NoteNextra-origin/pages/Math4111/Math4111_L15.md

Lecture 15

Review

Let (a_n)_{n=1}^\infty and (b_n)_{n=1}^\infty be sequence in \mathbb{R}. Let x_n=(a_n,b_n)\in \mathbb{R}^2, so (x_n)_{n=1}^\infty be a sequence in \mathbb{R}^2. Consider the following statement:

a_n\to a\textup{ and }\quad b_n\to b\iff x_n\to (a,b)

1.Prove the \impliedby direction. That means you should prove the two things:
(a) If x_n\to (a,b), then a_n\to a. (The proof of this begins: Suppose x_n\to (a,b). Let \epsilon>0 be arbitrary. Then \exists N such that \forall n\geq N)
We begins (with the goal \forall \epsilon>0,\exists N such that \forall n\geq N,|a_n-a|<\epsilon).
Proof:
Let \epsilon>0 be arbitrary, then \exists N such that \forall n\geq N,|a_n-a|<\epsilon. Then if n\geq N, |a_n-a|\leq \sqrt{|a_n-a|^2}\leq\sqrt{|a_n-a|^2+|b_n-b|^2}=|x_n-(a,b)|<\epsilon.
EOP
(b) If x_n\to (a,b), then b_n\to b. This follows from the same argument from (a) 2. Prove the \implies direction. Goal: \forall \epsilon>0,\exists N such that \forall n\geq N,|a_n-a|<\epsilon. Proof: Let \epsilon>0 be arbitrary.
Since a_n\to a, \exists N_1 such that \forall n\geq N_1,|a_n-a|<\epsilon.
Since b_n\to b, \exists N_2 such that \forall n\geq N_2,|b_n-b|<\epsilon.
Let N=\max\{N_1,N_2\}. Then if n\geq N, |a_n-a|<\epsilon and |b_n-b|<\sqrt{2}\epsilon.
Same as last time, we can choose any smaller epsilon.
Since a_n\to a, \exists N_1 such that \forall n\geq N_1,|a_n-a|<\frac{\epsilon}{\sqrt{2}}.
Since b_n\to b, \exists N_2 such that \forall n\geq N_2,|b_n-b|<\frac{\epsilon}{\sqrt{2}}.
Let N=\max\{N_1,N_2\}. Then if n\geq N, |a_n-a|<\epsilon and |b_n-b|<\sqrt{\frac{\epsilon^2}{2}+\frac{\epsilon^2}{2}}=\epsilon.
EOP

New Materials

Continue from Theorem 3.3

Suppose (s_n),(t_n) are sequences in \mathbb{C} and s_n\to s,t_n\to t. Then

(a) s_n+t_n\to s+t
(b) cs_n\to cs, c+s_n\to c+s
(c) s_nt_n\to st
(d) If \forall n\in \mathbb{N},s_n\neq 0, s\neq 0, then \frac{1}{s_n}\to \frac{1}{s}

Thought process for (d):

\left|\frac{1}{s_n}-\frac{1}{s}\right|=\left|\frac{s-s_n}{s_ns}\right|=\frac{|s-s_n|}{|s||s_n|}

We choose large enough N such that \forall n\geq N,|s_n-s|<\frac{|s|}{2}. Then by triangle inequality, |s_n|>\frac{|s|}{2}.

\begin{aligned}
|s|&=|s-s_n+s_n|\\
|s|&\leq |s-s_n|+|s_n|\\
|s|&<\frac{|s|}{2}+|s_n|\\
\frac{|s|}{2}&< |s_n|
\end{aligned}

So \frac{|s_n-s|}{|s||s_n|}<\frac{2|s_n-s|}{|s|^2}.

We choose n large enough such that

\frac{2|s_n-s|}{|s|^2}<\epsilon

Then |s_n-s|<\frac{\epsilon|s|^2}{2}.

Proof:

Let \epsilon>0, since s_n\to s

\exists N such that \forall n\geq N,|s_n-s|<\frac{|s|}{2}.

\exists N such that \forall n\geq N,|s_n-s|<\frac{\epsilon|s|^2}{2}.

Let N=\max\{N_1,N_2\}. Then if n\geq N,

\left|\frac{1}{s_n}-\frac{1}{s}\right|=\frac{|s-s_n|}{|s||s_n|}<\frac{\frac{\epsilon|s|^2}{2}}{|s|^2}=\epsilon

EOP

Subsequences

Definition 3.5

Given a sequence (p_n)_{n=1}^\infty, a sequence of (n_i)_{i=1}^\infty is strictly increasing sequence in \mathbb{N}. i.e. n_1<n_2<n_3<\cdots.

The sequence (p_{n_i})_{i=1}^\infty is called a subsequence of (p_n)_{n=1}^\infty.

Example:

p_n=\frac{1}{n},n_i=i^2, then the subsequence is (p_{n_i})_{i=1}^\infty=\left(\frac{1}{i^2}\right)_{i=1}^\infty. i.e. (\frac{1}{4},\frac{1}{9},\frac{1}{16},\cdots)

(p_n)_{n=1}^\infty converges to p if and only if every subsequence of (p_n)_{n=1}^\infty converges to p.

Proof:

\impliedby:

(p_{n_i})_{i=1}^\infty is a subsequence of (p_n)_{n=1}^\infty.

\implies:

Thought process: show what if the sequence does not converge to p, then there exists a subsequence that does not converge to p.

EOP

Theorem 3.6

(a) If (p_n) is a sequence in a compact metric space X, then (p_n) has a convergent subsequence converges to a point of X.

(b) If (p_n) is a bounded sequence in \mathbb{R}^k, then (p_n) has a convergent subsequence in \mathbb{R}^k.

Proof:

(a) Let E=\{p_n:n\in \mathbb{N}\}. Note that E is a set, not a sequence.

Case 1: E is finite.

Then some term appears infinitely many times. i.e \exists p\in E and subsequence (p_{n_i}) such that for all i, p_{n_i}=p.

Then (p_{n_i}) converges to p.

Case 2: E is infinite.

By Theorem 2.37, if E is an infinite subset of a compact set K, then E has a limit point in K.

p\in E'\implies \forall r>0, B_r(p)\cap E\backslash \{p\}\neq \phi

• Choose n_i such that p_{n_i}\in B_i(p)
• If n_1,\dots, n_{i-1} have bee chosen, choose n_i such that n_i>n_{i-1} and p_{n_i}\in B_{\frac{1}{i}}(p). Then p_{n_i}\to p

(b) Since (p_n) is bounded , \exists M such that \forall n\in N, p_n\in \overline{B_M(0)}=\{y\in\mathbb{R}^k:|y|\leq M\}

\overline{B_M(0)} is a closed and bounded set in \mathbb{R}^k.

Then by Theorem 2.41, \overline{B_M(0)} is compact.

By part (a), (p_n) has a subsequence (p_{n_i}) has a subsequence that converges to B_M(0).

Theorem 3.37

Let X be a metric space, (p_n) is a sequence in X.

Let E^*=\{p\in X:\exists\textup{ subsequence }(p_{n_i})\textup{ such that }p_{n_i}\to p\}.

Then E^* is closed in X.

Example:

X=\mathbb{R}

1. p_n=\frac{1}{n}, E^*=\{0\}. (Specifically, if p_n\to p, then E^*\to \{p\})
2. p_n=\begin{cases}1,n\textup{ is odd}\\ 0,n\textup{ is even}\end{cases}, E^*=\{0,1\}
3. p_n=n, E^*=\phi
4. p_n=\sin nx, E^*=\{0,1\}
5. p_n=\sin n, E^*=[0,1]