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Let me ask the following question: How Rudin applies Theorem 2.7 in the begining? He take some $x\in K$ then $x\in V_i$ where $i=i(x)$. What's next?

I thought on this about couple hours but no results.

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    $\begingroup$ He is taking $\{x\}$ to be the compact set of theorem $2.7$. Then $x \in V_i$ for some $i$. Therefore the $W_x$ exists. $\endgroup$
    – Maik Pickl
    Jun 19, 2016 at 13:03
  • $\begingroup$ @MaikPickl, Precisely $\{x\}\subset V_i$ where $\{x\}$ is compact since it's a one-point set. Using theorem 2.7 we get that exists neighborhood $W_x$ with compact closure such that $\{x\}\subset W_x \subset \overline W_x\subset V_i$. Right? $\endgroup$
    – ZFR
    Jun 19, 2016 at 13:06
  • $\begingroup$ That's the idea. You could argue that the notation is misleading since the $K$ in 2.13 is not the $K$ he is reffering to in 2.7., but after all it's just notation. :) $\endgroup$
    – Maik Pickl
    Jun 19, 2016 at 13:08
  • $\begingroup$ @MaikPickl, Yes you're right! But you can write you comment as answer! I'd accepted it as the best answer $\endgroup$
    – ZFR
    Jun 19, 2016 at 13:11

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Allright, just so the question doesn't appear as unanswered: He is taking $\{x\}$ as the compact set of theorem 2.7., since $x \in V_i$ for some $i$ you have $\{x\}\subset V_i$. Since $\{x\}$ is compact theorem 2.7. applies and you find an open $W_x$ with $\{x\} \subset W_x \subset \overline{W_x} \subset V_i$ where $\overline{W_x}$ is compact.

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  • $\begingroup$ That's better :) Thanks for answer! $\endgroup$
    – ZFR
    Jun 19, 2016 at 13:18

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