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Convergence in topological space

  1. Feb 3, 2010 #1
    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 realised 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?
     
  2. jcsd
  3. Feb 3, 2010 #2
    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.
     
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