Perfect sets and Cantor sets, II

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

  • Case 1. W contains some bounded closed interval [a,b].
  • Case 2. W does not any bounded closed interval.

If case 1 holds, then we can apply the middle third process on [a,b] 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 [a,b], there is some x \in [a,b] such that x \notin W.

Background Discussion
Let A \subset \mathbb{R} and p \in A. The point p is a right-sided limit point of A if for each open interval (a,b) containing p, the open interval (p,b) contains a point of A. The point p is a left-sided limit point of A if for each open interval (a,b) containing p, the open interval (a,p) contains a point of A. The point p is a two-sided limit point of A if it is both a right-sided limit point and a left-sided limit point of A. 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 X \subset \mathbb{R} is an uncountable set. Then X contains a two-sided limit point.

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

Lemma 2
Suppose E \subset \mathbb{R} is a nonempty perfect set that satisfies Case 2 indicated above. Suppose that for the closed interval [a,b], we have:

  • (a,b) \cap E \ne \phi,
  • the left endpoint a is a right-sided limit point of E,
  • the right endpoint b is a left-sided limit point of E.

Then we have a<a^*<b^*<b such that:

  • there are no points of E in the open interval (a^*,b^*),
  • the point a^* is a left-sided limit point of E,
  • the point b^* is a right-sided limit point of E.

Proof. Since Case 2 holds, for the closed interval [a,b] in question, there is some x \in (a,b) such that x \notin E. Then we can find an open interval (c,d) such that x \in (c,d) and a<c<d<b and (c,d) \cap E = \phi.

Any point in (c,d) is an upper bound of W_1=[a,c) \cap E. By the least upper bound property, W_1 has a least upper bound a^*. Any point in (c,d) is an lower bound of W_2=(d,b] \cap E. By the greatest lower bound property, W_2 has a greatest lower bound b^*. Then a^* and b^* satisfy the conclusion of the lemma. \blacksquare

Lemma 3
Suppose E \subset \mathbb{R} is a nonempty perfect set. Suppose we have a closed interval [s,t] such that the left endpoint s is a right-sided limit point of E and the right endpoint t is a left-sided limit points of E. Then we have s<s_*<t_*<t such that:

  • the open interval (s_*,t_*) contains points of E,
  • both endpoints s_* and t_* are two-sided limit points of E,
  • t_*-s_*<0.5(t-s).

Proof. Suppose we have a closed interval [s,t] as described in the lemma. Then E_1=[s,t] \cap E is a nonempty perfect set. Thus E_1 is uncountable. So pick p \in (s,t) such that p is a two-sided limit point of E_1.

Choose open interval (c,d) such that s<c<p<d<t and d-c<0.5(t-s). Since p is a two-sided limit point of E_1, choose s^* and t^* such that c<s^*<p, p<t^*<d and both s^* and t^* are two-sided limit points of E_1. It follows that s^* and t^* satisfy the conclusion of the lemma. \blacksquare

Lemma 4
Suppose E \subset \mathbb{R} is a nonempty perfect set that satisfies Case 2 indicated above. Suppose we have a closed interval [a,b] such that the endpoints are two-sided limit points of E. Then we have disjoint closed intervals K_0=[p_0,q_0] and K_1=[p_1,q_1] such that

  • K_0 \subset [a,b] and K_1 \subset [a,b],
  • the lengths of both K_0 and K_1 are less then 0.5(b-a),
  • for each of K_0 and K_1, the endpoints are two-sided limit points of E.

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 [a,b] and obtain the open interval (a^*,b^*). We remove the open interval (a^*,b^*) from [a,b] and obtain two disjoint closed intervals [a,a^*] and [b^*,b]. Each of these two subintervals contains points of the perfect set W since the endpoints are limit points of W in the correct direction.

Now apply Lemm 3 to shrink [a,a^*] to obtain a smaller subinterval K_0=[p_0,q_0] such that the length of K_0 is less than 0.5(a^*-a) and is thus less than 0.5(b-a). Likewise, apply Lemma 3 on [b^*,b] to obtain K_1=[p_1,q_1] such that the length of K_1 is less than 0.5(b-b^*) and is thus less than 0.5(b-a). Note that both K_0 and K_1 constain points of E, making both K_0 \cap E and K_1 \cap E perfect sets and compact sets. \blacksquare

Construction
Suppose W \subset \mathbb{R} is a nonempty perfect set that satisfies Case 2. Pick two two-sided limit a_0 and b_0 of W. Obtain B_0=K_0 and B_1=K_1 as a result of applying Lemma 4. Let A_1=B_0 \cup B_1.

Then we apply Lemma 4 on the closed interval B_0 and obtain closed intervals B_{00}, B_{01}. Likewise we apply Lemma 4 on the closed interval B_1 and obtain closed intervals B_{10}, B_{11}. Let A_2=B_{00} \cup B_{01} \cup B_{10} \cup B_{11}.

Continue this induction process. Let C=\bigcap \limits_{n=1}^{\infty} A_n. The set C 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 C \subset W. For any countably infinite sequence g of zeros and ones, let g_n be the first n terms in g. Let y \in C. Then \left\{y\right\}=\bigcap \limits_{n=1}^{\infty} B_{g_n} for some countably infinite sequence g of zeros and ones (see post #6 listed below). Then every open interval (a,b) containing y would contain some closed interval B_{g_n}. Thus y is a limit point of W. Hence y \in W.

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

Advertisements

One thought on “Perfect sets and Cantor sets, II

  1. 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.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s