$x\in \overline{A} \iff$ there exist a sequence $(x_n)\subset A$ which converge to $x$. 
Indicate if the following result is true or false and justify your answers:
Let $X$ be a topological space and let $A\subset X$. Then
$x\in \overline{A} \iff$ there exist a sequence $(x_n)\subset A$ which converge to $x$.

My attempt:
Let $(x_n)\subset A$ such that $(x_n)\to x$ then
$\forall V\in V(x), \exists N \in \Bbb{N}, \forall p\ge N, \ x_p \in V$
And since $(x_n)\subset A$ then $V \cap A \ne \emptyset$.
Hence $x\in \overline{A}$.
Therefore, if there exist a sequence $(x_n)\subset A$ such that $(x_n)\to x \implies x\in \overline{A}$
But what about:
$x\in \overline{A} \implies $ there exist a sequence $(x_n)\subset A$ which converge to $x$ ? Is it true or not?
 A: 
$x\in \overline A\implies{}$ there exists sequence $\{x_n\}\subseteq A$ which converges to $x$

is true only if $X$ is a first countable space. First countable means that for any point $p$ in the space, there is a countable set $\{U_1, U_2, \ldots\}$ of open neighbourhoods of $p$ such that any open neighborhood of $p$ contains at least one of the $U_i$ (i.e. each point has a countable basis of neighbourhoods).
As an example of a non-first countable space, take the ordinal $\omega_1 +1$ under the order topology, where $\omega_1$ is the first uncountable ordinal. Then if $A=[0, \omega_1)$ is the subset of countable ordinals, we have $\omega_1\in \overline A$, but no countable sequence of countable ordinals has $\omega_1$ as limit.
A: Topological spaces for which this equivalence holds are called Fréchet–Urysohn spaces. It is easy to show that first-countable spaces (and hence all mewtric/metrizable spaces) are Fréchet–Urysohn, however there are non-first-countable spaces which are Fréchet–Urysohn. 
One example is to take the co-finite topology on an uncountable set $X$.


*

*Fix $x \in X$ and suppose that $\{ U_n : n \in \mathbb{N} \}$ is a countable collection of open neighborhoods of $x$. Note that $X \setminus U_n$ is finite for each $n \in \mathbb{N}$, and so $X \setminus \bigcap_{n \in \mathbb{N}} = \bigcup_{n \in \mathbb{N}} ( X \setminus U_n )$ is countable. Since $X$ is uncountable there must be a $y \neq x$ in $X$ such that $y \in U_n$ for each $n$. But note that $V = X \setminus \{ y \}$ is an open neighborhood of $x$, and no $U_n$ is a subset of $V$, meaning that the collection cannot be a base of open neighborhoods of $x$.  Thus the space is not first-countable.

*Note that for $A \subseteq X$ the closure of $A$ is determined by $$ \overline{A} = \begin{cases}
A, &\text{if $A$ is finite} \\
X, &\text{if $A$ is infinite.}
\end{cases}
$$
Clearly if $A \subseteq X$ is finite and $x \in \overline{A} = A$ then the constant sequence $( x_n = x )_n$ is a sequence of points of $A$ converging to $x$.  If $A \subseteq X$ is infinite, then $\overline{A} = X$ and let $( x_n )_n $ be any one-to-one sequence of points in $A$. It can be shown that this sequence converges to every point of $X$: if $U$ is an open neighbourhood of some $x \in X$, then $X \setminus U$ is finite, and since the sequence is one-to-one each $y \in X \setminus U$ can appear at most once in the sequence, and since $X \setminus U$ is finite there must be an $N$ such that $x_n \in U$ for all $n \geq N$. It follows that the space is Fréchet–Urysohn.
A: This equivalence doesn't hold. On $\mathbb R$, let $\tau$ be the topology on $X$ for which the closed sets (other than $X$) are the sets which are finite or countable. Then $0\in\overline{\mathbb{R}\setminus\mathbb{Q}}$, but no sequence of elements of $\mathbb{R}\setminus\mathbb{Q}$ converges to $0$.
