Is a limit point in a sequential space the limit of some sequence? I am following this Wikipedia entry.
Let $X$ be a topological space and $S \subseteq X$ an arbitrary subset.
A point $x \in X$ is a limit point of $S$ if every neighborhood $U$ of $x$ contains a point of $S$ different from $x$ (that is $U \cap S \setminus \{ x \} \neq \emptyset$) or equivalently $x \in \textrm{cl}(S \setminus \{ x \})$.
Wikipedia states that:

Alternatively, if the space $X$ is sequential, we may say that $x \in X$ is a limit point of $S$ if and only if there is an $\omega$-sequence of points in $S \setminus \{ x \}$ whose limit is $x$; hence, $x$ is called a limit point.

Is this statement true?


*

*It is clearly true for first-countable spaces $X$: take a countable neighborhood base $U_n \subseteq X$ of $x$ and a point $x_n \in U_n \cap S \setminus \{ x \}$. Then we have the desired convergent sequence $x_n \to x$.

*More generally, the statement is true for Fréchet-Urysohn spaces $X$ since then the closure $\textrm{cl}$ equals the sequential closure $\textrm{scl}$, i.e. $\textrm{cl}(S \setminus \{ x \}) = \textrm{scl}(S \setminus \{ x \})$ implying that for the limit point $x$ of $S$ there is the desired sequence $x_n \in S \setminus \{ x \}$ with $x_n \to x$.

*If $X$ is merely sequential (i.e. a set is closed iff it is sequentially closed) then from $S \setminus \{ x \}$ being not closed we can only deduce that $S \setminus \{ x \}$ is not sequentially closed implying that there is a sequence $x_n \in S \setminus \{ x \}$ converging to some point $y \not\in S \setminus \{ x \}$. But $y$ needs not be equal to the given point $x$.
 A: This is not true.  For instance, take $X=\{p\}\cup(\mathbb{N}\cup\{\infty\})\times\mathbb{N}$, where $(\mathbb{N}\cup\{\infty\})\times\mathbb{N}$ is an open subspace with its usual topology and a set $U$ containing $p$ is open iff $U$ contains all but finitely many of the points $(\infty,n)$ and $U\cap (\mathbb{N}\cup\{\infty\})\times\mathbb{N}$ is open.  This is sequential: $(m,n)$ converges to $(\infty,n)$ as $m\to\infty$ for each $n$, and $(\infty,n)$ converges to $p$ as $n\to\infty$, and these convergent sequences generate the topology.
Now take $S=\mathbb{N}\times\mathbb{N}\subset X$.  Then $p\in\overline{S}$, since a neighborhood of $p$ must contain points of the form $(\infty,n)$ and thus contain points of the form $(m,n)$.  But no sequence in $S$ converges to $p$.  Indeed, given a sequence $(x_k,y_k)_{k\in\mathbb{N}}$ in $S$, there are two cases.  If $y_k$ takes some value $n$ infinitely often, then $X\setminus (\mathbb{N}\cup\{\infty\}\times\{n\})$ is a neighborhood of $p$ that is missing infinitely many points of our sequence.  If $y_k$ does not take any value infinitely many times, on the other hand, it is easy to see that $X\setminus\{(x_k,y_k)\}$ is open, and so is a neighborhood of $p$ containing no points of the sequence.
What's going on here is that $p$ is not the limit of a sequence in $S$, but is a limit of a sequence of limits of sequences in $S$.  More generally, to find the closure of a subset of a sequential space, you may need to iterate the process of taking limits of sequences up to $\omega_1$ times.
