Verify: Topology of Point-Set Question Needs to be First Countable

[Poopsilon]

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.

 

[Citan Uzuki]

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.

 

[Greg Bernhardt]

Yes, the statement does require the assumption that the topology is first countable. In a first countable space, every point has a countable neighborhood basis, meaning that for every point x, there exists a sequence of open sets U_n such that x is contained in each U_n and any open set containing x must contain one of the U_n. This is necessary in order for the statement to hold, as it allows us to construct a sequence x_n that converges to x and is contained in E, satisfying the condition that "for every sequence x_n → x, E contains all but finitely many x_n." Without this assumption, the statement may not hold in general.