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I'm having difficulty following this proof and was hoping someone could help give a clear picture of what Rudin is doing.

pf If no point of $K$ were a limit point of $E$, then each $q \in K$ would have a neighborhood $V_{q}$ which contains at most one point of $E$ (namely, q, if $q \in E$). It is clear that no finite subcollection of $\{V_{q}\}$ can cover $E$; and the same is true of $K$, since $E \subset K$. This contradicts the compactness of $K$.

I'm not exactly understanding the construction of $V_{q}$... if we have $q \in E$ then is $V_{q}$ by definition not a neighborhood by the construction of $V_{q}$? Ie. if $q' \in E$ and every point around it is in $E$, then $V_{q'}$ is not a neighborhood, but just a point?

zzz2991
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    The "at most one point" can be a little confusing. It'd be enough to remark that it would cotain at most a finite number of points of $;E;$. Now, what's the definition of limit point? Well: $;q\in K;$ is a limit point of $;E;$ if for any neighborhood $;V_q\subset K;$ of $;q;$ there exists $;e_q\in E;;s.t.;;e_q\in V_q;$ . Now, negate this definition...and you have what's described in the above proof. – DonAntonio May 31 '14 at 17:40
  • I think I see. There does not exist a point $q\in K$ which is a limit point of $E$. Then every $q$ is $isolated$ from $E$. Then, constructing $V_{q}$ to be neighborhoods of every $q$ such that at most one point is touched by $V_{q}$, namely $q$ if $q \in E$. Then ${V_{q}}$ is an open covering of $E$. But clearly there is no finite subcollection which covers $E$, given $E$ is infinite. $$.$$ I think it was my understanding of the negation of limit point (isolation point). So around every point we can create a neighborhood which contain no other points in $E$ not equal to itself. – zzz2991 May 31 '14 at 18:26
  • Still trying to get a grasp on the definitions... Thanks! – zzz2991 May 31 '14 at 18:30
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    @David Jhoo: Yes, this is a correct reasoning. You can now answer your own question and accept the answer. – Moishe Kohan May 31 '14 at 20:14

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