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:

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):

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

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

We choose $n$ large enough such that

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 $n\geq N$,

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. $p_n=\sin n$, $E^*=[0,1]$