I am trying to understand the proof given in the textbook of the theorem in the title. Here is what the book has:
Let $(x_n)$ be a dense sequence in $X$.
Let $\mathcal{F} = \{(n,k) \in \mathbb{N}^2 | k \geq 1, B(x_n, 1/k) \cap S \neq \emptyset\}$
For every $(n,k) \in \mathcal{F}$, we choose $y_{n,k} \in B(x_n ,1/k) \cap S$. The family
$\{y_{n,k} | (n,k) \in \mathcal{F}\}$
is countable and dense in S.
I am trying to understand the construction of $\mathcal{F}$. It seems as if we are picking indices for a new sequence $y_{n,k}$ so that this sequence is countable and dense. Is this the correct notion, or is there another way to view it?
Just for reference, the definition used for separable in the book is:
A metric space is separable if it contains a countable dense subset.
Any help with this would be appreciated.
Best Answer
Yes, $\mathcal{F}$ is just a set of indices for the (implicitly defined) set $\mathcal{B}$ below. $\mathcal{F}$ is obviously countable, and will provide an enumeration of $\mathcal{B}$.
definition of $\mathcal{B}$ :
For $n, k \in \mathbb{N}$, set $B_{n,k} := B(x_n,\frac{1}{k}) \cap S$, where $B(x_n,\frac{1}{k})$ is the open ball with center $x_n$ and radius $\frac{1}{k}$.
One wishes to pick one element in each $B_{n,k}$ but of course this is possible if and only if $B_{n,k} \neq \emptyset$. Let's keep only the $B_{n,k}$ that aren't empty!
$$\mathcal{B} := \{B_{n,k} \big| n,\,k \in \mathbb{N}, B_{n,k} \neq \emptyset\}$$
$\mathcal{B}$ is a set of non-empty subsets of $S$.
By axiom of choice, there is a function $f : \mathcal{B} \mapsto S$ such that $f(B) \in B$ for all $B \in \mathcal{B}$.
We can now define $y_{n,k} := f(B_{n,k})$ for all $(n,k) \in \mathcal{F}$. Call your metric $d$, the relevant property is that $d(y_{n,k},x_n) < \frac{1}{k}$.
$Y := \{y_{n,k} \big| \ (n,k)\in \mathcal{F}\}$ is countable. Let's check that $Y$ is dense in $S$ :
We need to show that for every $s \in S$, for every $\epsilon > 0$, there is $y \in Y$ with $d(y,s) < \epsilon$.
Chose $k$ such that $\frac{1}{k} < \frac{\epsilon}{2}$. Since $s \in X$, there is $n$ such that $d(s, x_n) < \frac{1}{k}$. We claim that $d(y_{n,k},s) < \epsilon$ : $$d(y_{n,k},s) \leqslant d(y_{n,k},x_n) + d(x_n, s) < \frac{1}{k} + \frac{1}{k} < \epsilon $$ QED