Accumulation point is a limit of some sequences in first-countable space with Axiom of Choice.

Question

I have a statement which says that

Let $X$ be a first-countable space and $E$ be a subset of $X$. If $p$ is an accumulation point of $E$, then there exists a sequence in $E$ that converges to $p$.

Now I have proved it. The problem is that I think that I had invoked the Axiom of Choice (AC) accidentally. I saw the similar proofs by other people here and they said nothing about AC (See links below). Thus my question is that does my proof here contain AC?.

Accumulation points of sequences as limits of subsequences?

In a first-countable space, an accumulation point of the set of terms in a sequence is also a limit-point of the sequence

Also, There were other people who also had the similar questions, but since they worked in metric space which is stricter than first-countable space, the answer was another proof that can avoided AC (See links below).

Proving that without the axiom of choice there is a set with an accumulation point that isn'tthe limit of a convergeant sequence

(ZF)subsequence convergent to a limit point of a sequence

I wrote my proof myself below. If you think it is too long, please see tl;dr

Let $X$ be a first-countable space, $E\subseteq X$, $p$ be an accumulation point of $E$. Since $X$ is first-countable, there exists a collection of open sets $(U_i)_{i\in\mathbb{N}}$ such that $p\in U_i$ for every $i\in\mathbb{N}$ and for every neighborhood $U$ of $p$, there exists a number $m\in\mathbb{N}$ such that $U_m\subseteq U$. Next, we let $S_n = \cap_{i=1}^n U_i$. Since each $U_i$ is an open neighborhood of $p$, this implies that $S_n$ is also an open neighborhood of $p$ and thus must contain a point in $E$ which is different from $p$ (i.e., $S_n-\lbrace p\rbrace$ is not empty). This is true for every $n\in\mathbb{N}$. We also have the fact that $S_n\subseteq U_n$. Thus we have the collection of $(S_i)_{i\in\mathbb{N}}$ such that for every neighborhood $U$ of $p$, there exists a number $m\in\mathbb{N}$ such that $S_m\subseteq U_m\subseteq U$. Now if we 'choose' a point $p_n\in S_n$ such that $p_n\in E$ for every $n\in\mathbb{N}$, we get a sequence $(p_i)_{i\in\mathbb{N}}$ such that it is in $E$. Now to show that this sequence converges to $p$, we recall that for every neighborhood $U$ of $p$, there exists a number $m$ such that $S_m\subseteq U$. Hence $p_m \in U$. For every other $n>m$, we note that $S_n\subseteq S_m$. Thus $p_n\in U$ and this implies that $(p_i)_{i\in\mathbb{N}}$ really converges to $p$.

tl;dr

First, I have an infinitely countable collection of non-empty sets. Then I choose an element of each set to construct an infinite sequence. Does this count as using the axiom of choice?

Answer 1

Your argument uses the axiom of countable choice, which says that every countable family of non-empty sets has a choice function. This question (one of those to which you linked) shows that some amount of choice is needed. The axiom of choice is usually not mentioned in such arguments simply because it’s generally taken for granted in doing topology.