NoteNextra-origin/pages/Math4111/Math4111_L14.md

Lecture 14

Review

Consider the following statement: If sequence (p_n) converges, then its bounded.

1. Will the proof involve an arbitrary \epsilon>0 (one that you, the prover, do nto get to choose) or a specific \epsilon>0 (on that you can choose)
   We can choose, for example \epsilon=1.
2. Give a proof of the statement.

Continue on sequence

Convergence

Theorem 3.2(c)

(p_n) converges \implies(p_n) is bounded.

Proof:

Suppose (p_n) converges, then \exists p\in X such that p_n\to p. Let \epsilon=1, then \exists N\in \mathbb{N} such that \forall n\geq N,d(p_n,p)<1. Let r=1+max\{1,d(p_1,p),d(p_2,p),\dots,d(p_{N-1},p)\}.

Then \forall n\in \mathbb{N}, d(p_n,p)\leq r.

Theorem 3.2

Let (p_n) be a sequence in (X,d)

(a) p_n\to p\iff \forall r>0,\{n\in \mathbb{N},p_n\notin B_r(p)\} is finite
(b) p_n\to p; p_n\to p'\implies p=p' (converging point is unique)
(c) (p_n) converges \implies(p_n) is bounded.
(d) If E\subset X and p\in \overline{E}, then \exist (p_n)\in E such that p_n\to p.

Proof:

(a) We need to show:

\forall \epsilon>0 \in N, \forall n\geq N,d(p_n,p)<\epsilon if and only if \forall r>0, \{n\in \mathbb{N}:p\notin B_r(p)\} is finite.

\implies

Suppose \forall \epsilon>0 \in N, \forall n\geq N,d(p_n,p)<\epsilon.

We start with arbitrary r>0. and choose \epsilon=n

\exists N such that \forall n\geq \mathbb{N},d(p_n,p)<r.

Then \{n\in \mathbb{N}:p\notin B_r(p)\}<\{1,2,\dots,N-1\} is finite.

\impliedby

Suppose \forall r>0, \{n\in \mathbb{N}:p\notin B_r(p)\} is finite. Choosing r=\epsilon. We choose r=\epsilon. \{n\in \mathbb{N}:p\notin B_\epsilon(p)\}<\{1,2,\dots,N-1\}.

Let N=1+max\{n\in \mathbb{N},p_n\notin B_\epsilon(p)\}

Then \forall n\geq \mathbb{N},p_n\leq B_\epsilon(p)

(b) We'll prove \forall \epsilon>0,d(p,p')<2\epsilon to prove it, let \epsilon >0. Then

p_n\to p\implies \exists N such that \forall n\geq \mathbb{N},d(p_n,p)<\epsilon
p_n\to p'\implies \exists N' such that \forall n\geq \mathbb{N},d(p_n,p')<\epsilon

Let n_0=max\{N,N'\}, then

d(p,p')\leq d(p_n,p_{n_0})+d(p_{n_0},p')<2\epsilon

And \forall \epsilon>0,d(p,p')<2\epsilon\implies d(p,p')=0. So p=p'

Remark: We can also prove this with contradiction. Idea p\neq p'\implies d(p,p')>0, let \epsilon=\frac{1}{2}d(p,q')\dots

(d) Suppose p\in \overline{E}. Then \forall n\in \mathbb{N}, B_{\frac{1}{n}}(p)\cap E\neq \phi. So \forall n\in \mathbb{N}, \exists p_n\in B_{\frac{1}{n}}(p)\cap E. We'll show p_n\to p.

Let \epsilon>0. Choose N\in \mathbb{N} such that N>\frac{1}{\epsilon}. Then if n\geq N, d(p_n,p)<\frac{1}{n}\leq \frac{1}{N}\leq \epsilon

EOP

Theorem 3.3

Let (s_n), (t_n) be sequence in \mathbb{C}. Suppose s_n\to s,t_n\to t

(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}

Proof:

(a) We want to prove \forall \epsilon>0, \exists N such that \forall n\geq N, |(s_n+t_n)-(s+t)|<\epsilon

Let \epsilon >0

s_n\to s\implies \exist N_s such that \forall n\geq N_s,|s_n-s|<\frac{\epsilon}{2}
t_n\to t\implies \exist N_t such that \forall n\geq N_t,|t_n-t|<\frac{\epsilon}{2}

Let N=\max\{N_t,N_s\}, then if n\geq N,

\begin{aligned}
|(s_n+t_n)-(s+t)|&=|(s_n+s)-(t_n-t)|\\
&\leq |s_n-s|+|t_n-t|\\
&< \frac{\epsilon}{2}+\frac{\epsilon}{2}\\
&<\epsilon
\end{aligned}

(b) exercise

(c) First we'll prove a special case.

s_n\to 0 \textup{ and }t_n\to 0\implies s_nt_n\to 0

Suppose s_n\to 0 and t_n\to 0.

Let \epsilon >0

s_n\to 0\implies \exist N_s such that \forall n\geq N_s,|s_n-s|<\sqrt{\epsilon}
t_n\to 0\implies \exist N_t such that \forall n\geq N_t,|t_n-t|<\sqrt{\epsilon}

Let N=\max\{N_t,N_s\}, then if n\geq N,

|s_n t_n|< \sqrt{\epsilon}^2=\epsilon

Now we prove the general case.

s_n\to s \textup{ and }t_n\to t\implies s_nt_n\to st

Since

s_n t_n=(s_n-s)(t_n-t)+s(t_n-t)+t(s_n-s)

So

\lim_{n\to \infty}(s_nt_n-st)=\lim_{n\to \infty}(s_n-s)(t_n-t)+\lim_{n\to \infty}s(t_n-t)+\lim_{n\to \infty}t(s_n-s)

\lim_{n\to \infty}(s_n-s)(t_n-t)=0 by special case

\lim_{n\to \infty}s(t_n-t)=0 by (b)

\lim_{n\to \infty}t(s_n-s)=0 by (b)

Thought process for (d)

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

If n is large enough, then...