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I am aware that in a first countable space (and thus any metric space) is completely determined by its convergent sequences and their limits, i.e.,

If $\tau_1$ and $\tau_2$ are two first countable topologies on a set $X$ such that $x_i\to c$ in $\tau_1$ iff $x_i\to c$ in $\tau_2$, then $\tau_1 = \tau_2$.

However, this raises the following question: If two metrics on a space have the same convergent sequences, will they have the same limits as well (and thus the same topology)?

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    $\begingroup$ maybe more accurate to say "two metrics on a space" rather than "two metric spaces", since problem doesn't make much sense if the underlying set isn't the same. $\endgroup$
    – M W
    Feb 18 at 5:29
  • $\begingroup$ @MW agreed and edited. $\endgroup$
    – Atom
    Feb 18 at 8:41
  • $\begingroup$ math.stackexchange.com/q/4554239/688539 $\endgroup$ Feb 18 at 11:06

2 Answers 2

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I believe this is true -- we can recover what the sequences converge to.

Say $(a_n)$ is a sequence in $X$ that we know converges, but we don't know what it converges to. There will be a unique $x \in X$ so that the new sequence $(a_1, x, a_2, x, a_3, x, a_4, x, \ldots)$ is convergent (which we can detect), and this $x$ is necessarily the limit of the sequence $(a_n)$.

Of course, this process is not "computable" in any sense. I have no idea how one would find such an $x$ in practice (though maybe it's doable in case $X$ is compact...) but, in the abstract, this shows that the convergent sequences remember limits.


I hope this helps ^_^

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    $\begingroup$ Maybe I am missing something, but is this an answer to OP's question? There are two metric spaces, for which say $a_n\to x_1$ (in $X_1$) and $a_n\to x_2$ (in $X_2$). $\endgroup$ Feb 18 at 3:15
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    $\begingroup$ @WishYouTheBest As the answer noted, the $x$ in the answer is unique. That is to say, if $x_1 \neq x_2$, then $(a_1, x_1, a_2, x_1, a_3, \cdots)$ is convergent in $X_1$ but not in $X_2$, so the two spaces do not share convergent sequences in that case. $\endgroup$
    – David Gao
    Feb 18 at 3:46
  • $\begingroup$ @DavidGao Thanks, I couldn't figure out your point! $\endgroup$ Feb 19 at 11:21
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This is just to expand a bit on HallaSurvivor's answer.

It turns out that limits of convergent sequences can be encoded in sequences themselves: In any $T_1$ space, $a_i\to x$ iff the sequence $a_1, x, a_2, x, a_3, x, a_4, \ldots$ converges. Hence the topology of a first countable $T_1$ space is completely determined by its convergent sequences.

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    $\begingroup$ Worth noting that $T_1$ is absolutely necessary here, since for $X=\{0,1\}$ the distinct topologies $\tau_i=\{\emptyset,\{i\},X\}$, for $i=0,1$, each have the property that every sequence converges (in $\tau_0$ they all converge to $1$, and in $\tau_1$ they all converge to $0$). As a sillier example, the trivial topology has this property as well. $\endgroup$
    – M W
    Feb 18 at 4:57
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    $\begingroup$ @MW I was just going to add that! :) $\endgroup$
    – Atom
    Feb 18 at 5:05
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    $\begingroup$ Here is an example to see that first countable is also necessary: the weak and norm topology on $\ell^1$ have the same convergent sequences (with the same limits) but are very different topologies $\endgroup$ Feb 18 at 8:45
  • $\begingroup$ @AlessandroCodenotti A more accessible example would be cocountable and discrete topologies on an uncountable set. :) $\endgroup$
    – Atom
    Feb 18 at 8:49

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