Show that every compact metrizable space has a countable basis

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.

Solution 1:

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.

Solution 2:

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 $B(x_{nm},1/n)\subset B(x,r)\subset U$.)