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I have a question about the following proof by contradiction.

Let $(X,d)$ be a metric space. If $K \subset X$ is compact, then it is closed.

Proof. We show if $K$ is not closed, then it is not compact. If $K$ is not closed, then there is $p \in \overline{K} \setminus K$. From $p \in \overline{K}$, we know for every $r>0$, $B_{r}(p) \cap K \ne \emptyset$. Consider for $n \in \mathbb{N}$

\begin{align*} U_{n} = (\overline{B}_{\frac{1}{n}}(p))' \end{align*}

then \begin{align*} \bigcup U_{n} = \bigg(\bigcap_{n \in \mathbb{N}} \overline{B}_{\frac{1}{n}}(p)\bigg) \end{align*}

So

\begin{align*} \bigcup_{n \in \mathbb{N}} U_{n} = X \setminus \{p\} \supset K \end{align*}

and we have an open cover $(U_{n})_{n \in \mathbb{N}}$ of $K$.

If $K$ were compact, we would have a finite subcover $(U_{n})_{n \in F}$, $|F| < \infty$, and we would have $N = \max \{n: n \in F \}$ with $U_{N} = \bigcup_{n \in F} U_{n}$

We see \begin{align*} B_{\frac{1}{N}}(p) &\subset \overline{B}_{\frac{1}{N}}(p) =(U_{N})' \end{align*}

This gives us a contradiction between subcover property \begin{align*} K \subset U_{N} \end{align*}

I was wondering if the contradiction has to do with $B_{r}(p) \cap K \ne \emptyset$ since there exists at least one element of $K$ in $B_{r}(p)$, but every element in $B_{\frac{1}{N}}(p)$ is an element of $(U_{N})'$. Then this implies that an element of $K$ is in $(U_{N})'$, which is the contradiction?

Skm
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1 Answers1

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The contradiction lies in that there is some $x \in K \cap B_{\frac1N}(p)$ as $p \in \overline{K}$. This $x$ does not lie in $U_N$ by definition, so $K \subseteq U_N$ does not hold, as $x \in K$ but $x \notin U_N$.

The proof could also be done following the general topology fact that $X$ is Hausdorff: let $p \notin K$. Then for each $x \in K$ we pick $r_x >0$ such that

$$B(x,r_x) \cap B(p, r_x) = \emptyset$$ (e.g. $r_x = \frac{d(x,p)}{2} >0$ will work)

And the $\{B(x,r_x): x \in K\}$ cover $K$, so finitely many do by compactness, say all balls $B(x,r_x)$ for $x \in F \subseteq K$, where $F$ is finite. Then check that $B(p, r) \cap K = \emptyset$ for $r=\min(r_x: x \in F\} >0$ (as we take a minimum of finitely many radii), and so $p \notin \overline{K}$.

Or use sequential compactness instead, if you've covered that...

Henno Brandsma
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