# Proof of “every countably compact metric space is totally bounded”.

My book has a proof that has some confusing definitions involved. And searching online only got me a proof for "compact spaces" rather than "countably compact spaces". So I've tried to construct a proof fishing points from here and there. Any help would be greaty appreciated.

Let us assume: $X$ is an uncountable countably compact space, and $\epsilon\in\Bbb{R}$ is the number for which $X$ is not totally bounded.

Take a point $x_{1}\in X$. $B(x_{1},\epsilon)$ does not cover $X$; i.e. there is a $x_{2}\in X$ such that $x_{2}\notin B(x_{1},\epsilon)$. Take $B(x_{1},\epsilon)\bigcup B(x_{2},\epsilon)$. There is a $x_{3}$ such that $x_{3}\notin B(x_{1},\epsilon)\bigcup B(x_{2},\epsilon)$. In this way, construct $x_{n+1}\notin \bigcup_{i=1}^{n}B(x_{i},\epsilon)$.

Clearly, $d(x_{i},x_{j})\geq\epsilon$, where $i\neq j$. But if we select a denumerable number of $B(x_{i},\epsilon)$ in the above construction, it will have a finite subcover. Hence, some $B(x_{i},\epsilon)$ will contain a $x_{j}$, where $i,j\in\Bbb{N}, i\neq j$, contradicting the fact that $d(x_{i},x_{j})\geq\epsilon$, for $i\neq j$. This shows that we can't select a denumerable number of such points from $X$. And if we can't select a denumerable number of such points, we definitely can't select an uncountable number of such points (as $\mathfrak{N_{0}}<c$). This shows that there are a finite number of points $x_{i}$ such that $\bigcup B(x_{i},\epsilon)$ covers $X$.

Is this reasoning correct?

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You don't need that $X$ is uncountable at all. It just muddies the argument, IMO. If you could construct an infinite sequence of $x_i$ in that way, you'd have a sequence in $X$ without cluster point. Hence $X$ isn't countably compact. – Daniel Fischer Jul 7 '13 at 18:39

There is no reason to assume that $X$ is uncountable: the cardinality of $X$ is irrelevant to the result and should play no rôle in the proof.

Just assume that there is some $\epsilon>0$ such that no finite set of $\epsilon$-balls covers $X$. Then carry out the recursive construction of $\{x_n:n\in\Bbb N\}$ as you did, always choosing $x_{n+1}\in X\setminus\bigcup_{k\le n}B(x_k,\epsilon)$. Let $D=\{x_n:n\in\Bbb N\}$, and let $U=X\setminus D$.

Suppose that $x\in U$. If there is an $n\in\Bbb N$ such that $x\in B(x_n,\epsilon)$, then $B(x_n,\epsilon)\setminus\{x_n\}$ is an open nbhd of $x$ disjoint from $D$. If there is no such $n$, then $d(x,x_n)\ge\epsilon$ for all $n\in\Bbb N$, and $B(x,\epsilon)$ is an open nbhd of $x$ disjoint from $D$. In all cases, then $x$ has an open nbhd contained in $U$, and $U$ is therefore open.

Let $\mathscr{U}=\{U\}\cup\{B(x_n,\epsilon):n\in\Bbb N\}$; clearly $\mathscr{U}$ is a countable open cover of $X$. Suppose that $\mathscr{V}\subseteq\mathscr{U}$ is finite. Then there is an $n\in\Bbb N$ such that $B(x_n,\epsilon)\notin\mathscr{V}$. And $B(x_n,\epsilon)$ is easily seen to be the only member of $\mathscr{U}$ containing $x_n$, so $x_n\notin\bigcup\mathscr{V}$, and $\mathscr{V}$ does not cover $X$. Thus, $\mathscr{U}$ has no finite subcover, and $X$ is not countably compact. It follows that every countably compact metric space is totally bounded.

(Another way to look at the argument is to note that we’ve shown that if $X$ is not totally bounded, it contains a countably infinite closed discrete subset: that’s the set $D$ above. The last paragraph of the proof is just the proof that no space containing a countably infinite closed discrete subset can be countably compact.)

Your argument does in fact have a significant hole: $\{B(x_n,\epsilon):n\in\Bbb N\}$ need not cover $X$, so you can’t apply countable compactness to conclude that some finite subset covers $X$. You really do need to show that what I called $D$ is closed, so that you can use $X\setminus D$ to cover everything that’s not covered by $\{B(x_n,\epsilon):n\in\Bbb N\}$.

There’s also no need for the sentence that begins ‘And if we can’t select ...’. First, your construction as given produces at most a countably infinite set of points $x_n$, though extending it to a transfinite recursion would be easy enough. More important, there’s no reason to extend it any further: once you have the countably infinite set $D$, you have what you need to show that $X$ can’t be countably compact.

I just realized that you may have been thinking of extending the recursive construction until it can’t be extended any further. In that case you have to use transfinite recursion or Zorn’s lemma or the like, but maximality of the set of $\epsilon$-balls will indeed allow you to conclude that the balls cover $X$, without any need to throw in an extra open set.

If you know about infinite ordinals, you can proceed as follows. Given an ordinal $\eta$ and the points $x_\xi$ for $\xi<\eta$, if $\bigcup_{\xi<\eta}B(x_\xi,\epsilon)=X$, stop; otherwise, choose $x_\eta\in X\setminus\bigcup_{\xi<\eta}B(x_\xi,\epsilon)$. When the construction stops at some stage $\eta$, let $D=\{x_\xi:\xi<\eta\}$. Then $\{B(x_\xi,\epsilon):\xi<\eta\}$ really will be an open cover of $X$, and there won’t be any need to drag in $X\setminus D$ to fill out the cover. However, you have to be a bit more careful in finishing off the argument:

If $\eta$ is finite, we’re done: $\{x_\xi:\xi<\eta\}$ is an $\epsilon$-net for $X$. Otherwise let $U=\bigcup_{\omega\le\xi<\eta}B(x_\xi,\epsilon)$, and let $\mathscr{U}=\{U\}\cup\{B(x_n,\epsilon):n<\omega\}$. Then $\mathscr{U}$ is a countable open cover of $X$ with no finite subcover, contradicting countable compactness of $X$. (The argument that $\mathscr{U}$ has no finite subcover is exactly as in my proof above.)

If you don’t know about infinite ordinals but are familiar with Zorn’s lemma, you can use it instead. Say that $A\subseteq X$ is a partial $\epsilon$-net if $B(x,\epsilon)\cap A=\{x\}$ for each $x\in A$. (Equivalently, if $x,y\in A$ and $x\ne y$, then $d(x,y)\ge\epsilon$.) Let $\mathscr{D}$ be the family of all partial $\epsilon$-nets in $X$. If $\mathscr{C}$ is a chain in the partial order $\langle\mathscr{D},\subseteq\rangle$, then $\bigcup\mathscr{C}\in\mathscr{D}$, so every chain in $\mathscr{D}$ has an upper bound in $\mathscr{D}$, and by Zorn’s lemma $\mathscr{D}$ has a maximal element $D$. The maximality of $D$ implies that $\bigcup_{x\in D}B(x,\epsilon)=X$. If $D$ is finite, we’re done: $D$ is the desired $\epsilon$-net. If $D$ is infinite, let $D_0$ be a countably infinite subset of $D$, and let $U=\bigcup_{x\in D\setminus D_0}B(x,\epsilon)$. Let $\mathscr{U}=\{U\}\cup\{B(x,\epsilon):x\in D_0\}$, and argue as before that $\mathscr{U}$ is a countable open cover of $X$ with no finite subcover.

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Brian, in the first part of your answer, isn't your contradiction based on the fact that you can't select a non-finite bunch of points $\{x_{n}\},n\in\Bbb{N}$ such that $x_{i+1}\notin \bigcup_{n\leq i}B(x_{n},\epsilon)$? I was hoping to emphasize this point by assuming an uncountable space. – Ayush Khaitan Jul 8 '13 at 3:21
I understand that without mention of the cardinality of the space, we're making a statement about both finite and non-finite spaces. If the number of points in $D$ is finite, then we already have a finite cover. – Ayush Khaitan Jul 8 '13 at 3:23
@Ayush: No, it’s not, because I’m proving the contrapositive: if $X$ is not totally bounded, then $X$ is not countably compact. I do construct the set $D$, and it is infinite. // But the argument that I gave is independent of the cardinality of the space. There’s no reason to complicate matters by introducing unnecessary case distinctions. – Brian M. Scott Jul 8 '13 at 3:24
@Ayush: If you do cast it as a proof by contradiction, you still don’t have to deal with a countable/uncountable distinction. – Brian M. Scott Jul 8 '13 at 3:27
And you're right, I haven't assumed that $\bigcup_{i=1}^{n}B(x_{i},\epsilon)$ covers $X$. I have just said that the open set $\bigcup_{i=1}^{n}B(x_{i},\epsilon)$ can't be covered by a finite subcover if $D$ is denumerable, which is a contradiction as the space is countably compact. – Ayush Khaitan Jul 8 '13 at 3:27

There's one small detail, just for the sake of the proof being precise. The set of theese $\epsilon$-balls might not cover the whole $X$, so you have to add $U = X\backslash \{x_1,...x_n,...\}$ to your cover. The set $U$ is open, since if you take any $x\in U$ and consider $B(x,{\epsilon}{2})$, the latter set has at most one point in common with $\{x_1,...x_n,...\}$ - say $x_{i_0}$ (it can't have two points, for example, since that would imply that the distance between them, by the triangle inequality, is not greater than $\epsilon$). So now consider $B(x,d(x,x_{i_0}))$. This ball contains no points from $\{x_1,...x_n,...\}$. So $U$ is indeed open. Now you proceed with the countable cover $\{U,B(x_i,\epsilon)|i\in \mathbb{N}\}$... Perhaps my arguments were a kind of a mess, but I hope that they'll be of some use to you...

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