# Math4121 Lecture 18

## Continue

### Small sets

A set that is nowhere dense, has zero outer content yet is uncountable.

By modifying this example, we can find similar with any outer content between 0 and 1.

#### Definition: Perfect Set

$S\subseteq[0,1]$ is perfect if $S=S'$.

<details>
<summary>Examples of perfect set</summary>

- $[0,1]$ is perfect
  - perfect sets are closed
- Finite collection of points is not perfect because they do not have limit points.
  - perfect sets are uncountable (no countable sets can be perfect)

</details>

#### Middle third Cantor set

We construct the set by removing the middle third of the interval.

Let $C_0=[0,1]$, $C_1=[0,\frac{1}{3}]\cup[\frac{2}{3}]$ ...

Continuing this process indefinitely, we define the Cantor set as

$$
C=\bigcap_{n=0}^{\infty}C_n
$$

1. $C_n\subseteq C_{n-1}$
2. $\ell(C_n)=\ell(C_{n-1})$
3. Each $C_n$ is closed.

> The algebraic expression for $C_n$, where $a\in[0,1]$, we write as a decimal expansion in base $3$.
>
> $$ a=\sum_{n=1}^{\infty} \frac{a_n}{3^n}$$, where $a_n\in\{0,1,2\}$.
>
> In this case, $C_0\to C_1$ means deleting all numbers with $a_1=1$. (the same as deleting the interval $[\frac{1}{3},\frac{2}{3}]$)
>
> $C_1\to C_2$ means deleting all the numbers with $a_2=1$.$
>
> So we can write the set as $$C=\left\{\sum_{n=1}^{\infty}\frac{a_n}{3^n},a_n\in\{0,2\}\right\}$$

#### Proposition 4.1

$C$ is perfect and nowhere dense, and outer content is 0.

<details>
<summary>Proof</summary>

(i) $c_e(C)=0$

Let $\epsilon>0$, then $\exists n$ such that $\left(\frac{2}{3}\right)<\epsilon$. Then $C_n$ is a cover of $C$, and $\ell(C_n)<\epsilon$.

(ii) $C$ is perfect

Since $C_n$ is closed, $C$ is closed (any intersection of closed set is closed) so $C'\subseteq C$.

Let $a\in C$, and we need to show $a$ is a limit point. Let $\epsilon>0$, and we need to find $a^*\in C\setminus\{a\}$ and $|a^* - a| < \epsilon$. Suppose $a=\sum_{n=1}^{\infty} \frac{a_n}{3^n}, a_n \in \{0, 2\}$, Notive that if $a^*\in C$ has the expansion as $a$ except the k-th term.

So $|a^*-a|=\frac{2}{3^k}$, which can be made arbitrarily small by choosing a sufficiently large $k$. Thus, $a$ is a limit point of $C$, proving that $C$ is perfect.

(iii) $C$ is nowhere dense

It is sufficient to show $C$ contains no intervals.

Any open intervals has a real number with 1 in it's base 3 decimal expansion (proof in homework)

_take some interval in $(a,b)$ we can change the digits that is small enough and keep the element still in the set_
</details>