Accumulation without converging subsequence

“Any sequence with an accumulation point also has a subsequence converging to that point” — right? I believed this until yesterday when I stumbled over the following curious, and countable, Hausdorff space.

Endow \omega\times\omega with the following topology:

  • (\omega\times\omega)\setminus\{(0,0)\} is given the discrete topology;
  • U \subseteq \omega\times \omega is a neighborhood of (0,0) if (0,0)\in U and for almost every n\in \omega the set \bar{U}_n := \{k\in\omega: (n,k) \notin U\} is finite.
  • It is routine to verify that this is a topology on \omega\times\omega.

    Observation 1: No sequence in (\omega\times\omega)\setminus\{(0,0)\} converges to (0,0).
    Proof: Let a:\omega \to (\omega\times\omega)\setminus\{(0,0)\} be any sequence. We distinguish two cases:
    Case 1: For all n\in \omega the set \textrm{im}(a) \cap (\{n\}\times \omega) is finite. Then we set U = (\omega\times\omega) \setminus \textrm{im}(a). This is easily seen to be an open nbhood of (0,0) and clearly a doesn’t converge to (0,0).
    Case 2: There is n_0\in \omega such that V:=\textrm{im}(a) \cap (\{n_0\}\times \omega) is infinite. Then U = (\omega\times\omega) \setminus V is an open nbhood of (0,0), because for all n\in \omega with n\neq n_0 we have \bar{U}_n = \emptyset. Moreover, by the conditition described in the statement of Case 2, the sequence a keeps “popping out” of U and therefore does not converge to (0,0).

    Observation 2: There is a sequence b in (\omega\times\omega)\setminus\{(0,0)\} such that (0,0) is an accumulation point of b.
    Proof: Since (\omega\times\omega)\setminus\{(0,0)\} is countable, there is a bijection from \omega to (\omega\times\omega)\setminus\{(0,0)\} and we let b be that bijection.

    So the sequence b we constructed in Observation 2 has (0,0) as an accumulation point, but no subsequence of b converges to (0,0) because of Observation 1!

    About dominiczypen

    I'm interested in general topology, order theory, and graph theory. This link takes you to my preprints on arXiv.
    This entry was posted in Point-set topology. Bookmark the permalink.

    2 Responses to Accumulation without converging subsequence

    1. I’m a little bit confused when I try to verify your discription: if U is the nbhood of (0,0), then clearly, (\omega\times\omega)\backslash U is an element of the discrete topology in the first discription. Hence, (\omega\times\omega)\backslash U is open and closed. Since U is not empty, then U should be \omega\times\omega and is the only open set containing (0,0). If there was nothing wrong in my deduction, the observation 1 did’t hold.

    2. dominiczypen says:

      I think the deduction “Since U is not empty, then…” isn’t valid.

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