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The Cantor set is closed, so its complement is open. So the complement can be written as a countable union of disjoint open intervals. Why can we not just enumerate all endpoints of the countably many intervals, and conclude the Cantor set is countable?

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For an extremely convoluted madness on a similar topic, see my answer here and the comments that ensued. – Asaf Karagila Oct 10 '12 at 22:13
@Asaf: No-one should be subjected to the reading of the comments on that answer. – Arthur Fischer♦ Oct 10 '12 at 22:30
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The set of endpoints of these countably many intervals is strictly contained in the Cantor set. The Cantor set is perfect and therefore uncountable. – leo Oct 10 '12 at 22:31
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@Brian: If you think that would help I'd make an exception. But the good Herr Professor Doktor likely has them all enraptured by his "revolutionary" outsider status within the mathematics community. He doesn't exactly hide his "idio(t)syncratic" beliefs and I cannot imagine that the administration is ignorant of the fact that his take on mathematics stands in direct opposition to the overwhelming consensus of the vast majority within the community. (Either that or he will claim that someone is fraudulently using his name to discredit him.) – Arthur Fischer♦ Oct 11 '12 at 4:25
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The Cantor set is not only closed but also has empty interior. Maybe you had that in mind because this means that Cantor set equals it's own boundary and is thus equal to the boundary of the complement, a countable union of disjoint open intervals. The point is that the closure of this union is different from the union of the closures, and therefore the boundary of the complement, i.e. the Cantor set, contains many more points than the boundary points of those intervals. – Stefan H. Oct 11 '12 at 12:32
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3 Answers

up vote 12 down vote

Because the Cantor set includes numbers which are not the endpoints of any intervals removed. For example, the number $\frac{1}{4}$ (0.02020202020... in ternary) belongs to the Cantor set, but is not an endpoint of any interval removed.

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And $3/10$ is another. In the Online Encyclopedia of Integer Sequences, check out the denominators of rational numbers in the Cantor set. If I recall correctly, besides powers of $3$, and first ones are $4$ and $10$. – Michael Hardy Oct 11 '12 at 2:21
up vote 11 down vote

You cannot do that because a countable set can have an uncountably many limit points. The points in the Cantor set are limit points of these endpoints.

For example, the real numbers are all limit points of the rational numbers. If between every two real numbers there is a rational number, but we still can't establish that the real numbers are countable.

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up vote 11 down vote

The reason that does not work is that there are some points in the Cantor set which are not the endpoint of any interval that is removed during the construction of the Cantor set. In the proof that is suggested in the question, it would be necessary to show that every point in the set is one of these endpoints, but that just isn't true. The proof that the Cantor set is uncountable already shows there has to be at least one such "non-endpoint" point, because the set of endpoints is countable, as the question above points out. In fact we can give a specific example: the number $1/4$ is in the usual middle-thirds Cantor set, but it is not an endpoint of any interval that is removed, because all those endpoints are rationals whose denominator can be written as a power of $3$.

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