Convergence in topological space

[Cairo]

Let X be an infinite set and p be a point in X, chosen once and for all. Let T be the collection of open subsets V of X for which either p is not a member of V, or p is a member of V and its complement ~V is finite.

Now, let (a_n) be a sequence in X (that is, for all n in N, a_n in X) such that the set of the sequences, {a_n : n in N}, is infinite. Using the definition of convergence in topological spaces, prove that (a_n) has a subsequence which converges to p.

I thought I could do this by showing that if a sequence (a_n) converges to p, then so does every subsequence. But then realized that (a_n) is ANY sequence, so my proof would not hold. I'm also not sure if a sequence even does converge to p!

Any ideas how to prove this result?

 

[JSuarez]

There are few things that are unclear in your post, but I think what you menat is this:

(1) The set T is the topology of the space (X,p), that is, its elements are the open sets of (X,p).

(2) In a general topological space, sequence convergence is defined by: $a_n \in X$ converges to $a \in X$ iff:

$$\forall O \in T\exists n_0 \in \mathbb N \left(a \in O \wedge n > n_0 \rightarrow a_n \in O\right)$$

This means that, for any (open) neighborhood of $a$, all terms of $a_n$, for $n > n_0$ will belong to this neighborhood.

(3) Now, you want to prove that, for any sequence, such that its set of terms [itex]$$\left\{a_n:n \in \mathbb N\right\}$$ is infinite will have a convergent subsequence to p. For this topology, note that any neighborhood $O_p$ of p will be a set such that its complement is finite; this implies that there exists an $n_0$, such that, for all $n> n_0$, $a_n \in O_p$. From this, you may extract a subsequence $a_{n_k}$, convergent, in the above sense, to p.[/itex]