A compact subset $A\subseteq X$ of a metric space $X$ is closed and bounded. The contrary does not hold.

I want to proof this statement. Showing that $A$ is closed, is not hard.

A is closed iff $A^c$ is open. Let $\{U_i\}$ be an open cover of $A$. Then there exists a finite set of indices $J$ such that $\bigcup_{i\in J} U_i = A$.

Now we have that $A^c=X\setminus \bigcup_{i\in J} U_i=X\cap\bigcap_{i\in J} U_i^c$ what is closed.

Now I want to show that $A$ is bounded. So I have to show that it exists $x\in X$ and a finite $r>0$ such that for every $a\in A$ it is $d(x,a)<r$.

The definition is taken from here: https://en.wikipedia.org/wiki/Bounded_set#Metric_space

Well, A is compact, so I take for every $a\in A$ an $\varepsilon_a>0$. Then $\bigcup_{a\in A} B_{\varepsilon_a(a)}$ is an open cover of A, and thus has a finite supcover. So let $B_{\varepsilon_{a_1}}(a_1),\dotso, B_{\varepsilon_{a_n}}(a_n)$ be this subcover.

Then $A=\bigcup_{i=1}^n B_{\varepsilon_{a_i}}(a_i)$ and $0<r:=\sum_{i=1}^n \varepsilon_{a_i}<\infty$

Now I set $x=a_1$. Then $d(a_1,a)<r$ for every $a\in A$, since there is an ball which contains $a$ and thus $d(a_1,a)<r$ should trivially hold.

Unfortunatly I can not give a crystal clear proof for this, as it seems obvious. The triangle inequality should do the trick, but I just can not figure out a good way to proof it rigorously.

For a counterexample I took simply $(\mathbb{Q},|\cdot|)$

Then $M:=[0,1]\cap\mathbb{Q}$ is closed and bounded, but not compact.

If we take for every $q\in M$ an $\varepsilon_q>0$, then $\bigcup_{q\in M} B_{\varepsilon_q}(q)$ is an open cover, but has not a finite subcover, since $\mathbb{Q}$ is dense in $\mathbb{R}$, so there are always irrational points contained in the subcover and there can not be equality of sets.

Thanks in advance.

  • 1
    $\begingroup$ "$A$ is closed iff $A^c$ is bounded." Absolutely not. $\endgroup$
    – Ivo Terek
    Nov 10 '19 at 23:37
  • 2
    $\begingroup$ @IvoTerek Thanks for pointing that out. It is a typo. I mean $A$ is closed iff $A^c$ is open. $\endgroup$
    – Cornman
    Nov 10 '19 at 23:38
  • 1
    $\begingroup$ Oh, I just noticed, that I did not even showed that $A$ is closed... Wait what... $\endgroup$
    – Cornman
    Nov 10 '19 at 23:43

Suppose that $A$ is compact.

  • $A$ is closed because $A^c$ is open. The reason is as follows: if $x \not\in A$, for every $a \in A$ there is $r_a>0$ such that $B(a,r_a)\cap B(x,r_a) = \varnothing$. So $\{B(a,r_a)\}_{a \in A}$ is an open cover of $A$. Extract a finite subcover $\{B(a_i,r_{a_i})\}_{i=1}^n$. Then $x \in \bigcap_{i=1}^n B(x,r_{a_i}) \subseteq A^c$, and this intersection is open. This shows that $A^c$ is open.

  • $A$ is bounded: take a finite subcover of $\{B(x,n)\}_{n \geq 1}$, where $x \in X$ is any point chosen a priori. The union of this finite subcover is in fact one of the balls $B(x,N)$ for some $N \geq 1$, meaning that $A\subseteq B(x,N)$ is bounded.

For your counter-example, you have to look at $[0,1]\cap \mathbb{Q}$ instead of $[0,1]$, as $[0,1]$ is not a subset of $\mathbb{Q}$. The correct way of saying "in $\mathbb{Q}$" is writing this intersection.

  • $\begingroup$ Thanks. The $\subseteq$ was a typo too... I meant $\cap$ lol. I think I should go to sleep. xD $\endgroup$
    – Cornman
    Nov 10 '19 at 23:57

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