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Point-Set Question

  1. Oct 30, 2011 #1
    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.

  2. jcsd
  3. Oct 30, 2011 #2
    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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