Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

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 $\mathbb N$:$$\tau_1=\{X\subseteq\mathbb N:1\notin X\}\cup\{X\subseteq\mathbb N:|\mathbb N\setminus X|\lt\aleph_0\};$$$$\tau_2=\{\emptyset\}\cup\{X\subseteq\mathbb N:|\mathbb N\setminus X|\lt\aleph_0\}\cup\{X\subseteq\mathbb N:1\notin X\text{ and }\sum_{n\in\mathbb N\setminus X}\frac1n\lt\infty\}.$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.