Show that every compact metrizable space has a countable basis.

My try:

Let $X$ be a compact space and metrizable. Now for each $n\in \Bbb N$; I can consider the open cover $\{B(x,\frac{1}{n}):x\in X\}$ of $X$ .As $X$ is compact we can find finite number of $x_i;1\leq i\le n$ corresponding to each $n$ .

$B_n=\{B(x_i,\frac{1}{n});1\leq i\le n\}$ .Now $\Bbb B=\{B_n:n\in \Bbb N\}$ is a countable collection .

It remains to show that $\Bbb B$ is a basis of $X$.

Let $U$ be an open set in $X$.Let $x\in U\implies \exists r>0$ such that $B(x,r)\in U$.Then we have for some $n;\frac{1}{n}<r\implies x\in B(x,\frac{1}{n})\subset B(x,r)\subset U$

The problem is I can't show the existence of a member of $\Bbb B$ say $B_n$ such that $x\in B_n\subset U$.

Can someone please help to complete the above proof ? I will be grateful if done so.

  • 4
    $\begingroup$ "As $X$ is compact we can find finite number of $x_i;1\leq i\le n$ corresponding to each $n$ ." I don't quite follow why $1\le i\le n$. I think it's only reasonable to enumerate these representatives as $$x_{n1},x_{n2},\cdots,x_{nk_n}$$ since the finite cover corresponding to $n$ depends on $n$. You need to change your description of $\Bbb B$. $\endgroup$
    – Vim
    Feb 6, 2016 at 6:02
  • $\begingroup$ isn't $\displaystyle U \subset B_n$ ? $\endgroup$
    – reuns
    Feb 6, 2016 at 6:05
  • $\begingroup$ Okay Vim but I don't think that's going to affect the proof much;though technically you are right $\endgroup$
    – Learnmore
    Feb 6, 2016 at 6:31
  • $\begingroup$ @learnmore Since you are probably the user who created (countability-axioms) tag, I thought it might be useful to let you know that there is an ongoing discussion on meta related to this tag. $\endgroup$ Dec 9, 2016 at 12:42
  • $\begingroup$ @MartinSleziak;Thank you very much;I did have a look at it $\endgroup$
    – Learnmore
    Dec 9, 2016 at 13:58

4 Answers 4


You have pretty much got it though there are some minor errors in your proof.

For each $n$ we consider the open cover $C_n=\{B(x, 1/n):\ x\in X\}$ of $X$. By compactness, there is a finite subset $X_n$ of $X$ such that $B_n:=\{B(x, 1/n):\ x\in X_n\}$ covers $X$.
(Here you made a mistake in your proof by implicitly assuming that $X_n$ can be chosen to have cardinality $n$).

Now we define $B=\bigcup_{n\in \mathbf N}B_n$.
(Here too there is a minor error in your proof. You define $B=\{B_n:n\in \mathbf N\}$. Note that the way I defined $B$ is different than yours.)

We want to show that $B$ forms a basis of $X$.

To show this, it suffices to show that every open set in $X$ can be written as a union of some members of $B$. So let $U$ be an open set in $X$ and $x\in U$ be chosen arbitrarily.

Then let $k>0$ be an integer such that $B(x, 1/k)$ is contained in $U$. Now by our construction, we know that $B_{2k}$ is an open cover of $X$. So some member of $B_{2k}$ contains $x$. There there is $x_0\in X_{2k}$ be such that $x\in B(x_0, 1/2k)$.

By triangle inequality, we have $B(x_0, 1/2k)\subseteq B(x, 1/k)\subseteq U$.

Thus we have found a member $B(x_0, 1/2k)$ of $B$ which contains $x$ and is contained in $U$, showing that $U$ can be written as a union of some of the members of $B$ and we are done.

  • 1
    $\begingroup$ Towards the ending, was finding $B(x_0, 1/2k)$ really necessary? Doesn't $B(x, 1/k)$ give us what we want since it belongs to our basis $B$? $\endgroup$
    – John11
    Jun 4, 2018 at 14:36
  • 2
    $\begingroup$ The point $x$ may not be in $X_k$. Do you see the point of going to $1/2k$ now? $\endgroup$ Jun 5, 2018 at 3:50

You need to redefine your $\Bbb B$.

For each $n\in \Bbb N^+$ the corresponding open cover $B(x,1/n),x\in X$ admits a finite subcover $$B_n:=\{B(x_{ni},1/n),i=1,2,\cdots,k_n\}$$ due to compactness of $X$.

Now you want to show $\Bbb B:=\cup B_n$ is a desired basis.

Countableness has been shown in your attempt, it only suffices to show $\Bbb B$ is a basis indeed. That's to say, for all open subset $U\subset X$ and for all $x\in U$, there exists a member $B\in \Bbb B$ such that $$x\in B\subset U$$ Since there exists $r>0$ such that $B(x,r)\subset U$, we need only to find a $B$ that's completely included in $B(x,r)$. For $n$ so large that $1/n<r/4$, find some $m\in \{1,2,\cdots,k_n\}$ such that $x\in B(x_{nm},1/n)$, then it is readily apparent that this ball can serve as the desired $B$. (For all $y\in B(x_{nm},1/n)$, we have $$d(y,x)\le d(x,x_{nm})+d(y,x_{nm})<r/2<r$$ which says $B(x_{nm},1/n)\subset B(x,r)\subset U$.)

  • $\begingroup$ @hardmath Yeah nice catch. Thanks. (I've always been confusing the two :) $\endgroup$
    – Vim
    Feb 6, 2016 at 16:16

First of all, Vim's comment is absolutely right: you might need more than $n$-many balls of radius ${1\over n}$ to cover your space. But of course this doesn't affect the argument, which only needs there to be finitely many. So let's let $x_{n, i}$ ($i\le k_n$) be the corresponding centers.

Hint: If $U$ is open and $x\in U$, then there is some $s$ such that $B(x, s)\subseteq U$. Now pick $n$ "large enough" and $x_{n, i}$ such that $x\in B(x_{n, i}, {1\over n})$; what can you say about $B(x_{n, i}, {1\over n})$ and $U$?

An interesting side note: this argument also shows that any compact metrizable space is separable (=has a countable dense subset), and it is easy to show that if a metrizable space is separable then it has a countable basis. In general topological spaces, this doesn't work: for instance the Stone-Cech Compactification of $\mathbb{N}$ (or, the usual topology on the set of ultrafilters on $\mathbb{N}$) has a countable dense subset but no countable basis. Metric spaces are special.


I think you need a somewhat different proof. For each $n$ consider $\{B(x,1/n):x\in X\}$. Refine it to a finite cover using compactness. The union of those finite covers for all $n$ is a countable set of balls. Let the $x_i$ be the centers of those balls. For every $x$ and every $n$, $x$ has to be in the finite cover corresponding to $n$, and thus there is an $x_i$ such that $d(x_i,x)<1/n$.

Now, given $U$ and $x\in U$ choose $n$ such that $B(x,1/n)\subset U$, and choose $x_i$ such that $d(x_i,x)<1/(2n)$. If $d(x_i,y)<1/(2n)$ then $d(x,y)\le d(x_i,x)+d(x_i,y)<2/(2n)=1/n$. Thus $B(x_i,1/(2n))\subset B(x,1/n)\subset U$ as desired.


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