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All of the famous examples for sequentially indistinguishable topologies on a set $X$ are provided on an uncountable set $X$ (an uncountable set $X$ with discrete and cocountable topology or the $l^1(\mathbb{N})$-space with norm and weak topology, see this thread).

A natural question arises: Does there exist on a countable set $X$ two topologies that have the same convergent sequences? (Note that the countability of $X$ does not has a large influence on the sequentiality of the topology, e.g. the Arens-Fort space is countable and not sequential.) If this is so, can such topologies be explicitely given?

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    $\begingroup$ Can you have a topology on $\mathbb Z$, other than the discrete topology, where the only convergent sequences are eventually constant? $\endgroup$
    – Gerald Edgar
    Jan 11, 2016 at 15:14
  • $\begingroup$ I see, so you mean that we can take the Arens-Fort space in which all subsets are sequentially open or equivalently all convergent sequences are eventually constant. In other words, the family of sequentially open sets which forms a topology is the strictly finer discrete topology. $\endgroup$
    – yada
    Jan 11, 2016 at 16:00
  • $\begingroup$ See also this post at math.SE: Example of different topologies with same convergent sequences $\endgroup$
    – Martin Sleziak
    Dec 6, 2016 at 3:30

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There are plenty of topologies on a countable set for which all convergent sequences are eventually constant.

The most constructive example I know is the Arens-Fort space, given as example 26 in Steen and Seebach's Counterexamples in Topology.

This topology is constructed on the set $\mathbb{N}^2 \cup \{\infty\}$. Every point of $\mathbb{N}^2$ is isolated, and open neighborhoods of $\infty$ are defined to contain all but finitely many points from all but finitely many columns of $\mathbb{N} \times \mathbb{N}$ (it helps to draw a picture).

Obviously no nontrivial sequence converges to a point of $\mathbb{N}^2$. Suppose $(x_n)$ is a sequence in $\mathbb{N}^2$: we will show it does not converge to $\infty$. Either (a) it has points in only finitely many columns, so some neighborhood of $\infty$ misses the whole sequence, or (b) it has points in infinitely many columns, and we can find a infinite subsequence with one point in each column, which shows that some neighborhood of $\infty$ misses infinitely many of the $x_n$.

A little less constructively, any countable subset of the Stone-Cech compactification of the integers will have the property that every convergent sequence is eventually constant. Using this fact, you obtain $2^{2^{\aleph_0}}$ non-homeomorphic countable spaces, none of which have any nontrivial convergent sequences.

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  • $\begingroup$ Thanks. I didn't know this property of the Stone-Cech compactification. $\endgroup$
    – yada
    Jan 11, 2016 at 16:04
  • $\begingroup$ For a good explanation of why it's true, see Joseph van Name's answer to this question: mathoverflow.net/questions/138161/… $\endgroup$
    – Will Brian
    Jan 11, 2016 at 16:06

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