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JSuarez
JSuarez is offline
#2
Feb3-10, 04:36 PM
P: 403
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: [itex]a_n \in X[/itex] converges to [itex]a \in X[/itex] iff:

[tex]
\forall O \in T\exists n_0 \in \mathbb N \left(a \in O \wedge n > n_0 \rightarrow a_n \in O\right)
[/tex]

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

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