1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Eric.Nice. Rudin's $q \in K, $ I would read as : and let q be in K.(Nitpicking:)) Greetings. $\endgroup$ – Peter Szilas Feb 18 at 21:37
  • $\begingroup$ 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 $K_c$? $\endgroup$ – paul Feb 18 at 23:05
  • $\begingroup$ @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 $p$. $\endgroup$ – Eric Wofsey Feb 18 at 23:12
  • $\begingroup$ 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 $K$? $\endgroup$ – paul Feb 19 at 12:27
  • $\begingroup$ No, the step "there are finitely many points..." may fail if $K$ is not compact. $\endgroup$ – Eric Wofsey Feb 19 at 15:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.