Lecture 19 - Cantor Set

We construct the Cantor Set in the following way:

C_{0} = [0, 1]

C_{1} = [0, \frac{1}{3}] \cup [\frac{2}{3}, 1] = C_{0} ∖ (\frac{1}{3}, \frac{2}{3})

C_{2} = C_{1} ∖ ((\frac{1}{9}, \frac{2}{9}) \cup (\frac{7}{9}, \frac{8}{9}))

And so on:

Notice that:

C_{0} \supset C_{1} \supset C_{2} \supset \dots

We define the final set as:

C = ⋂_{i = 0}^{\infty} C_{i}

What is in $C$ ? Well notice that:

0, 1, \frac{1}{3}, \frac{2}{3}, \dots \in C

$C$ has all the endpoints each of the form:

\frac{m}{3^{n}} \in C

So does this imply that $C$ is countably infinite? Consider instead asking how much length of the interval is there. We have $C_{0}$ has a length of 1:

The first step removes 1/3 of the length.
The second step removes 2/9 of the length
The next removes 4/27 of the length
...
We remove:

\frac{1}{3} + \frac{2}{9} + \dots = \frac{1}{3} (1 + \frac{2}{3} + {(\frac{2}{3})}^{2} + \dots)

This is a geometric series:

= \frac{1}{3} \cdot \frac{1}{1 - \frac{2}{3}} = 1

So this implies that $C$ should have no length. Maybe it is indeed countably infinite?

Another Perspective

Let $x \in C$ . Construct a sequence of $0$ 's and $1$ 's as follows (the example uses the case where $x = \frac{7}{9} \in C$ ):

\frac{7}{9} : 10 \overset{―}{1} = 101111111 \dots

Where the first digit says:

1 if it is in the right sub-interval after the first cut
0 if it is in the left sub-interval after the first cut

We repeat this process for each cut, adding digits. As another example:

\frac{1}{9} : 00 \overset{―}{1} = 0011111111 \dots

Now consider an arbitrary sequence of these digits. We want to do Cantor's Diagonalization Method to try to construct some new digit that isn't in the list:

(b_{1}, b_{2}, b_{3}, \dots)

We can change the first digit of $b_{1}$ , the second of $b_{2}$ , the third of $b_{3}$ , and so on to make a new digit $x$ .

Another way of thinking of this is that each $C_{i} \supseteq C_{i + 1}$ . This implies that we can use the $C_{i}$ 's with the Nested Interval Property to show that:

\exists x \in ⋂_{i = 1}^{\infty} C_{i} = C

But in both cases, there are uncountable many such sequences! So then $C$ must not be countable.

Dimensions of A Set

To set the stage, consider a 0-dimensional object, 1-dimensional object, ...

The idea from geometry is if you multiply an $n$ -dimensional object by $c$ , then the area/volume/... of that object is scaled by $c^{n}$ . In this case $c = 3$ so it magnifies by $3^{n}$ .

For the case of Cantor's set is if we scale each line by 3:

So somehow we had some $n$ -dimensional object, scaled by $3$ , where the area/volume/length here got scaled by 2! Not 3 as we expected. Here:

size \cdot 2 = 3^{n} \Rightarrow n = \log_{3} (2)

So somehow we have fractional dimension. This is where the word fractal comes from (fracttional dimension). This brings us into talking about the start of the new section:

Topology of $R$

Recall the definition of the Epsilon-Neighborhood of $x$ :

ε

-neighborhood of

a

$V_{ε} (a) = (a - ε, a + ε) = {x \in R | | x - a | < ε}$

open set

Let $U \subseteq R$ . Then $U$ is open if:
$\forall x \in U \exists ε > 0 (V_{ε} (x) \subseteq U)$
ie: each $x \in U$ has some $ε$ -neighborhood contained in $U$ .

Example: $(a, b)$

$(a, b)$ is open.

Since choose $ε = min (| x - a |, | x - b |)$ implies that $V_{ε} (x) \subseteq (a, b)$ .

By constrast if we considered even a one-sided interval like $(a, b]$ then it is NOT open:

Notice if $x = b$ then any $\varepsilon > 0$ would have some element in $V_\varepsilon(b)$ be outside the interval.

Connecting it to Sequences

Definition

Let $A \subseteq R$ and $x \in R$ . Then
$x$ is a limit point of $A$ if every Epsilon-Neighborhood of $x$ intersects $A$ at a point other than $x$ . Namely:
$\forall ε > 0 (V_{ε} (x) \cap A \neq {x} \land \emptyset)$

For example, if $A = (a, b]$ , then what are the limit points of $A$ ? The limit points of $A$ are:
$= a, b$

Alternative Definition of Limit Point

Let $A \subset R$ and $x \in R$ . Then $x$ is a limit point of $A$ iff $\exists (a_{n})$ :
$a_{n} \in A ∖ {x}$
for all $n$ and $a_{n} \to x$ .

Proof

( $\Rightarrow$ ): Suppose $x$ is a limit point of $A$ . Let's apply a sequence of neighborhoods such that we can construct $a_{n}$ . To do this, let $n \in N$ . Then choose:
$a_{n} \in (V_{\frac{1}{n}} (x) ∖ {x}) \cap A$
we claim that $a_{n} \to x$ . To prove this via the definition let $ϵ > 0$ . Choose $N \in N$ such that $\frac{1}{N} < ϵ$ . Then suppose $n \geq N$ then:
$\begin{aligned} | a_{n} - x | & < \frac{1}{n} & (a_{n} \in V_{\frac{1}{n}} (x)) \\ \leq \frac{1}{N} \\ < ϵ \end{aligned}$
( $\Leftarrow$ ): Suppose $\exists a_{n}$ where $a_{n} \in A - {x}$ for all $n$ and that $a_{n} \to x$ . Let $ϵ > 0$ so that $V_{ϵ} (x)$ is an epsilon-neighborhood of $x$ . We want to show that $(V_{ϵ} (x) - {x}) \cap A \neq \emptyset$ . We can choose a $N \in N$ such that:
$| a_{n} - x | < ϵ$
by the definition and that $a_{n} \to x$ . Then definitely $| a_{N} - x | < ϵ$ , so then $N \in V_{ϵ} (x)$ . Furthermore clearly $a_{N} \in A$ while $a_{N} \neq x$ so then $a_{N} \in (V_{ϵ} (x) - {x}) \cap A$ is not empty as desired.

☐

To illustrate this, consider:
$A = {\frac{1}{n} | n \in N}$
Then the set of all limit points of $A$ is just ${0}$ . Notice other points like $\frac{1}{2}$ are not in this set since there's no sequence of numbers constructing from the set cannot approach $\frac{1}{2}$ .

Another example is when $A = Q$ . The set of limit points of $A$ are now $R$ .

Another Perspective

Dimensions of A Set

Topology of R

Example: (a,b)

Connecting it to Sequences

Alternative Definition of Limit Point

Topology of $R$

Example: $(a, b)$