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

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?

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.
• Cool. This conditions on the set $A_\epsilon$ seems highly non-obvious. Is there some instace in which they appear that they must be considered "natural" to choose them? Or is it really that after Zorn's lemma that they become obvious? Commented Oct 20, 2020 at 20:57
• @user2820579: That conditions ensures that for each $n\in\Bbb Z^+$ there is a finite $A_{1/n}\subseteq X$ such that every point of $x$ is within $\frac1n$ some point of $A_{1/n}$. Thus, $\bigcup_{n\in\Bbb Z^+}A_{1/n}$ is countable (because each $A_{1/n}$ is finite) and dense in $X$. Metric spaces that have such a set $A_\epsilon$ for each $\epsilon>0$ are said to be totally bounded, and it turns out that a metric space is compact iff it is complete and totally bounded. Commented Oct 20, 2020 at 23:16