TITLE: (Baby Rudin) Thm 2.34: Compact subsets of metric spaces are closed.
QUESTION [0 upvotes]: At the beginning of proof, $V_q$ should be $V_p$, so it seems to be just typo. Am I right? 
And I have understood up to the part $V=V_{q1} \cap \cdots \cap V_{qn}$ but cannot understand why the $V$ is a "neighborhood" of some $p$. Even though $V_{qi}$'s for $1\leq i \leq n$ are neighborhoods of $qi$'s, I thought the intersection of them doesn't have to be a neighborhood of some $p \in X $, $p \notin K$. 
I think the last part of proof should be like "If $V=V_{q1} \cap \cdots \cap V_{qn}$, then since $V$ is the intersection of the neighborhoods, which are open sets, $V$ is an open set, which consists of interior points. Because $V$ does not intersect $W$ by definition, $V \subset K^{c}$, so that every point in $V$ is an interior point of $K^{c}$. The Theorem follows."
Thank you for reading this!

REPLY [0 votes]: $V_q$ is a neighbourhood of $p$ (not a typo). We see from the letter $V$ that it is. But it does depend on $q$, as its radius is $\frac{d(p,q)}{2}$, which only is done to ensure that $V_q \cap W_q = \emptyset$ for all $q \in K$ (this is all we will need). So we know that the $V$ neighbourhood of $p$ and the corresponding $W_q$ neighbourhood (i.e. with the same subscript $q$) of $q$ are disjoint.
All of the sets $V_{q_1}, V_{q_2},\ldots, V_{q_n}$ are balls around $p$ (the fixed $p \notin K$ we start with!) of varying radii. So their intersection $V$ is also an open ball (of radius equal the minimum of these finitely many radii) and thus an open neighbourhood of $p$ as well.
If now $x \in K$, then $x \in W_{q_i}$ for some $q_i \in \{q_1, \ldots, q_n\}$ (these are finitely many distinct points of $K$ such that their corresponding $W_{q_i}$ open balls are a finite cover of $K$ (the existence of those is ensured by $K$ being compact) and then $V = V_{q_1} \cap \ldots V_{q_n} \subseteq V_{q_i}$ and $V_{q_i}$ and $W_{q_i}$ (same index point!) are disjoint so $x  \notin V$. As $x \in K$ was arbitrary, $K \cap V = \emptyset$.
So we see that each $p \notin K$ is an interior point of $K^c$, so $K^c$ is open and so $K$ is closed.
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