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Are there two topologies on the same underlying set, exactly one of which is metrizable, which share the same convergent sequences with the same limits but are not the same?

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up vote 6 down vote accepted

Let $E$ be an uncountable set, let $\tau_1=\mathcal P(E)$, and let $\tau_2=\{X\in\mathcal P(E):X=\emptyset\text{ or }|E\setminus X|\le\aleph_0\}$.

In either topology, a sequence $(x_1,x_2,x_3,\dots)$ converges to a point $x$ if and only if $x_n=x$ for all sufficiently large $n$. The discrete topology $\tau_1$ is metrizable, e.g., let $d(x,y)=1$ whenever $x\ne y$. The cocountable topology $\tau_2$ is not even Hausdorff, let alone metrizable.

P.S. If you want a countable example with some nontrivial convergent sequences, consider the following two topologies on $\mathbb N$:$$\tau_1=\{X\subseteq\mathbb N:1\notin X\}\cup\{X\subseteq\mathbb N:|\mathbb N\setminus X|\lt\aleph_0\};$$$$\tau_2=\{\emptyset\}\cup\{X\subseteq\mathbb N:|\mathbb N\setminus X|\lt\aleph_0\}\cup\{X\subseteq\mathbb N:1\notin X\text{ and }\sum_{n\in\mathbb N\setminus X}\frac1n\lt\infty\}.$$

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