Let $X$ be a metric space which is totally bounded. Show that $X$ is separable.
A metric space is called separable if it contains a countable dense subset.
A metric space is called totally bounded if for every $\delta>0$ there exists $N_{\delta}\in \mathbb{N}$ and $x_i\in X, \, 1\le i\le N_{\delta}$ such that $$X \subseteq \bigcup_{i=1}^{N_\delta} B(x_i,\delta).$$
My attempt :
Since $X$ is totally bounded for each $\delta_n = 1/n$ there exists a finite subset $A_n$ of $X$ containing $N_{\delta_n}(\in \mathbb{N})$ elements such that $$X \subseteq \bigcup_{x\in A_n} B(x,1/n).$$
Let $$A = \bigcup_{i=1}^{\infty} A_i.$$
Claim : A is a countable dense subset of $X.$
$A$ is dense :
Consider $B(x,\epsilon)$ for any $x\in X$ where $\epsilon>0.$ Since there exists some $p\in A_k$ such that $x\in B(p,1/k)$ and $k>1/\epsilon$ implies that $x$ is either a limit point of $A$ or a point of $A.$ Hence $A$ is dense
$A$ is countable :
Now this is where my trouble begins. Since Rudin defines that a set is countable if it is in one-one correspondence with the positive integers. Also he defines a set to be at most countable if it is finite or countable.
Here since $A$ is a union of finite sets, $A$ is at most countable.
Hence to show that $A$ is countable I need only show that $A$ cannot be finite.
So this how I proceeded . . .
Suppose $A$ is finite. This would mean (or would this ?)$$A = A_{m_1}\cup A_{m_2}\cup \cdots \cup A_{m_k} (?)$$
Now I think maybe by considering some large $m$ and invoking Archimedean property, I might get the contradiction I'm looking for.
Can anyone please help me complete this?
Best Answer
As I noted in the comments:
But to answer your followup question . . .
Yes, if $X$ is infinite, then the set $A$ will also be infinite.
Suppose instead that $A$ is finite.
Choose $y\in X{\,\setminus\,}A$.
Since $A$ is dense, there is an infinite sequence of elements of $A$ which approach $y$.
But then since $A$ is finite, some element of the sequence has minimum distance to $y$, contradicting convergence.