9

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?

Dreamer
  • 1,952

2 Answers2

6

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.

Eric Wofsey
  • 330,363
6

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 $$\tau=\left\{A\subseteq\mathbb N:1\notin A\text{ or }\sum_{n\in\mathbb N\setminus A}\frac1n\lt\infty\right\}$$

bof
  • 78,265