Theorem 2.34 Compact subsets of a metric space are closed.
Proof.
Suppose $K\subseteq X$, $K$ compact. Let $p\in K^c$, $q\in K$. Let $V_q,W_q$ be neighborhoods of $p$ and $q$ with radius less than $\frac 1 2 d(p,q)$.
Since $K$ is compact, we have $K\subseteq W_{q_1}\cup \cdots\cup W_{q_n}=W$ for some $q_1,…,q_n\in K$.
If V=$V_{q_1}\cap \cdots V_{q_n}$, $V$ is a neighborhood of p which does not intersect $W$, then $V\subseteq K^c$ so $p$ is an interior point of $K^c$. QED.
My understading of this proof is that $V$ is actually
$$V= \text {'$\min$'}\left\{V_{q_i}\right\}$$
Where the minimum is to be taken in terms of inclusion.
Also, we have that $V$ doesn't intersect $W$ because $V_{q_i}$ doesn't intersect $W_{q_i}$ for every $i$, right (How do I prove that)?
I'm uploading a picture of my understanding of the situation, am I understanding this correctly?
Best Answer
Since each $V_{q}$ is a neighborhood of $p$, any finite intersection of them is again a neighborhood of $p$; yes, $V=V_{q_1}\cap\dots\cap V_{q_n}$ is a ball with the radius which is the minimum, but it's mostly irrelevant: the important fact is that it is a neighborhood of $p$.
Now, since $V\subseteq V_{q_i}$, we have by construction that $$ V\cap W_{q_i}\subseteq V_{q_i}\cap W_{q_i}=\emptyset $$ and therefore $$ V\cap W=V\cap(W_{q_1}\cup\dots\cup W_{q_n})= (V\cap W_{q_1})\cup\dots\cup(V\cap W_{q_n})=\emptyset $$ Hence $V\subseteq W^c\subseteq K^c$.