Lecture 23 - Perfect Sets and Connected Sets

Definition

Given $P \subseteq R$ then $P$ is perfect if $P$ is closed and $P$ has no Isolated Points.

For example, if $a < b$ :

$P = [a, b]$ is a perfect set.
In contrast if $P = [a, b] \cup {c}$ ( $c > b$ ), then this set is not perfect since it has an isolated point.

See its use in Perfect Sets are Uncountable.

Using this we can try to show a cool theorem:

Theorem

If $P \subseteq R$ is perfect then $P$ is uncountable.

Proof

We'll use the Nested Compact Set Property to show this, but let's prove it by contradiction.

Let $P$ be perfect and for the sake of contradiction let $P$ be countable (obviously it can't be infinite since then any point is isolated, so then $P$ wouldn't be perfect). Then:
$P = {x_{1}, x_{2}, x_{3}, \dots}$
We'll construct a sequence of compact subsets $K_{n} \subseteq P$ such that we'll have $x_{1} \notin K_{2}$ , $x_{2} \notin K_{3}$ , and so on. Some care should be taken to show that each $K_{n}$ is nonempty, but using the Nested Compact Set Property to produce an:
$x \in ⋂_{n = 1}^{\infty} K_{n} \subseteq P$
that cannot be on the list ${x_{1}, x_{2}, x_{3}, \dots}$ .

Let $I_{1}$ be a closed interval that contains $x_{1}$ . Then $x_{1}$ is not an endpoint of $I_{1}$ so then $x_{1}$ is not isolated so $\exists y_{1} \neq x_{1}$ where $y_{1} \in P$ such that $y_{1} \in I_{1}$ .

then let $I_{2}$ is the closed subinterval centered on $y_{2}$ of $I_{1}$ . Namely if $I_{1} = [a, b]$ let:
$ε = min {y_{1} - a, b - y_{1}, | x_{1} - y_{1} |}$
then the interval $I_{2} = [y_{1} - \frac{ε}{2}, y_{1} + \frac{ε}{2}]$ has the desired properties.

Repeat this process. Because $y_{1}$ is not isolated then $\exists y_{2}$ in the interior of $I_{2}$ and we can say $y_{2} \neq x_{2}$ . Now construct $I_{3}$ centered on $y_{2}$ and small enough so that $x_{2} \notin I_{3}$ and $I_{3} \subseteq I_{2}$ . Observe that $I_{3} \cap P \neq \emptyset$ because the set contains $y_{2}$ .

Carrying out the construction inductively, the result is a sequence of $I_

The Cantor set is perfect.

Recall the cantor set $C$ (see also Lecture 19 - Cantor Set):

$C_{0} = [0, 1]$
$C_{1} = [0, \frac{1}{3}] \cup [\frac{2}{3}, 1]$
$C_{2} = [0, \frac{1}{9}] \cup [\frac{2}{9}, \frac{1}{3}] \cup [\frac{2}{3}, \frac{7}{9}] \cup [\frac{8}{9}, 1]$
...

We'll show that the Cantor set is perfect.

Proof

Let $x \in C$ . For each $n \in N$ if $x \in C_{n}$ then $\exists I_{n}$ (some subinterval) of length $\frac{1}{3^{n}}$ in which $x$ lies.

Now let $x_{n}$ be an endpoint of $I_{n}$ such that $x \neq x_{n}$ . Now note that $x_{n} \in C$ (since $C$ is closed) but then $x_{n} \to x$ so the $x$ is not isolated. As a result then $C$ must be perfect.

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C

can have no interval as a subset.

Separated, Disconnected

$A, B$ are separated if both:
$\overset{―}{A} \cap B = \emptyset, A \cap \overset{―}{B} = \emptyset$
If $E \subseteq R$ can be written as $E = A \cup B$ where $A, B \neq \emptyset$ and $A, B$ are separated, then $E$ is disconnected.

This is an implication definition: Suppose

E = A \cup B

where

A, B

nomempty and separated implies that

E

is disconnected.

The idea here is that the limit points of $A$ aren't in $B$ , and vice versa.

connected

A set is connected if it is not disconnected. Namely, $E$ is connected for any $A, B \neq \emptyset$ and $A, B$ are not separated ( $\overset{―}{A} \cap B \neq \emptyset$ or $A \cap \overset{―}{B} \neq \emptyset$ ) then $E \neq A \cup B$ .

Examples

$A = (0, 1), B (1, 2)$ then these two sets are separated (the closure of $A$ throws on $0, 1 \notin B$ and the closure of $B$ with the limit points $1, 2$ are not in $A$ neither). Thus $E = A \cup B = (0, 1) \cup (1, 2)$ must be a disconnected set (by construction).

Note that

A, B

being disjoint is not equivalent. For example

A = (0, 1), B = [1, 2)

are not separated since the limit point

1

from

A

is in

B

. Namely:

$\overset{―}{A} \cap B = {1} \neq \emptyset$

Let $E = Q$ . Is $E$ disconnected or connected? Since $Q$ is dense in $R$ , then it is disconnected via the following construction. Let $A = {x \in Q | x < \sqrt{2}}$ and $B = {x \in Q : x > \sqrt{2}}$ . Here $A, B \neq \emptyset$ and $A, B$ are indeed separated. However, $\overset{―}{A} = (- \infty, \sqrt{2}]$ and $\overset{―}{B} = [\sqrt{2}, \infty)$ but $A \cap \overset{―}{B} = \overset{―}{A} \cap B = \emptyset$ showing that $E = A \cup B$ must be disconnected.

We can prove an interesting theorem for being a connected set:

Condition for Connected

Let $E \subseteq R$ . $E$ is connected iff $\forall a < b \in E$ and $\forall c \in R$ if $a < c < b$ then $c \in E$ .

Really this is saying that $E$ is an 'interval', in the more traditional sense.

Proof

#todo/school : Read this proof up 📅 2024-11-05 ✅ 2024-11-06

( $\Rightarrow$ ): Suppose $E$ is connected, and let $a, b \in E$ and $a < c < b$ . Set:
$A = (- \infty, c) \cap E; B = (c, \infty) \cap E$
Because $a \in A$ and $b \in B$ , then neither set is empty and neither set contains a limit point of the other (see Separated, Disconnected, Connected Sets#Examples to see how that works, namely example 2). If $E = A \cup B$ , then we would have that $E$ is disconnected, which it is not (by our given). It must then be that $A \cup B$ is missing some element of $E$ and $c$ is the only possibility. Thus, $c \in E$ .

( $\Leftarrow$ ): Suppose $\forall a < b \in E$ and $\forall c \in R$ that $(a < c < b \to c \in E)$ . To show that $E$ is connected, assume $E = A \cup B$ for contradiction, where $A, B$ are nonempty and disjoint (being separated allows us to get disjoint for free). We need to show that one of these sets contains a limit point of the other. Pick $a_{0} \in A$ and $b_{0} \in B$ and, suppose (for the sake of argument) that $a_{0} < b_{0}$ . Because $E$ itself is an interval, then interval $I_{0} = [a_{0}, b_{0}] \subseteq E$ . Now, bisect $I_{0}$ into two equal halves. The midpoint of $I_{0}$ must either be in $A$ or $B$ , so choose $I_{1} = [a_{1}, b_{1}] \subseteq E$ to be the half that allows us to have $a_{1} \in A$ and $b_{1} \in B$ . Continue this process, where each $I_{n} = [a_{n}, b_{n}] \subseteq E$ where $a_{n} \in A$ and $b_{n} \in B$ and the length $(b_{n} - a_{n}) \to 0$ . By the Nested Interval Property then $\exists x$ where:
$x \in ⋂_{n = 0}^{\infty} I_{n}$
It is straightforward to see that $a_{n} \to x$ and $b_{n} \to x$ . But now $x \in E$ must belong to either $A$ or $B$ , thus making it a limit point of the other, contradicting $A, B$ being separated. As such, then $E \neq A \cup B$ so then $E$ must be connected.

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s that are closed where: 1. $I_{n+1} \subseteq I_{n}$ 2. $x_{n} \not \in I_{n+1}$ 3. $I_{n} \cap P \neq \emptyset$

To finis the proof we let $K_{n} = I_{n} \cap P$ . For each $n \in N$ we have $K_{n}$ is closed because it is the intersection of closed sets, and bounded because it is contained in the bounded set $I_{n}$ . Hence $K_{n}$ is compact, so by construction then $K_{n}$ is not empty and $K_{n + 1} \subseteq K_{n}$ . Thus using the Nested Compact Set Property then:
$⋂_{n = 1}^{\infty} K_{n} \neq \emptyset$
But since each $K_{n} \subseteq P$ and since $x_{n} \notin I_{n + 1}$ then $⋂_{n = 1}^{\infty} K_{n} = \emptyset$ which is our contradiction.

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!The Cantor Set is Perfect

!Separated, Disconnected, Connected Sets

We can prove an interesting theorem for being a connected set:

!Condition for Connected