Closed iff sequentially closed in metric spaces and the Axiom of Countable Choice

Question

Since I couldn't find no similar question already asked here, I launch this question (I'm sorry if what I'm going to ask has been already posted. I would like to know if so in order to delete this post).

My question is regarding the well-known characterization of the closure of a set in metric spaces,

Let $(X,d)$ be a metric space and $A\subset X$. The following are equivalent:

1. $a\in \overline{A}$.
2. There exists a sequence $\{a_n\}\subset A$ converging to $a$.

(such result I think is the same as saying that closedness is equivalent to sequential closedness in metric spaces).

Now, what I'm concerned about is the relation that this result has with the Axiom of Choice. Specifically, I'm concerned about its proof and the usage of the Axiom of Countable Choice.

The proof that I know for 1. $\Rightarrow$ 2. goes as following,

Let $a\in \overline{A}$. Then $B(a,\frac{1}{n})\cap A\neq \varnothing$ for every $n\in \mathbb{N}$ (where $B(x,r)$ is the open ball of center $x\in X$ and radius $r>0$ for the metric $d$). Now, for each $n\in \mathbb{N}$, pick some $a_n \in B(a,\frac{1}{n})\cap A$. It is now easy to show that $\{a_n\}\subset A$ converges to $a$.

I don't know about foundations nor set theory, but I think that what is being used in the choice of the $a_n$ is the Axiom of Countable Choice (on the other hand, the 2. $\Rightarrow$ 1. part present no problematic whatsoever regarding AC and it is a general result valid for arbitrary topological spaces).

My questions are:

• Can 1. $\Rightarrow$ 2. be proven without ACC?
• If not, could it even be that such statement is equivalent to ACC? (1. $\Rightarrow$ 2.) $\iff$ ACC.

Any kind of help will be appreciated.

Answer 1

In the book "Consequences of the Axiom of Choice" (Howard and Rubin) they mention a lot of different forms of AC. [8] is the ACC (any countable family of non-empty sets has a choice function) and [73] is:

for $A \subseteq \Bbb R$ it is equivalent that $x \in \overline{A}$ and "$\exists (x_n)_n \text{ in } A: x_n \to x$".

Then they reference Sierpinski (1918) for [8] implies [73] at the end, which seems obvious enough, using the countable family $\{B(x,\frac{1}{n})\cap A\mid n \in \Bbb N\}$.

In the models section, page 218, they mention a model of ZF (De la Cruz/Di Prisco model) in which 73 holds, but 8 is false. This doesn't yet rule out the general metric case, which I could not find as a separate form of AC. Asaf might know.