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.
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    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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