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Let $\mathscr T_1$ and $\mathscr T_2$ be two topologies on space $X$. Assume that $(X,\mathscr T_1)$ is metrizable, and any sequence in $X$ that converges in one of the two topologies must also converge in the other topologies, i.e., $$(\,\forall\{x_n\}_{n\in\mathbb N}\subset X\,)\Big(\big(x_n\xrightarrow[]{\mathscr T_1} x\big) \Leftrightarrow \big(x_n\xrightarrow[]{\mathscr T_2} x\big)\Big).$$

The question is, whether $\mathscr T_1$ and $\mathscr T_2$ are the same topology on $X$?

The answer should be affirmative, since this argument is used in many analysis books without hesitation, such as the proof of metrizable of locally convex spaces in Conway's book of functional analysis. But I don't know how to prove.

Any comments or hints will be appreciated!

Edit: Very sorry! I'll explain here the details I thought about the link of Conway's book. The motivation is from the proof of Proposition IV.2.1 in that book as linked.

To prove the necessity part of the proposition, that is,

If a local compact space $X$ is metrizable, then its topology is determined by a countable family of seminorms.

the author construct a countable family of seminorms $\{p_n\}$, and then show that $\{p_n\}$ determine the same sequential convergence as a given metric $\rho$ of $X$, that is, $$(\,\forall\{x_n\}_{n\in\mathbb N}\subset X\,)\Big(\big(x_n\xrightarrow[]{\{p_n\}} x\big) \Leftrightarrow \big(x_n\xrightarrow[]{\rho} x\big)\Big).$$ But how to assert from the preceding statement that $\{p_n\}$ determine the same topology as the original one?

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    $\begingroup$ I don't see where this statement is used in the material you linked. Can you explain in greater detail how you think it is used? $\endgroup$ Commented Oct 9, 2017 at 3:49
  • $\begingroup$ You can find some examples here: Example of different topologies with same convergent sequences and Is a topology determined by its convergent sequences? (Maybe not all of those examples have one of the topologies metrizable, as you require, but at least the ones with discrete the ones with discrete topology and with topology derived from a norm are such examples.) $\endgroup$ Commented Oct 9, 2017 at 4:47
  • $\begingroup$ @Eric Wofsey Sorry and I have reedited my question to add the details. And thanks for your counterexample. $\endgroup$ Commented Oct 9, 2017 at 5:22
  • $\begingroup$ @Martin Sleziak Nice references! I have reedited to add the motivation of my question. $\endgroup$ Commented Oct 9, 2017 at 5:26
  • $\begingroup$ I think the proof works equally for nets, though his notation "$x_i\to0$ as $j\to\infty$" suggests otherwise. So there really is no problem of first-countability here. $\endgroup$ Commented Feb 2, 2019 at 1:03

2 Answers 2

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No, this is not true in general. For instance, let $F$ be a nonprincipal ultrafilter on $X=\mathbb{N}$. Let $\mathscr{T}_1$ be the discrete topology, and let $\mathscr{T}_2$ consist of all sets that either are in $F$ or do not contain $0$. In both these topologies, a sequence converges iff it is eventually constant, and $\mathscr{T}_1$ is metrizable.

In the proof of Proposition IV.2.1, $\mathscr{T}_2$ is additionally known to be first-countable, since it is generated by countably many seminorms.

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  • $\begingroup$ Ok, I see... I just... So stupid I am... $\endgroup$ Commented Oct 9, 2017 at 5:52
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Inasmuch as a discrete space is metrizable and has no nontrivial convergent sequences, all you need for a counterexample is any nondiscrete space with no nontrivial convergent sequences. There are lots of examples; here are two that don't involve the axiom of choice.

Example 1. Let $X$ be an uncountable set, choose an element $p\in X,$ and let $$\tau=\{A\subseteq X:p\notin A\text{ or }|X\setminus A|\le\aleph_0\}.$$

Example 2. Let $X=\mathbb N=\{1,2,3,\dots\}$ and let $$\tau=\left\{A\subseteq\mathbb N:1\notin A\text{ or }\sum_{n\in\mathbb N\setminus A}\frac1n\lt\infty\right\}$$

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    $\begingroup$ Cool examples! Why is the special point p necessary? I would expect the cocoa table sets to be enough since there are no nontrivial convergent sequences. $\endgroup$ Commented Oct 9, 2017 at 16:21
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    $\begingroup$ @ZachBoyd Although it was not required by the question I preferred to give examples of Hausdorff spaces. $\endgroup$ Commented Oct 9, 2017 at 19:17
  • $\begingroup$ Makes sense. Thanks. $\endgroup$ Commented Oct 9, 2017 at 19:28

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