This is post #9 of the series on the Euclidean topology of the real line. See the links at the bottom for other posts in the series.

Recall that a subset of the real line is a perfect set if is closed in Euclidean topology of the real line and that every point of is a limit point of . Any closed and bounded interval is a perfect set. The Cantor sets (the middle third version and other variations) are perfect sets (see the links #7 and #8 below). It turns out that any nonempty perfect set contains a Cantor set. In this series of posts on Euclidean topology of the real line, by Cantor sets we mean any set that can be constructed by a binary process of splitting closed intervals into two halves at each stage (see links for #6 and #8 below). We demonstrate the algorithm of constructing a Cantor set from any perfect set. This post (part I) shows that any nonempty perfect set is uncountable. Knowing that a perfect is uncountable will simplify the construction process (next post).

Suppose is a nonempty perfect subset. We show that is uncountable. Since has at least one point and every point is a limit point, is infinite. The key to showing is uncountable is that every nested decreasing sequence of compact subsets of the real line (actually in any topological space) has nonempty intersection. If happens to be countable, we can define a nested sequence of compact subsets of with empty intersection. Thus cannot be countable.

The following lemma is a corollary to Theorem 3 in the post # 4 listed below. The lemma applies to any abstract spaces where compactness can be defined. We state the lemma in terms of the real line since this is our focus.

**Lemma**

Suppose are compact subsets of the real line such that

.

Then .

To make the argument that is uncountable more precise, suppose that is countable. Then we can enumerate in a sequence indexed by the positive integers. We have:

Pick a bounded open interval such that . Next, pick an open interval such that and and .

In the stage where , pick an open interval such that and and . Since is a perfect set, the induction step can continue at every stage.

Now, let . Note that is a compact set since is compact. By the lemma, the intersection of the must be nonempty. By the induction steps, no point of belongs to all the sets , implying the intersection of is empty, a contradiction. Thus must be uncountable.

**Remark**

As a corollary, the real line and the unit intervals are uncountable. A more interesting corollary is that any nonempty perfect set has a two-sided limit point. In fact all but countably many points of a nonempty perfect set are two sided limit points. See the post The Lindelof property of the real line for a proof that any uncountable subset of the real line has a two sided limit point. This fact will simplify the construction of a Cantor set from a perfect set.

* 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

*Reference*

- Rudin, W.,
*Principles of Mathematical Analysis, Third Edition*, 1976, McGraw-Hill, Inc, New York.

Interesting proof, but I have a question, does exist a direct proof of the fact “perfect sets are uncountable”? I mean, show a bijection between a perfect set X and real numbers.