Math4121 Lecture 18

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

Example:

[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)

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

C_n\subseteq C_{n-1}
\ell(C_n)=\ell(C_{n-1})
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.

Proof:

(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

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