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First Countable

  1. May 21, 2010 #1
    In a first countable space any point that is adherent to a set S is also the limit of a sequence in S.

    In my head, this seems obvious, but I can't seem to get it on paper.. I know that is has to do with inverse functions preserving unions and intersections, but can't seem to write the proof out.
  2. jcsd
  3. May 22, 2010 #2
    We may as well consider the neighbourhood base to consist of open sets; let the open neighbourhood base of the adherence point be [itex]\{ G_n : n\in \mathbb{N} \}[/itex].Put
    [itex]B_1 = G_1[/itex]
    [itex]B_n = G_1\cap \ldots \cap G_n[/itex]
    Then in each [itex]B_n[/itex] we have a point [itex]s_n[/itex] of S. Then the sequence [itex](s_n)[/itex] converges to the adherence point.
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