Theorem: Every compact metric space $K$ has a countable base and therefore $K$ is separable.
A: Since every infinite subset of $K$ has a limit point in $K$ as $K$ is a compact metric space, it follows that $K$ is separable and since every separable metric space has a countable base so does $K$. This proves the theorem.
Now, let's say that this A is not known to us so we proceed as below:
Choose $\delta \gt 0$,
Clearly $K\subseteq \cup_{x\in K} N_\delta (x)$. Since, $K$ is compact, there must exist a finite subset $K_f\subset K$ such that $K\subseteq \cup_{x\in K_f} N_\delta (x)=B$
Let $G\subseteq K$ be any open subset of $K$. For some $t\in G$, $\exists \epsilon_t\gt 0$ such that $N_{\epsilon_{t}}(t)\subset G$.
Clearly, $t\in B$ and hence $d(t,x)\lt \delta$ for some $x\in K_f$
If $a\in N_\delta (x)$ then $d(a,x)\lt \delta$
Hence, $d(a,t)\le d(a,x)+d(x,t)\lt 2\delta$
Since $\delta\gt 0$ is arbitrary, we choose $\delta =\epsilon_t/2\implies d(a,t)\lt \epsilon_t\implies a\in G$. Thus we have proved that for any $x\in G, x \in N_\delta (x) \text{(for some $\delta \gt 0$)}\subset G$. Hence, $B$ is a countable base of $K$.
I am stuck at proving (if this is true at all):
$K$ has a countable base $\implies K$ is separable. If this statement is not true, then $A$ can be employed to prove $K$ is separable.
Please help. Thanks.
Best Answer
There are multiple problems with what you’ve written. I’ll leave the ones in A for the end.
It is true that $\{N_\delta(x):x\in K\}$ is an open cover of $K$ and that there is a finite $K_f\subseteq K$ such that $K\subseteq\bigcup_{x\in K_f}N_\delta(x)$. However, all of the sets $N_\delta(x)$ are subsets of $K$, so in fact $\bigcup_{x\in K_f}N_\delta(x)=K$: the set that you called $B$ is simply $K$. Thus, it makes no sense later on to say that $B$ is a countable base for $K$: $B$ is just the set $K$ and is not a base for $K$ at all.
What you have actually shown here is that for each $\delta>0$ there is a finite $K_\delta\subseteq K$ such that for each $x\in K$ there is at least one $y\in K_\delta$ such that $d(y,x)<\delta$. You could now use this directly to get a countable dense subset of $K$, as follows. For $n\in\Bbb Z^+$ let $D_n=K_{1/n}$, and let $D=\bigcup_{n\in\Bbb Z^+}D_n$. $D$ is the union of countably many finite sets, so $D$ is countable.
To show that $D$ is dense in $K$, let $x\in K$ and $\epsilon>0$. There is an $n\in\Bbb Z^+$ such that $\frac1n<\epsilon$, and there is a $y\in D_n$ such that $d(x,y)<\frac1n$, so $y\in D\cap N_\epsilon(x)$. Thus, $x\in\operatorname{cl}D$, and $D$ is dense in $K$.
Thus, you don’t need to show that every space with a countable base is separable, though this is actually very easy. Suppose that $\mathscr{B}$ is a countable base for a space $X$. For each $B\in\mathscr{B}$ let $x_B\in B$, and let $D=\{x_B:B\in\mathscr{B}\}$. Clearly $D$ is countable. If $x\in X$ and $U$ is any open nbhd of $x$, there is a $B\in\mathscr{B}$ such that $x\in B\subseteq U$, so $x_B\in D\cap U$, $x\in\operatorname{cl}D$, and $D$ is dense in $X$.
Showing that every separable metric space has a countable base is actually harder. It looks like you may have been trying to do something like that in the two paragraphs after you took the finite subcover $K_f$, but what you actually have there simply doesn’t make much sense. (E.g., for any $x\in K$ and $\delta>0$ it’s always true that $x\in N_\delta(x)$; this is not something true just for points $x$ in some open set $G$ and does not require any argument at all, since it’s true by definition.)
One way to prove that result is to show that $\left\{B_{1/n}(x):x\in D\text{ and }n\in\Bbb Z^+\right\}$ is a countable base for the topology. You can use a similar idea to prove directly from compactness that $K$ has a countable base, as the author of this question did.
Finally we come to A. It is true that the compactness of $K$ implies that every infinite subset of $K$ has a limit point, but the fact that every infinite subset of the space has a limit point does not by itself imply that the space is separable: there are non-separable spaces in which every infinite subset has a limit point. (The space of countable ordinals is one.) You really do need to use the fact that $K$ is metric to show that it is separable.