# 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 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 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 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 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, $p 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 $y \in W$.