Metric space, countable dense subset and motivation for finite subsets.

Question

Assertion: Every sequence in a metric space $X$ has a subsequence that converges to a point of $X$

Lemma: Any metric space satisfying 'Assertion' has a countable dense subset.

Proof: Suppose that $X$ is a metric space satisfying 'Assertion' and let $\epsilon>0$ be given. We assert that there exists a finite set $A_\epsilon \subset X$ satisfying

$$ \rho(a,b)\geq \epsilon, \quad {\rm for}\quad a\neq b\quad {\rm in}\quad A_\epsilon,\\ B_\epsilon (x) \cap A_\epsilon \neq \emptyset \quad {\rm for \quad each}\quad x\in X $$

(proof continues)

I've seen that this statement has to be fulfilled using Zorn's lemma in here and here. However, I'm puzzled because my book does not uses this lemma at all and just throws this statement about $A_\epsilon$. Is there a way one can understand and justify the existence of $A_\epsilon$ without resorting to a technicality like Zorn's lemma?

Answer 1

You can do it with the recursive construction of a sequence that has no convergent subsequence. This actually does use a very weak form of the axiom of choice, but what it uses is much weaker than Zorn’s lemma, and the construction is probably a bit more intuitive.

Suppose that no such $A_\epsilon$ exists. Let $x_0\in X$ be arbitrary. Suppose that for some $n$ we have chosen points $x_0,\ldots,x_n\in X$ so that $\rho(x_k,x_\ell)\ge\epsilon$ whenever $0\le k<\ell\le n$; by hypothesis there is a point $x_{n+1}\in X$ such that

$$B_\epsilon(x_{n+1})\cap\{x_0,\ldots,x_n\}=\varnothing\,,$$

so $\rho(x_k,x_{n+1})\ge\epsilon$ for $k=0,\ldots,n$, and we can continue the recursive construction to get a sequence $\langle x_n:n\ge 0\rangle$ in $X$ such that $\rho(x_k,x_\ell)\ge\epsilon$ whenever $0\le k<\ell$. This sequence clearly has no convergent subsequence, contradicting the Assertion.