This is post #11 of the series on the Euclidean topology of the real line. See the links at the bottom for other posts in the series.

In the previous post Perfect sets and Cantor sets, I, we show that every nonempty perfect set is uncountable. We now show that any perfect contains a Cantor set. Hence the cardinality of any perfect set is continuum.

Given a perfect set , we construct a Cantor set within . Consider the following cases:

- Case 1. contains some bounded closed interval .
- Case 2. does not any bounded closed interval.

If case 1 holds, then we can apply the middle third process on and produce a Cantor set. So in the remaining dicussion of this post, we assume case 2 holds. This means that for each closed interval , there is some such that .

**Background Discussion**

Let and . The point is a right-sided limit point of if for each open interval containing , the open interval contains a point of . The point is a left-sided limit point of if for each open interval containing , the open interval contains a point of . The point is a two-sided limit point of if it is both a right-sided limit point and a left-sided limit point of . For the proof of the following lemma, see the post labeled #10 listed below.

In Lemma 2 below, we apply the least upper bound property and the greatest lower bound property. See the post labeled #4 listed below.

**Key Lemmas for Construction**

**Lemma 1**

Suppose that is an uncountable set. Then contains a two-sided limit point.

As a corollary to the lemma 1, for the perfect set in question, all but countably many points of are two-sided limit points of .

**Lemma 2**

Suppose is a nonempty perfect set that satisfies Case 2 indicated above. Suppose that for the closed interval , we have:

- ,
- the left endpoint is a right-sided limit point of ,
- the right endpoint is a left-sided limit point of .

Then we have such that:

- there are no points of in the open interval ,
- the point is a left-sided limit point of ,
- the point is a right-sided limit point of .

* Proof*. Since Case 2 holds, for the closed interval in question, there is some such that . Then we can find an open interval such that and and .

Any point in is an upper bound of . By the least upper bound property, has a least upper bound . Any point in is an lower bound of . By the greatest lower bound property, has a greatest lower bound . Then and satisfy the conclusion of the lemma.

**Lemma 3**

Suppose is a nonempty perfect set. Suppose we have a closed interval such that the left endpoint is a right-sided limit point of and the right endpoint is a left-sided limit points of . Then we have such that:

- the open interval contains points of ,
- both endpoints and are two-sided limit points of ,
- .

* Proof*. Suppose we have a closed interval as described in the lemma. Then is a nonempty perfect set. Thus is uncountable. So pick such that is a two-sided limit point of .

Choose open interval such that and . Since is a two-sided limit point of , choose and such that , and both and are two-sided limit points of . It follows that and satisfy the conclusion of the lemma.

**Lemma 4**

Suppose is a nonempty perfect set that satisfies Case 2 indicated above. Suppose we have a closed interval such that the endpoints are two-sided limit points of . Then we have disjoint closed intervals and such that

- and ,
- the lengths of both and are less then ,
- for each of and , the endpoints are two-sided limit points of .

* Proof*. This is the crux of the construction of a Cantor set from a perfect set and is the result of applying Lemma 2 and Lemma 3.

Applying Lemma 2 on and obtain the open interval . We remove the open interval from and obtain two disjoint closed intervals and . Each of these two subintervals contains points of the perfect set since the endpoints are limit points of in the correct direction.

Now apply Lemm 3 to shrink to obtain a smaller subinterval such that the length of is less than and is thus less than . Likewise, apply Lemma 3 on to obtain such that the length of is less than and is thus less than . Note that both and constain points of , making both and perfect sets and compact sets.

**Construction**

Suppose is a nonempty perfect set that satisfies Case 2. Pick two two-sided limit and of . Obtain and as a result of applying Lemma 4. Let .

Then we apply Lemma 4 on the closed interval and obtain closed intervals , . Likewise we apply Lemma 4 on the closed interval and obtain closed intervals , . Let .

Continue this induction process. Let . The set is a Cantor set and has all the properties discussed in the posts labled #6 and #7 lised at the end of this post.

We claim that . For any countably infinite sequence of zeros and ones, let be the first terms in . Let . Then for some countably infinite sequence of zeros and ones (see post #6 listed below). Then every open interval containing would contain some closed interval . Thus is a limit point of . Hence .

* Links to previous posts on the topology of the real line*:

1. The Euclidean topology of the real line (1)

2. The Euclidean topology of the real line (2)

3. The Euclidean topology of the real line (3) – Completeness

4. The Euclidean topology of the real line (4) – Compactness

5. The Cantor bus tour

6. The Cantor set, I

7. The Cantor set, II

8. The Cantor set, III

9. Perfect sets and Cantor sets, I

10. The Lindelof property of the real line

Does anyone know how to prove the following: Any closed interval is the union of c pairwise disjoint perfect sets, where c is the power of the continuum? Would such sets need to be Cantor-like sets? This exercise was given in Natanson’s Volume 1 on Real Analysis.

Pingback: The cardinality of compact first countable spaces, I | Dan Ma's Topology Blog

Pingback: Counterexamples to the conjecture of Menger | Dan Ma's Topology Blog