We know by BCT that in a complete metric space the countable intersection of open dense sets is dense. Will it also be open?
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No.
Take the irrational numbers $\Bbb{I}$ on the real line.
We have that $\Bbb{I}=\bigcap_{n=1}^{\infty}O_n$ where $O_n$ are dense open sets.
But $\Bbb{I}$ is not open.
answered Dec 4 '19 at 18:20
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1$\begingroup$ To make things explicit, we can take $O_n=\Bbb R\setminus\frac1n\Bbb Z$ $\endgroup$ – Hagen von Eitzen Dec 4 '19 at 18:22
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1$\begingroup$ @Angry_Math_Person yes it was a typo...thank you. $\endgroup$ – Marios Gretsas Dec 4 '19 at 18:23
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$\begingroup$ Ok just wondering is it possible the intersection is countable dense? $\endgroup$ – Angry_Math_Person Dec 4 '19 at 18:24
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1$\begingroup$ @Angry_Math_Person No, not unless there are isolated points in the whole space. If $X$ has no isolated points and is complete metric, then the intersection will be uncountable. $\endgroup$ – Henno Brandsma Dec 4 '19 at 22:59
