## Example 5.2.13(b): Properties of the Cantor Set

The definition of the Cantor set is as follows: let*A*_{0}= [0, 1]

*n*, the sets

*A*recursively as

_{n}*A*_{n}= A_{n-1}\

**C**= A_{n}

**C**is closed. Next, we need to show that every point in the Cantor set is a limit point.

One way to do this is to note that each of the sets
*A _{n}* can be written as a finite union of

*2*closed intervals, each of which has a length of

^{n}*1 / 3*, as follows:

^{n}*A*_{0}= [0, 1]*A*_{1}= [0, 1/3] [2/3, 1]*A*_{2}= [0, 1/9] [2/9, 3/9] [6/9, 7/9] [8/9, 1]- ...

*x*Then

**C**= A_{n}*x*is in

*A*for all

_{n}*n*. If

*x*is in

*A*, then

_{n}*x*must be contained in one of the

*2*intervals that comprise the set

^{n}*A*. Define

_{n}*x*to be the left endpoint of that subinterval (if

_{n}*x*is equal to that endpoint, then let

*x*be equal to the right endpoint of that subinterval). Since each subinterval has length

_{n}*1 / 3*, we have:

^{n}*| x - x*_{n}| < 1 / 3^{n}

*{ x*converges to

_{n}}*x*, and since all endpoints of the subintervals are contained in the Cantor set, we have found a sequence of numbers contained in

**C**that converges to

*x*. Therefore,

*x*is a limit point of

**C**. But since

*x*was arbitrary, every point of

**C**is a limit point. Since

**C**is also closed, it is then perfect.

Note that this proof is not yet complete. One still has to prove the
assertion that each set
*A _{n}* is indeed comprised of

*2*closed subintervals, with all endpoints being part of the Cantor set. But that is left as an exercise.

^{n}Since every perfect set is uncountable, so is the Cantor.