# Point-Set Question

## Main Question or Discussion Point

I'm fairly certain that the following statement requires the assumption that the topology is first countable, can someone verify this?

If for every sequence x_n → x, E contains all but finitely many x_n, than x is in the interior of E.

Thanks.

## Answers and Replies

Related Topology and Analysis News on Phys.org
Yes, this requires first countability. Here's a counterexample -- Let $X=[0, \omega_{1}]$ where $\omega_{1}$ is the first uncountable ordinal, and give X the order topology. Let $E=\{\omega_{1}\}$. Suppose $x_n\rightarrow \omega_{1}$. If it is not the case that cofinitely many of the x_n are in E, then the subsequence of all x_n not in E would be a sequence of countable ordinals converging to $\omega_{1}$, which is impossible. So for every sequence $x_n\rightarrow \omega_{1}$, all but finitely many of the x_n are in E. But $\omega_{1}$ is not in the interior of E. So the statement does not hold for spaces that are not first-countable.