Note: I originally submitted this proof on StackOverflow. Reposting here to test out .
Let be a nonempty perfect set in . Then is uncountable.
Proof: Since is non-empty, and every point of is a limit point, contains at least one limit point. Hence, is infinite (by 2.20).
Suppose is countable and denote the points of by
We can construct a sequence of neighborhoods as follows:
Let be any neighborhood of . Then is non-empty, because is a limit point of .
If consists of all such that , the closure of is (note: I will not prove this rigorously, but see Lemma 1 below for a sketch of the proof).
Suppose has been constructed so that is not empty and for some . Since every point of is a limit point of , there is a neighborhood such that
(iii) is not empty, and for some point in .
By (iii), satisfies our induction hypothesis, and the construction can proceed.
We now show how we can always construct such a neighborhood . Let's say that for some .
Just a quick note: it's a common point of confusion that people believe that must be in . While , we cannot assume that this holds true for any other , and at no point does the proof assume this (or need to assume this).
Let be some point in , such that (a.) , (b.) , and (c.) . It follows from Lemmas 2 and 3 that such a point exists.
Lemma 2: For , there exists some point such that , and such that and . Suppose not. Note that for , we have that and (by our choice of ). Therefore, and . So there is some neighborhood of (namely, the neighborhood with radius , where ) such that the only point of that neighborhood in is the point , which contradicts our assumption that is a limit point of .
Lemma 3: For , there exists some point such that , . Suppose not. Note that . Therefore, . So there is some neighborhood of (namely, the neighborhood with radius , such that the only point of that neighborhood in is the point , which contradicts our assumption that is a limit point of .
Let with chosen subject to the following conditions.
(1.) , and
By our choice of , and we have the following:
(I) satisfies (i):
If , then
[by the properties of a metric space]
[by our choice of ]
Thus, . Hence, satisfies (i).
(II) satisfies (ii):
If , then . Thus, . Hence, satisfies (ii).
(III) satisfies (iii):
Because was chosen to be in , we have that is non-empty for all neighborhoods of . Thus, satisfies (iii).
Let . Since is closed and bounded in , is compact (by 2.41). is closed (because is perfect). Thus, is closed (by 2.24(b)). Hence is compact (by 2.35, because is a closed subset of a compact set).
Since , no point of lies in (this is implied by the fact that is countable, hence for every , there is a that excludes from ). But , so this implies that is empty.
But each is nonempty (by (iii), and (by (i)). But this contradicts the corollary to 2.36. The theorem follows.
Lemma 1: is a limit point of if and only if or . Proof Sketch: suppose . Then for some . So and is not a limit point of . Now suppose . Then every neighborhood of contains a point in , so is a limit point of ).