\documentclass[12pt]{article}
\usepackage{amsmath,amsthm,amsfonts,amssymb,latexsym, hyperref}
\begin{document}

\begin{center}
\textbf{Analysis II; Background Notes and Definitions.}
\end{center}

In regard to the definitions, the a, b or a, b, c of the definitions are meant to be equivalent definitions for the same concept.  In each case it would be a theorem that each pair is equivalent.

\

Notation.

\qquad $\mathbb{R}$ denotes the real numbers.

Assume for this sections that the space under consideration is always the reals.

\qquad $\mathbb{N}$ denotes the positive integers.

\qquad $(a,b)$ denotes the set $\{ x | a<x<b \}$; this set is called a segment or an open interval.

\qquad $[a,b]$ denotes the set $\{ x | a \le x \le b \}$; this set is called an interval or (sometimes for emphasis) a closed interval.

\

Definition 1.  The subset $U$ of $\mathbb{R}$ is said to be \textit{open} if and only if $U = \emptyset$ or:

a. for each point $p \in U$ there exists a segment $(a,b)$ so that $p \in (a,b) \subset U$.

b. for each point $p \in U$ there exists a number $\delta$ so that $ \{ x | \ |x-p| < \delta \} \subset U$.

\

Basic Theorem about open sets:  The common part of two open sets is open and the union of an arbitrary collection of open sets is open.

\

Definition 2.  The point $p$ is a \textit{limit point} of the set $M$.  means that:

a.  If $S = (a,b)$ is a segment containing $p$ then $S$ contains a point of $M$ distinct from $p$.

b.  If $U$ is an open set containing $p$ then $U$ contains a point of $M$ distinct from $p$.

c.  If $\epsilon > 0$ then there exists a point $x \in M$ so that $0 < |p-x| < \epsilon$.

\

Definition 3.  Let $f: D \rightarrow \mathbb{R}$ be a function.  Then $$\lim_{x \rightarrow a} f(x) = L $$ means that $a$ is a limit point of $D$ and

a.  if $U$ is an open set containing $L$ then there is an open set $V$ containing $a$ so that $f((V - \{a\})\cap D) \subset U$.

b.  if $\epsilon > 0$ there exists a number $\delta$ so that if $x \in D$ and $0<|x-a| < \delta$ then $| f(x) - L| < \epsilon$.

\

Definition 4.  Let $f: D \rightarrow \mathbb{R}$ be a function.  Then $$\lim_{x \rightarrow \infty} f(x) = L $$ means that $D$ does not have an upper bound and

a.  if $U$ is an open set containing $L$ then there is a number $B$ so that $f(\{x| x> B\}\cap D) \subset U$.

b.  if $\epsilon > 0$ there exists a number $B$ so that if $x \in D$ and $x > B$ then $| f(x) - L| < \epsilon$.

\

Observation.  Suppose that $s_1, s_2, s_3 , \ldots = \{s_n\}_{n=1}^{\infty}$ is a sequence; then the sequential limit of the sequence is defined by considering the function $s: \mathbb{N} \rightarrow \mathbb{R}$ defined by $s(n) = s_n$ and using the above definition of limit.  Equivalently, the sequence $\{s_n\}_{n=1}^{\infty}$ has sequential limit $L$ means that

a.  if $U$ is an open set containing $L$ then there is a number $N \in \mathbb{N}$ so that if $n> N$ then $s_n \in U$.

b.  if $\epsilon > 0$ then there is a number $N \in \mathbb{N}$ so that if $n> N$ then
$|s_n - L| < \epsilon$.

\

Definition 5.  Suppose $\{s_n\}_{n =1}^{\infty}$ is a sequence of numbers then the series $\sum_{n =1}^{\infty} s_n$ is said to \textit{converge} to the number $L$ means that the sequence $\{\sum_{n =1}^{k} s_n\}_{k=1}^{\infty}$ has sequential limit $L$.

\

Definition 6.  Suppose that $f$ is a function with domain $D \subset \mathbb{R}$ and range a subset of $\mathbb{R}$.  Then $f$ is said to be \textit{continuous at the point $p$} means that:

a.  if $U$ is an open set containing $f(p)$ then there is an open set $V$ containing $p$ so that if $x \in D \cap V$ then $f(x) \in U$.

b.  if $\epsilon >0$ then there exists a number $\delta$ so that if $x \in D$ and $|x-p| < \delta$ then $|f(x) - f(p)|< \epsilon$.

c.  $\lim_{x \rightarrow p}f(x) = f(p)$.

\

Definition 7.  A function is said to be \textit{continuous} if it is continuous at each point of its domain.

\

\textbf{Warning!}  There is at least one statement below denoted as a theorem which is not a theorem.  (I.e. the statement is false.)  Find it and construct a counter example.

\

Theorem 0.0 A.  Suppose that $f: \mathbb{R} \rightarrow \mathbb{R}$ is a function.  Then $f$ is continuous if and only if for each open set $U$, $f^{-1}(U)$ is open.

\

Theorem 0.0 B.  Suppose that $f: \mathbb{R} \rightarrow \mathbb{R}$ is a function.  Then $f$ is continuous if and only if for each segment $(a,b)$, $f^{-1}((a,b))$ is open.

\

Theorem 0.1 A. Suppose that $\{a_n\}_{n=1}^{\infty}$ is a sequence with sequential limit $a$ and $\{b_n\}_{n=1}^{\infty}$ is a sequence with sequential limit $b$.  Then the sequence $\{a_n+b_n\}_{n=1}^{\infty}$ has sequential limit $a+b$.

\

Theorem 0.1 B. Suppose that the series $\sum_{n=1}^{\infty} a_n$ converges to $a$ and the series $\sum_{n=1}^{\infty} b_n$ converges to $b$.  Then the series $\sum_{n=1}^{\infty} (a_n + b_n)$ converges to $a+b$.

\

Theorem 0.1 C. Suppose that each of $f$ and $g$ is a function with common domain $D$ and that both $f$ and $g$ are continuous at the point $p$.  Then the function $f+g$ is continuous at the point $p$.

\

Theorem 0.2 A. Suppose that $\{a_n\}_{n=1}^{\infty}$ is a sequence with sequential limit $a$ and $\{b_n\}_{n=1}^{\infty}$ is a sequence with sequential limit $b$.  Then the sequence $\{a_nb_n\}_{n=1}^{\infty}$ has sequential limit $ab$.

\

Theorem 0.2 B. Suppose that the series $\sum_{n=1}^{\infty} a_n$ converges to $a$ and the series $\sum_{n=1}^{\infty} b_n$ converges to $b$.  Then the series $\sum_{n=1}^{\infty} a_n b_n$ converges to $ab$.

\

Theorem 0.2 C. Suppose that each of $f$ and $g$ is a function with common domain $D$ and that both $f$ and $g$ are continuous at the point $p$.  Then the function $f\cdot g$ is continuous at the point $p$.

\

Theorem 0.3 A. Suppose that $\{a_n\}_{n=1}^{\infty}$ is a sequence with sequential limit $a$, and $a \ne 0$.  Then there is an integer $N$ so that the sequence $\{\frac{1}{a_n}\}_{n=N}^{\infty}$ has sequential limit $\frac 1a$.

\

Theorem 0.3 B. Suppose that the series $\sum_{n=1}^{\infty} a_n$ converges to $a$, $a \ne 0$, and $a_n \ne 0$ for all $n \in \mathbb{N}$.  Then the series $\sum_{n=1}^{\infty} \frac{1}{a_n}$ converges to $\frac 1a$.

\

Theorem 0.3 C. Suppose that $f$ is a function with domain $D$, that $f(x) \ne 0$ for all $x \in D$ and that $f$ is continuous at the point $p$.  Then the function $\frac 1f$ is continuous at the point $p$.

\

Definition 7.  A set is said to be \textit{closed} if and only if its complement is open.

\

Definition 8.  The collection of sets $G$ is said to \textit{cover} the set $M$ if and only if every point in $M$ lies in some element of $G$.

\

Definition 9.  The set $M$ is said to be \textit{compact} if and only if whenever $G$ is a collection of open sets covering $M$ then there is a finite subcollection of $G$ that covers $M$.


\

Theorem 0.4.  The set $M$ is closed if and only if it contains all of its limit points.

\

Theorem 0.5 (Bolzano-Weierstrass).  If $M$ is a bounded infinite set then there is a point that is a limit point of $M$.

\

Theorem 0.6 (Heine-Borel). The set $M$ is compact if and only if it is closed and bounded.

\

Theorem 0.7.  If $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous and $M$ is compact, then $f(M)$ is compact.

\

Theorem 0.8 (Intermediate value theorem). Suppose that $f: [a,b] \rightarrow \mathbb{R}$ is continuous and $c$ is between $f(a)$ and $f(b)$.  Then there is a point $z$ so that $a<z<b$ and $f(z) = c$.

\

Comment: Theorems 0.5, 0.6 and 0.8 require the least upper bound axiom or some equivalent statement.

\

Theorem 0.9 (High point theorem).  If $M$ is compact and $f: M \rightarrow \mathbb{R}$ is continuous, then there is a number $c \in M$ so that if $x \in M$ then $f(x) \le f(c)$.


\end{document}