# Is the Cantor set made of interval endpoints?

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?

• For an extremely convoluted madness on a similar topic, see my answer here and the comments that ensued. Oct 10, 2012 at 22:13
• @Asaf: No-one should be subjected to the reading of the comments on that answer. Oct 10, 2012 at 22:30
• 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, 2012 at 22:31
• @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.) Oct 11, 2012 at 4:25
• 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. Oct 11, 2012 at 12:32

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.

• 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 $10$. Oct 11, 2012 at 2:21
• @MichaelHardy I presume you mean OEIS A054591 Apr 30, 2021 at 15:08

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.

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 $3$.

There are a couple other ways to conclude that the Cantor set is uncountable, which you can read more about in Charles Pugh's "Real Mathematical Analysis" 2nd edition. Since the arguments clarified a lot for me I figured it would be good to put them in here for others to see since they have not been mentioned already.

While this theorem does not give much intuition about the points included in the Cantor set which make it uncountable, we can use the fact that every complete, perfect, and nonempty metric space is uncountable, and show that the Cantor set meets these criteria to show that it is uncountable.

Now for a more illuminating explanation that relates to the expression in base $3$ argument, we create an address for each point in the Cantor set which consists of an infinite string of $0$'s and $2$'s which is determined as follows.

We know that $C=\bigcap_{n=1}^{\infty}C^n$ where each $C^n$ consists of the $2^n$ subdivisions of $[0,1]$ each of length $\frac{1}{3^n}$. So we create an address system which indicates to which of the $2^n$ subintervals a given point $p$ belongs, where the string up to the $n$-th digit will tell us the specific subinterval in $C^n$. The general idea is that a $0$ indicates belonging to the left subinterval after the removal of the middle third , while a $2$ indicates belonging to the right subinterval after the removal of the middle third of a given interval of $C^n$.

Let's consider the example with $p=\frac{1}{4}$. We know that the first splitting of the interval $[0,1]$ in $C^1$ consists of the two subintervals $C_0= [0, \frac{1}{3}]$ and $C_2=[\frac{2}{3}, 1]$. and that $\frac{1}{4} \in [0, \frac{1}{3}]$. Therefore the first number of the address string for $\frac{1}{4}$ is a $0$.

Now for $C^2$, we know this consists of $4$ subintervals of $[0,1]$, namely $C_{00} =[0, \frac{1}{9}]$, $C_{02}=[\frac{2}{9}, \frac{1}{3}]$, $C_{20}=[\frac{2}{3}, \frac{7}{9}]$, and $C_{22}=[\frac{8}{9},1]$. We know that $\frac{2}{9}<\frac{1}{4}< \frac{1}{3}$, we have that $\frac{1}{4}\in C_{02}=[\frac{2}{9}, \frac{1}{3}]$, making our address for $\frac{1}{4}$ is $02....$. Continuing on to $C^3$, $C^4$,... we get an infinite address string for $\frac{1}{4}$.

The set of all of these infinite address strings can be shown to be uncountable, and it is in bijection with the Cantor set. By the continuum hypothesis, any uncountable subset of $\mathbb{R}$ has a cardinality equal to $\mathbb{R}$. Therefore the Cantor set is uncountable.

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.