This problem deals with sequences.
I have found many definitions for cluster point and there equivalences on MSE and beyond.
Here is the one my text uses:
Let X be a topological space and A $\subset$ X. A point p$\in$X is a cluster point of A if every neighbourhood of p meets A in at least one point other than p.
I am trying to wrap my head around this def.
Could it be this statement : let p$\in X$ ,N$\cap A\setminus${p}$\ne\emptyset$
If not what how can it be stated?
A limit point of $A$ is a point $p$ such that every neighbourhood $U$ of $p$ intersects/meets $A$ in some point different from $p$.
We can also state this in the way you mention (using the right quantifiers too, and using $\mathcal N_p$ for the set of neighbourhoods of $p$):
$$\forall U \in \mathcal{N}_p: U \cap (A\setminus \{p\}) \neq \emptyset$$
which is the same as
$$\forall U \in \mathcal{N}_p: (U\setminus \{p\}) \cap A \neq \emptyset$$
of course, or
$$\forall U \in \mathcal{N}_p: \exists q \neq p: q \in U \cap A $$
Take your pick...
$p$ is a cluster point of the sequence $(x_n)_n$ iff
$$\forall U \in \mathcal{N}_p: \forall n \in \Bbb N: \exists m \in \Bbb N: (m > n) \land x_m \in U$$
so that's a different notion altogether.
“Could it be this statement : let p$\in X$ ,N$\cap A\setminus${p}$\ne\emptyset$“ - You are almost correct. You need to be little bit more precise. Mostly with the notation and order of quantifies. Rather than using $N$, to describe the neighbourhood of $p$, you can use $N_p$, instead. As Henno mentioned, you can use $\mathcal{N}_p$ to describe the set of all neighbourhoods of $p$. When we use the word “meet” in set theory, we usually mean two(in this case) sets have point(s) in common. So we implicitly talk about intersection of two sets. Now the most important part, don’t confuse yourself we “limit point(or cluster point or accumulation point)” and “limit of sequence”. Limit point are defined on set and limit of sequence are defined on sequence. Both have different definitions. In the beginning you said “This problem deals with sequence”, let me tell you, you’re not technically wrong. Because there is a theorem which takes you from limit point of a set to computing limit of a sequence(at least in metric space, I don’t know exactly for topological space). Theorem: If $A \subset X$ and if $p$ is a limit point of $A$, then there is a sequence {$p_n$} in $A$ such that $p=\lim_{n \to \infty} p_n$. Proof: I will give proof of this in metric space setting, because in topological space there is no notation of “distance”. Since every neighbourhood of $p$ contains a point of $A$ other than $p$ itself, i.e. $\forall r \gt 0, N_p(r) \cap A \setminus \{p\} \neq \phi$, $r$ is the radius of the neighbourhood. So for each natural number $n$, take point $p_n \in A$ such that $d(p,p_n) \lt \frac {1}{n}$. Given $\epsilon \gt 0$, choose $N$ so that $N\epsilon \gt 1$. If $N \lt n$, it follows that $d(p,p_n) \lt \frac {1}{n} \lt \frac {1}{N} \lt \epsilon$. Hence $p_n \to p$.
Edit: look lemma 21.2 of Munkres’ Topology.
