all $W_{q_i}$ (and hence to $K$): A point that is in the intersection is in all of the $V_{q_i}$ hence in none of the $W_{q_i}$, hence not in the union of the $W_{q_i}$.

Note that compactness (i.e. the fact that you need only finitely many $q_i$) is used to show that the intersection $V$ is open: The intersection of finitely many open sets is open.

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why the proof doesn't consider union rather? Would it be different? –  Mathematics Sep 19 '12 at 14:49
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It would not work. $V_{q_1}$ is guaranteed not to contaiun $q_i$ but may well intersect $K$. Even worse for the union. And you can't take anything but the intersection o fthe $V_{q_i}$ because the definition of compactness forces you to use the union o fthe $W_{q_i}$. –  Hagen von Eitzen Sep 19 '12 at 14:50
    
The union of the $V_{q_i}$ is guaranteed to be disjoint from $W$ only if you let $W=\bigcap W_{q_i}$, which is probably much smaller than $K$. Recall deMorgan's law. –  Hagen von Eitzen Sep 19 '12 at 14:53
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In this question, it is quite sensible to discuss V. We assume that we fixed a point p which is not in K, and for each q in K, we define a new open ball and hence the interception is a smaller neighborhood of p which doesn't intercept with W (Noted that V is a smaller neighborhood among the neighborhood set $V_{q1}...$ ) and hence for any p there always exist a open ball. Comment me if there are anything wrong

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i think i made some mistake, it should be fine now. –  ABC Sep 20 '12 at 7:54
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