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

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 $n\geq N$,

(b) exercise

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

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 $n\geq N$,

Now we prove the general case.

Since

So

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

If $n$ is large enough, then...