## 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!

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## 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.