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

QED

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

QED

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)
