Is the Cantor set made of interval endpoints?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $3$.

Answer 4

Just because the open intervals approach a number, doesn't mean the number is the endpoint of one of the intervals. Consider:

$$D_1= \bigcup\limits_{\substack{n~\in~\mathbb{N},~n~\text{odd} \\}} \left(\frac{1}{n+1}, \frac{1}{n}\right). $$

This looks like:

$$$$ Next, let $$D_2= \bigcup\limits_{\substack{n~\in~\mathbb{N},~n~\text{odd} \\}} \left(\frac{-1}{n}, \frac{-1}{n+1}\right). $$

This looks like:

Now let $D = D_1 \cup D_2.\ $ Then $0 \in D^c\ $ and $0$ is a limit point of $D^c$. But $0$ is not the end-point of any of the intervals in $D$. The same goes for members of the Cantor set- most members of the Cantor set, it turns out.

Just because the intervals "close in" on $0\ $ doesn't make $0$ an endpoint of one of the intervals.