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For $m \in \mathbb{N} $ and $U \subset \mathbb{N} \times \mathbb{N}$ define the following two sets: \begin{align} S^U_m = \{n \in \mathbb{N} | (m,n) \notin U \} \end{align} and \begin{align} T^U = \{m \in \mathbb{N} | (m,0) \notin U \}. \end{align}

Then

$\mathcal{T} := \{ U \subset \mathbb{N} \times \mathbb{N} \ | \ (0,0) \in U \implies T^U \text{ is finite and } (m,0) \in U \implies S^U_m \text{ is finite } \}$

will be a topology.

I need to show that there cannot exist a sequence $(x_n)_n$ in $\mathbb{N}_0 \times \mathbb{N}_0$ such that it converges to $(0,0)$. I already saw the solution for other similar questions on this site, but they are all about the Arens-Fort space:https://en.wikipedia.org/wiki/Arens%E2%80%93Fort_space. However, this one is slightly different, there is this extra $T^U$ in the definition of the topology $\mathcal{T}$, which messes things up when I wanted to prove it using a similar method as in: No series converges towards $(0,0) $ in $(\mathbb{N}\times\mathbb{N})\setminus (0,0)$ with a non standard topology.

Einsteinwasmyfather
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1 Answers1

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I am assuming that you require the sequence elements to have nonzero coordinates (otherwise, $(0,n)$ converges to $(0,0)$).

Assume by contradiction that $S=(p_x)_{x \in \Bbb N}$ is a sequence of elements in $\Bbb N \setminus \{0\} \times \Bbb N \setminus \{0\}$ which converges to $(0,0)$. This means that every open subset containing $(0,0)$ hosts all elements of $S$ after some point.

Let $U_k = \{(m,n) \; | \; m=0 \text{ or } m \ge k, \; n \in \Bbb N\}$, it is an open neighborhood of $(0,0)$. This implies that there are only finitely many $p_x$ having first coordinate less thank any given $k \in \Bbb N$. In other words: $$A_k=\{n \in \Bbb N \; | \; (k,n) \text{ is in the sequence } S\}$$ is finite for $k \ge 1$ (and $A_0$ is empty). Now, the following set $W$ is an open neighborhood of $(0,0)$ which does not contain any element of $S$: $$W = \{(k,n) \; | \; n \not \in A_k\}.$$ Indeed, $W$ contains all $(0,n)$, $(m,0)$, and for a fixed $m \in \Bbb N$, only finitely many $(m,n)$ do not belong to $W$.

François Gatine
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  • Thank you very much! Any idea or intuition on how you came up with this neighborhood of (0,0)? I also tried constructing one by removing/adding finite/infinite elements but they always seem to contradict the condition on either $T^U$ or $S^U_m$. – Einsteinwasmyfather Aug 15 '24 at 11:27
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    Sure ! Any neighborhood $U$ of $(0,0)$ contains almost all $(m,0)$, so the simplest nontrivial example to consider is to say "what if its missing the first $k-1$ ?" Then, if $m \ge k$, $(m,0) \in U$, so $U$ has to contain almost all $(m,n)$: simplest example is "what if it has all of them ?". This gives you $U_k$, which gives you the finiteness condition on $S$ as discussed. Then, it's just a matter of playing around (use a grid to visualize if you need it) to see how one can avoid the finite terms of the sequence. I hope this answers your question ! – François Gatine Aug 15 '24 at 17:41