2.34 Theorem Compact subsets of metric spaces are closed.


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?


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.

  • 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

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