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:

<br /> \forall O \in T\exists n_0 \in \mathbb N \left(a \in O \wedge n &gt; n_0 \rightarrow a_n \in O\right)<br />

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

(3) Now, you want to prove that, for any sequence, such that its set of terms \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&amp;gt; n_0, a_n \in O_p. From this, you may extract a subsequence a_{n_k}, convergent, in the above sense, to p.