Cluster point of a sequence and limit point of some subsequence In a topological space $X$, quoted from Wikipedia:

A point $x ∈ X$ is a cluster point of a sequence $(x_n)_{n ∈ N}$ if,
  for every neighbourhood $V$ of $x$, there are infinitely many natural
  numbers $n$ such that $x_n ∈ V$. If the space is sequential, this is
  equivalent to the assertion that $x$ is a limit of some subsequence of
  the sequence $(x_n)_{n ∈ N}$.

I was wondering when a topological space is not necessarily sequential, what is the relation between cluster point of a sequence and limit point of some subsequence of the sequence?
When the topological space is sequential space, why are the two equivalent?
Thanks and regards!
 A: In general, if $p$ is the limit of a subsequence $(x_{n_k})$ of the sequence $(x_n)$ in $X$ then $p$ is a cluster point of the sequence: let $O$ be a neighbourhood of $p$, then as $(x_{n_k})$ converges to $p$, there is some $K \in \mathbb{N}$ such that for all $k \ge K$ we have that $x_{n_k} \in O$, and $\{ n_k \mid k \ge K \}$ then are infinitely many natural numbers such that $x_{n_k}$ is in $O$. 
For sequential spaces it does not to hold that a cluster point of a sequence is the limit of a subsequence of it. This will hold in $T_1$ Fréchet-Urysohn spaces (see below) and certainly in first countable spaces. The Arens space is a counterexample: it consists of all pairs $(n,m)$ in $\mathbb{N}^{+} \times \mathbb{N}^{+}$ (which are all isolated points), all $n \in \mathbb{N}^{+}$, which have basic neighbourhoods of the form $B(n,N) = \{ n \} \cup \{ (n,m) \mid m \ge N \}$, for $N \in \mathbb{N}^{+}$, so that $ ((n,m))_m \rightarrow n$ for all $n$, and the point $0$, which has basic neighbourhoods of the form that contain $0$, all but finitely many $n$ and for each of these $n$ a neighbourhood $B(n, N(n))$. One checks that this space is sequential and normal. Also, no sequence from $\mathbb{N}^{+} \times \mathbb{N}^{+}$ converges to $0$, but $0$ is in the closure of it. So if we index the set of all pairs as a sequence, $0$ is a cluster point of it, but not a subsequential limit.
For Fréchet-Urysohn spaces the set of all subsequential limits from a set $A$ equals the closure of $A$ (this is one of the definitions of being Fréchet-Urysohn), and if we have a sequence $(x_n)$ in a $T_1$ Fréchet-Urysohn space with a cluster point $p$, then either $p$ appears infinitely any times in the sequence, and this gives us a convergent subsequence to $p$, or we can assume $p$ does not appear at all in $A = \{ x_n \mid n \in \mathbb{N} \}$, and $p$ is in the closure of $A$, so is a subsequential limit from $A$, which allows us to define a subsequence from $(x_n)$ that converges to $p$ as well, using $T_1$ to get higher and higher indices from the sequence.       
