Prove: $F$ is countable and closed $\implies F$ has an isolated point. Let $F$ be a closed countable non-empty subset of $\Bbb R$. Prove that $F$ has an isolated point.
 A: 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
A: Hint: Every complete metric space without isolated points is uncountable.
See, Elementary Real and Complex Analysis, P 87.
A: 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.
A: 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.
