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Zheyuan Wu 717520624d upgrade structures and migrate to nextra v4

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Lecture 17

Review

Given a sequence (a_n) in \mathbb{R}, let E_n=\{a_k:k\geq n\}. Calculate diam E_1, diam E_2, diam E_3... for the following sequences:

a_n=0: E_n=\{0\}, diam E_1=0, diam E_2=0, diam E_3=0, \ldots
a_n=n: E_n=\{n\}, diam E_1=\infty, diam E_2=\infty, diam E_3=\infty, \ldots
a_n=(-1)^n: E_n=\{-1,1\}, diam E_1=2, diam E_2=2, diam E_3=2, \ldots
a_n=1/n: E_n=\{1/n,1/(n+1),\dots\}, diam E_1=\frac{1}{2}, diam E_2=\frac{1}{3}, diam E_3=\frac{1}{4}, \ldots
a_n=\frac{(-1)^n}{n}: E_n=\{-1/n,1/n,\dots\}, diam E_1=\frac{1}{1}+\frac{1}{2}, diam E_2=\frac{1}{2}+\frac{1}{3}, diam E_3=\frac{1}{3}+\frac{1}{4}, \ldots

New materials

Cauchy sequence

Theorem 3.11

(b) If X is a compact metric space, then every Cauchy sequence (p_n) in X converges.

Proof:

(b) Let E_N=\{p_n:n\geq N\}. Since (p_n) is Cauchy, \lim_{N\to\infty} diam E_N=0. By Theorem 3.10 (a), \lim_{N\to\infty} diam \overline{E_N}=0.

Since X is compact, and \overline{E_N} is closed, by Theorem 2.35, \overline{E_N} is compact.

Since E_1\supset E_2\supset E_3\supset\cdots, \overline{E_1}\supset \overline{E_2}\supset \overline{E_3}\supset\cdots. By Theorem 3.10(b), \exists p\in X such that p\in\bigcap_{N=1}^{\infty}\overline{E_N}.

We claim that (p_n) converges to p. Let \epsilon>0, there exists N_0 such that \forall N\geq N_0, diam \overline{E_N}<\epsilon.

For any n\geq N_0, p_n\in \overline{E_{N_0}}.

So d(p_n,p)\leq diam \overline{E_{N_0}}<\epsilon, by definition of diameter.

Therefore, (p_n) converges to p.

By Theorem 3.9, (p_n) is bounded. So \exists R>0 such that p_n\in B(0,R) for all n. Moreover p_n\in \overline{B(0,R)}. and \overline{B(0,R)} is closed and bounded. Thus by Theorem 2.41, \overline{B(0,R)} is compact.

Note that Theorem 2.41 only works for \mathbb{R}^k.

So by (b), (p_n) converges to some p\in \overline{B(0,R)}.

QED

Definition 3.12

Let X be a metric space. We say X is complete if every Cauchy sequence in X converges.

Theorem 3.11(b) can also be rephrased as:

X is a compact metric space \implies X is complete.

Theorem 3.11(c) can also be rephrased as:

\mathbb{R}^k is complete.

Note: completeness is a property of the "universe" X, not a property of any particular sequence in X.

\mathbb{Q} is not complete. \{3,3.1,3.14,3.141,3.1415,\dots\} is a Cauchy sequence in \mathbb{Q} but it does not converge in \mathbb{Q}.

Fact: If X is complete and E is a closed subset of X, then E is complete.

Definition 3.13

A sequence (s_n) of real numbers is said to be

monotone increasing if s_n\leq s_{n+1} for all n.
monotone decreasing if s_n\geq s_{n+1} for all n.
strictly monotone increasing if s_n<s_{n+1} for all n.
strictly monotone decreasing if s_n>s_{n+1} for all n.
monotone if it is either monotone increasing or monotone decreasing.

Example:

s_n=1/n is strictly monotone decreasing.
s_n=(-1)^n is neither monotone increasing nor monotone decreasing.

Theorem 3.14

Suppose (s_n) is monotonic. Then (s_n) converges \iff (s_n) is bounded.

Proof:

If (s_n) is monotonic and bounded, then by previous result, (s_n) converges.

If (s_n) is monotonic and converges, then by Theorem 3.2(c), (s_n) is bounded.

QED

Upper and lower limits

Definition 3.15 (Divergence to `\infty` or `-\infty`)

Let (s_n) be a sequence of real numbers with the following properties:

For every real number M there is an integer N such that n\geq N implies s_n>M. We then write s_n\to\infty

For every real number M there is an integer N such that n\geq N implies s_n<M. We then write s_n\to-\infty.

for every real number, we can find a element in the sequence that is greater than or less than it

Definition 3.16

Let (s_n) be a sequence of real numbers.

Let E_n=\{x\in[-\infty,\infty]:\exists \textup{ subsequence } (s_{n_k})\textup{ of } (s_n)\textup{ such that } s_{n_k}\to x\}.

Let S^*=\sup E_n, S_*=\inf E_n.

We define \limsup_{n\to\infty} s_n=S^* and \liminf_{n\to\infty} s_n=S_*

Informally, S^* is the largest possible value that a subsequence of (s_n) can converge to.