Math4121 Lecture 20

Continue on Chapter 4

Properties of the Cantor Set

Monotonicity: If $S\subseteq T$ , then $c_e(S)\leq c_e(T)$ .

Sub-additivity: $c_e(S\cup T)\leq c_e(S)+c_e(T)$ .

Example: $S=\mathbb{Q}\cap[0,1]$ , $T=[0,1]\setminus\mathbb{Q}$ .

Then $c_e(S)=1$ , $c_e(T)=1$ , even though $S\cap T=\emptyset$ .

$S\cup T=[0,1]$ , $c_e(S\cup T)=1\leq 1+1=c_e(S)+c_e(T)$

The above example shows that:

The following is not true: $c_e(S\cup T)=c_e(S)+c_e(T)$ if $S\cap T=\emptyset$ .

However, the following is true:

(In $\mathbb{R}$ )

If $S=\bigcup_{n=1}^{\infty} I_n$ , $T=\bigcup_{n=1}^{\infty} J_n$ , where $I_n$ and $J_n$ are intervals, and $S\cap T=\emptyset$ , then $c_e(S\cup T)=c_e(S)+c_e(T)$ .

Back to Osgood’s Lemma

Osgood’s Lemma

Let $S$ be a closed, bounded set in $\mathbb{R}$ , and $S_1\subseteq S_2\subseteq \ldots$ , and $S=\bigcup_{n=1}^{\infty} S_n$ . Then $\lim_{k\to\infty} c_e(S_k)=c_e(S)$ .

Proof of Osgood's Lemma

Trivial that $c_e(S_k)\leq c_e(S)$ .

We need to show that $\forall \epsilon>0, \exists K$ such that $c_e(S_k)>c_e(S)-\epsilon$ for all $k\geq K$ .

Let $U_k$ be finite union of open intervals containing $S_k$ such that $c_e(U_k)<c_e(S_k)+\frac{\epsilon}{2^k}$ .

So $\{U_k\}_{k=1}^{\infty}$ are an open cover of $S$ .

Since $S$ is closed and bounded in $\mathbb{R}$ , it is compact.

So, $\exists N$ such that $S\subseteq \bigcup_{k=1}^{N} U_k$ .

Then we split the $U_k$ into two parts:

$U_k=(U_k\cap U_N)\cup (U_k\setminus U_N)$ , we denote $U_k^{(1)}=U_k\cap U_N$ , $U_k^{(2)}=U_k\setminus U_N$ , for $k\leq N$ .

So, since $U_k^{(1)}, U_k^{(2)}$ disjoint intervals, and $S_k\subseteq U_k^{(1)}$ , we have

\begin{aligned} c_e(U_k^{(1)})+c_e(U_k^{(2)})&=c_e(U_k)\\ c_e(S_k)+c_e(U_k^{(2)})&<c_e(S_k)+\frac{\epsilon}{2^k}\\ c_e(U_k^{(2)})&<\frac{\epsilon}{2^k}\\ \end{aligned}

So,

\begin{aligned} c_e(S)&\leq c_e(U)\\ &\leq c_e(U_N)+\sum_{k=1}^{N-1} c_e(U_k^{(2)})\\ &<c_e(S_N)+\frac{\epsilon}{2^{N}}+\sum_{k=1}^{N-1}\frac{\epsilon}{2^k}\\ &\leq c_e(S_N)+\epsilon\\ &<c_e(S_N) \end{aligned}

Convergence Theorems for sequences of functions

\lim_{n\to\infty}\int f_n(x)\ dx=\int \lim_{n\to\infty} f_n(x)\ dx

Yes when $f_n\to f$ uniformly.

Uniform convergence also means $\lim_{n\to\infty} \sup_{x\in [a,b]}|f_n(x)-f(x)|=0$ .

But there exists some cases that does not converge to the limit but still satisfies the above condition.

Theorem 4.5 (Arzela-Osgood Theorem)

If $\{f_n\}_{n=1}^{\infty}$ is a sequence of continuous, uniformly bounded function and $f(x)=\lim_{n\to\infty} f_n(x)$ exists for all $x\in [a,b]$ (pointwise convergence), then

\lim_{n\to\infty}\int_a^b f_n(x)\ dx=\int_a^b f(x)\ dx

Proof of Arzela-Osgood Theorem (incomplete)

Define $\Gamma_{\alpha}=\{x:\forall m\in \mathbb{N} \textup{ and }\forall \delta>0, \exists n\geq m \textup{ s.t. } |y-x|<\delta \textup{ and } |f_n(y)-f_m(y)|>\alpha\}$ .

$\Gamma_{\alpha}$ is the negation of $(\alpha,\delta)$ definition of limit.

$\Gamma_{\alpha}$ is closed and nowhere dense.

Continue on next lecture.