NoteNextra-origin/content/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.

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)