Point-Set Question

  • Thread starter Poopsilon
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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

  • #2
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Yes, this requires first countability. Here's a counterexample -- Let [itex]X=[0, \omega_{1}][/itex] where [itex]\omega_{1}[/itex] is the first uncountable ordinal, and give X the order topology. Let [itex]E=\{\omega_{1}\}[/itex]. Suppose [itex]x_n\rightarrow \omega_{1}[/itex]. 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 [itex]\omega_{1}[/itex], which is impossible. So for every sequence [itex]x_n\rightarrow \omega_{1}[/itex], all but finitely many of the x_n are in E. But [itex]\omega_{1}[/itex] is not in the interior of E. So the statement does not hold for spaces that are not first-countable.
 

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