Is AC necessary to show that in metric spaces $x\in\operatorname{closure}(A)$ implies $\exists\{a_n\}_{n=1}^\infty\subseteq A$ s.t. $\lim a_n=x$?

Question

Let $(X,d)$ be a metric space.

Let $x\in\operatorname{closure}(A)$, where $A\subseteq X$.

Then for each $n\in\mathbb{N},\exists x_n\in B_{\frac{1}{n}}(x)\cap A$, where $B_\varepsilon(x)$ is the open ball around $x$ of radius $\varepsilon$.

Question: have I used the axiom of choice in the above conclusion?

I would say I did, because the sequence $(x_n)_{n\in\mathbb{N}}$ is an element of the Cartesian product $\prod_{n\in\mathbb{N}} (B_{\frac{1}{n}}(x)\cap A)$, following the definition here.

I just wanted to make sure that is correct.

Answer 1

Yes. You have used the axiom of countable choice to choose $x_n$ from each $B_{\frac1n}(x)\cap A$.

To wit, in Cohen's first model there exists a dense subset $D$ of $\Bbb R$ which is Dedekind-finite. Namely $D$ has no countably infinite subset. It is not hard to see, if so, that any $x\in\Bbb R\setminus D$ is in the closure of $D$, but no sequence in $D$ converges to $x$.