Let $F$ be a closed countable non-empty subset of $\Bbb R$. Prove that $F$ has an isolated point.
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Could anyone tell me what is the Negation of the statement? – Myshkin Feb 08 '13 at 06:57
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Here is a proof. (A perfect set is a closed set with no isolated points, so, requirement, you want to prove a perfect set of reals cannot be countable.) – Omar Antolín-Camarena Feb 08 '13 at 07:15
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@CityOfGod If $F$ has no isolated points then $F$ is not closed or $F$ is not countable. – JSchlather Feb 08 '13 at 07:45
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@JacobSchlather Thank you! – Myshkin Feb 08 '13 at 08:16
4 Answers
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If X is a countable complete metric space with no isolated points, then each singleton {x} in X is nowhere dense, and so X is of first category in itself. See wikipedia
Emanuele Paolini
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Hint: Every complete metric space without isolated points is uncountable.
See, Elementary Real and Complex Analysis, P 87.
M.Sina
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A perfect set is a closed set without isolated points. This answer shows that a (non-empty) perfect subset of $\Bbb R$ has $2^\omega=\mathfrak c$ points. In particular, it must be uncountable. Thus, a countable, closed subset of $\Bbb R$ cannot be perfect and must have an isolated point.
Brian M. Scott
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I would like to think in another direction.If there is a closed countable non-empty subset of R without any isolated point called F.Lets choose one point of it and then we can say there is a B(x, Δ) which has elements from F.We can then find a sequence of {Δ_x} and in the B_x=B(x, Δ_x) there always are points from F.we can let Δ_x from R,and then we can make a map from R to points in F.Because R is a uncountable set then F is an uncountable set.
xiaotong
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