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

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?

## 1 Answer

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.

• Eric.Nice. Rudin's $q \in K,$ I would read as : and let q be in K.(Nitpicking:)) Greetings. – Peter Szilas Feb 18 at 21:37
• The issue that I have is that I cannot see how this proof shows that if $p\in K^c$ then $p$ is an internal point. I could have equally constructed an open cover of $K$, say, $\{X\}$, so that $W=X$, and then in this case, $V$ does intersect $W$, so we cannot conclude that $V\subset K^c$. Why is it that by finding a set of $W$ that works, we can say that this $p$ is an interior point of $K_c$? – paul Feb 18 at 23:05
• @paul: By definition, $p$ is an interior point of $K^c$ if there exists a neighborhood $V$ of $p$ such that $V\subseteq K^c$. We don't need it to be true for every neighborhood of $p$. – Eric Wofsey Feb 18 at 23:12
• Thanks for your response. I’ve been thinking about it overnight, and perhaps my question can be answered if I phrased it this way. Let’s assume that $K$ is a set that is not compact. Then, I cannot guarantee that every of $K$ contains a finite subcover of $K$. Am I not able to go through each line of reasoning, and still conclude that I can find a neighbourhood of $p$ that is interior to the complement of $K$? – paul Feb 19 at 12:27
• No, the step "there are finitely many points..." may fail if $K$ is not compact. – Eric Wofsey Feb 19 at 15:39