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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.

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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 $y\in B(x_n,1/n)$, $d(x,y)\leq d(x,x_n)+d(x_n,y)\leq {2\over n}<r/2$ since ${1\over n}<{r\over 4}$.We deduce that $y\in B(x,r)$ and $B(x_n,1/n)\subset B(x,r)\subset U$.

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

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