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?
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).
(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?
Best Answer
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.