NoteNextra-origin/pages/Math4111/Math4111_L22.md

Lecture 22

Review

Let (a_n) be a sequence, and let b_n = a_{f(n)} be a rearrangement, where f is given by the following:

n	1	2	3	4	5	6	7	8	9	\dots
f(n)	1	2	4	3	6	8	5	10	12	\dots

(The pattern is "odd,even,even,") Defined the partial sums s_n = \sum_{k=1}^n a_k and t_n = \sum_{k=1}^n b_k.

1. In terms of a_1,a_2,\ldots, determine s_n-t_n for n=1,2,3,4,5,6,7. (For example, s_3-t_3 = a_3-a_4.)
   s_1 - t_1 = a_1 - a_1 = 0
s_2 - t_2 = a_2 - a_2 = 0
s_3 - t_3 = a_3 - a_4
s_4 - t_4 = a_4 - a_4 = 0
s_5 - t_5 = a_5 - a_6
s_6 - t_6 = a_6 - a_8
s_7 - t_7 = a_7 - a_8

2. What is the smallest n so that s_n - t_n does not contain any of the terms a_1,\dots, a_5?
   n=7

3. What is the smallest n so that s_n - t_n does not contain any of the terms a_1,\dots, a_{10}?
   n=13

New Material

Absolute Convergence

Theorem 3.55

Let (a_n) be a sequence in \mathbb{C} such that \sum_{n=1}^\infty |a_n| converges. If (b_n) is a rearrangement of (a_n), then \sum_{n=1}^\infty a_n = \sum_{n=1}^\infty b_n.

Proof:

Let f:\mathbb{N}\to\mathbb{N} be a bijection and let b_n = a_{f(n)}.

Let I(n)=\{1,2,\dots,n\}, J(n)=\{f(1),f(2),\dots,f(n)\}.

Then s_n = \sum_{k\in I(n)} a_k and t_n = \sum_{k\in J(n)} a_k.

\begin{aligned}
s_n - t_n &= \sum_{k=1}^n a_k - \sum_{k=1}^n b_k \\
&= \sum_{k\in I(n)} a_k - \sum_{k\in J(n)} a_{f(k)} \\
&= \sum_{k\in I(n)\backslash J(n)} a_k - \sum_{k\in J(n)\backslash I(n)} a_{f(k)} \\
|s_n - t_n|&\leq \sum_{i\in (I(n)\backslash J(n))\cup(J(n)\backslash I(n))} |a_i|
\end{aligned}

We will show that \lim_{n\to\infty} |s_n - t_n| = 0.

Let \epsilon > 0.

By the Cauchy criterion applied to \sum_{n=1}^\infty |a_n|, there exists N such that if m,n\geq N, then \sum_{k=m+1}^n |a_k| < \epsilon.

Then we choose p\in\mathbb{N} such that I(n)\subset J(p) (\{1,2,\dots,N\}\subset\{f(1),f(2),\dots,f(p)\}). p=\max\{f^{-1}(1),f^{-1}(2),\dots,f^{-1}(N)\}.

Note: This implies that p is at least N.

If n\geq p, then I(n)\subset J(p)\subset I(n)\cap J(n) so s_n = t_n.

So,

|s_n - t_n| \leq \sum_{i=N+1}^{\max J(n)} |a_i| < \epsilon

Back to the example of the review question.

I(9)=\{1,2,\dots,9\}, J(9)=\{1,2,4,3,6,8,5,10,12\}, I(9)\backslash J(9)=\{7,9\}, J(9)\backslash I(9)=\{10,12\}.

|s_9 - t_9| \leq |a_7|+|a_9|+|a_{10}|+|a_{12}| \leq \sum_{k=7}^{12} |a_k| < \epsilon

This proves that $\lim_{n\to\infty} |s_n - t_n| = 0$.

Since $\lim_{n\to\infty} s_n$ exists, $\lim_{n\to\infty} s_n = \lim_{n\to\infty} t_n$.

EOP

#### Theorem 3.54

Special case of Riemann rearrangement theorem

Suppose $a_n\in \mathbb{R}$, $\sum_{n=1}^\infty a_n$ converges conditionally. (i.e. $\sum_{n=1}^\infty a_n$ converges but $\sum_{n=1}^\infty |a_n|$ diverges.) Then $\forall \alpha\in\mathbb{R}$, there exists a rearrangement $(b_n)$ of $(a_n)$ such that $\sum_{n=1}^\infty b_n = \alpha$.

## Chapter 4 Continuity

### Limits of Functions

#### Definition 4.1

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces and $E\subset X$, $p\in E'$, $q\in Y$. Let $f:E\to Y$. We say that $\lim_{x\to p} f(x) = q$ if $\forall \epsilon > 0$, $\exists \delta > 0$ such that $\forall x\in E$, if $0 < d_X(x,p) < \delta$, then $d_Y(f(x),q) < \epsilon$.

This is same as:

If $\lim_{x\to p} f(x)=q$, then

\forall \epsilon > 0, \exists \delta > 0, \forall x\in E, 0 < d_X(x,p) < \delta \implies d_Y(f(x),q) < \epsilon

In many definitions, $E=X$

#### Theorem 4.2

$\lim_{x\to p} f(x) = q \iff$ forall sequence $(p_n)$ in $E\backslash\{p\}$ with $p_n\to p$, $f(p_n)\to q$.

Proof:

$\implies$

Suppose $\lim_{x\to p} f(x) = q$.

Let $(p_n)$ be a sequence in $E\backslash\{p\}$ with $p_n\to p$.

Let $\epsilon > 0$.

By definition of limit of function, $\exists \delta > 0$ such that if $0 < d_X(x,p) < \delta$, then $d_Y(f(x),q) < \epsilon$.

Since $p_n\to p$, $\exists N$ such that if $n\geq N$, then $0 < d_X(p_n,p) < \delta$. So $f(p_n)\in B_\epsilon(q)$.

Thus, we showed $\forall \epsilon > 0$, $\exists N$ such that if $n\geq N$, then $f(p_n)\in B_\epsilon(q)$.

$\impliedby$

We proceed by contrapositive.

Suppose $\lim_{x\to p} f(x) \neq q$, then $\exists$ sequence $(p_n)$ in $E\backslash\{p\}$ with $p_n\to p$ such that $f(p_n)\cancel{\to} q$.

Suppose $\lim_{n\to\infty} f(p_n) \neq q$, then $\exists \epsilon > 0$ such that for all $\delta > 0$, there exists $x\in E\cap B_\delta(p)\backslash\{p\}$ such that $f(x)\notin B_\epsilon(q)$.

For $n\in\mathbb{N}$, if we apply this with $\delta = \frac{1}{n}$, we get $p_n\in E\cap B_{\frac{1}{n}}(p)\backslash\{p\}$ such that $f(p_n)\notin B_\epsilon(q)$.

Then: $(p_n)$ is a sequence in $E\backslash\{p\}$ with $d_X(p_n,p) = \frac{1}{n}\to 0$ so that as $n\to\infty$, $f(p_n)\notin B_\epsilon(q)$.

So $\lim_{n\to\infty} f(p_n) \neq q$.

EOP

With this theorem, we can use the properties of limit of sequences to study limits of functions.

To prove things about limits of functions, we can use sequences.

- If $\lim_{x\to p} f(x) = q$, and $\lim_{x\to p} f(x)=q'$, then $q=q'$.
- If $\lim_{x\to p} f(x) = A$ and $\lim_{x\to p} g(x) = B$, then $\lim_{x\to p} f(x) + g(x) = A+B$.

### Continuous functions

#### Definition 4.5

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces and $E\subset X$, $p\in E$. Let $f:E\to Y$. We say that $f$ is continuous at $p$ if $\forall \epsilon > 0$, $\exists \delta > 0$ such that $f(E\cap B_\delta(p))\subset B_\epsilon(f(p))$.

- We say that $f$ is continuous on $E$ if $f$ is continuous at every $p\in E$.

> Remark: For $p\in E$, there are two possibilities.
>
> - $p$ is an isolated point of $E$.
> - $p$ is a limit point of $E$.
>
> In the first case, $f$ is automatically continuous at $p$. ($E\cap B_\delta(p)=\{p\}$ for all $\delta > 0$.)
>
> In the second case, $f$ is continuous at $p$ if and only if $\lim_{x\to p} f(x) = f(p)$.

#### Theorem 4.8

A function $f:E\to Y$ is continuous at $p\in E$ if the pre-image of every open set is open.

Two consequences if $f:E\to Y$ is continuous:

- Image of compact set is compact. (Implies Extreme Value Theorem)
- Image of connected set is connected. (Implies Intermediate Value Theorem)