# Proof Verification: Show that every compact metrizable space has a countable basis

It would be appreciated if someone could review my proof for accuracy. Thanks!

Show that every compact metrizable space has a countable basis

Proof:

Let X be a compact metrizable space. Then let d be a metric that induces the topology on X. Let A$$_n$$ = { B$$_d$$(x,1\n) | x $$\in$$ X, n$$\in$$ Z$$_+$$ }. Then each A$$_n$$ is an open covering of X. Since X is compact, each A$$_n$$ has a finite subcovering, call this subcovering A$$'_n$$.

Then $$B'$$ = $$\bigcup$$A$$'_n$$ is a countable set as it is the countable union of finite sets. $$B'$$ is also our desired countable basis since:

Given the original, uncountable basis, $$B$$, we have that for any basis element $$B_i$$ $$\in$$ $$B$$, and for any x $$\in$$ $$B_i$$, there must be some open $$\epsilon$$-ball such that B(x,$$\epsilon$$) $$\subset$$ $$B_i$$.

Now we can take $$N$$ such that 1/$$N$$ $$\lt$$ $$\epsilon$$/2. Then for all $$A'_n$$ for n $$\gt$$ N we have that each $$A'_n$$ must be an open covering of X and hence must have some open ball centered around some y $$\in$$ X such that x$$\in$$ B(y,1/n). Then B(y,1/n) $$\subset$$ $$B_i$$ since suppose some z $$\in$$ X - $$B_i$$ was contained within B(y,1/n). Then we would have:

$$d(x,z) \le d(x,y) + d(y,z)$$ $$= d(x,z) \lt \epsilon/2 + \epsilon/2 = \epsilon$$

But B(x,$$\epsilon$$) was chosen to be completely contained within $$B_i$$. Hence z cannot be an element of B(x,$$\epsilon$$), so we have a contradiction and so z cannot be an element of B(y,1/n). Hence B(y,1/n) $$\subset$$ $$B_i$$.

So, $$B'$$ is finer than the original basis $$B$$. The reverse inclusion is clear since the collection of $$\epsilon$$ balls for $$\epsilon$$ $$\lt$$ 1 form an equivalent basis for X. These $$\epsilon$$ balls are clearly contained within our proposed countable basis $$B'$$.

Hence $$B'$$ is a countable basis for X.

## 1 Answer

I dont understand why you use the original $$B$$. You have to show that $$\bigcup_nA'_n$$ is a basis. This is equivalent to saying that for every open subset $$U$$ and $$x\in U$$, there exists $$n$$ such that $$x\in B(x_n,1/n)$$ where $$B(x_n,1/n)$$ is an element of $$A'_n$$ and $$B(x_n,1/n)\subset U$$.

To show that since $$U$$ is open, there exists $$r>0$$ such that $$B(x,r)\subset U$$, take $${1\over{n}}<{r\over 4}$$. There exists $$B(x_n,1/n)$$ such that $$x\in B(x_n,1/n)$$. For every $$y\in B(x_n,1/n)$$, $$d(x,y)\leq d(x,x_n)+d(x_n,y)\leq {2\over n} since $${1\over n}<{r\over 4}$$.We deduce that $$y\in B(x,r)$$ and $$B(x_n,1/n)\subset B(x,r)\subset U$$.

• I use the original basis B to show they generate the same basis elements, hence establishing equivalent topologies – H_1317 Apr 21 at 22:47