# Is the metric topology determined by its convergent sequences?

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)?

• 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.
– M W
Feb 18 at 5:29
• @MW agreed and edited.
– Atom
Feb 18 at 8:41
• math.stackexchange.com/q/4554239/688539 Feb 18 at 11:06

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 ^_^

• 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 $X_2$). Feb 18 at 3:15
• @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. Feb 18 at 3:46
• @DavidGao Thanks, I couldn't figure out your point! Feb 19 at 11:21

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.

• 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.
– M W
Feb 18 at 4:57
• @MW I was just going to add that! :)
– Atom
Feb 18 at 5:05
• 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 Feb 18 at 8:45
• @AlessandroCodenotti A more accessible example would be cocountable and discrete topologies on an uncountable set. :)
– Atom
Feb 18 at 8:49