Theorem 2.34 Baby Rudin (Compact subsets of metric spaces are closed): How is his proof general enough?

Question

2.34 Theorem Compact subsets of metric spaces are closed.

Proof.

Let $K$ be a compact subset of metric space $X$. 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, there are finitely many points $q_1,...,q_n\in K$ such that $K\subset W_{q_1}\cup \cdots\cup W_{q_n}=W$.

If $V=V_{q_1}\cap \cdots \cap V_{q_n}$, then $V$ is a neighborhood of $p$ which does not intersect $W$. Hence $V\subset K^c$ so that $p$ is an interior point of $K^c$.

The way I understand it, he has just shown that it is possible to construct a finite open cover of $K$ such that $p$ is in interior point in $K^c$, but it does not show that all finite open covers of $K$ have this property.

Is this proof actually complete?

Answer 1

To prove $K^c$ is open, it suffices to show every point of $K^c$ is an interior point of $K^c$. This is what the proof has done: it started with an arbitrary point $p\in K^c$, and proved it is an interior point. So, the proof is complete. It is totally irrelevant whether something holds for "all finitie open covers of $K$" (and in any case I am not sure what it is you wish to hold) since that is not what is being proved.

The proof is a bit unclear about this setup, particularly in the line

Let $p\in K^c$, $q\in K$.

I would rephrase that instead as:

Let $p\in K^c$. Then for each $q\in K$, ...

This makes it clear that $p$ is fixed at the start and so the argument really does show an arbitrary $p\in K^c$ is an interior point.