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1. Review
Ross Ch.1 Introduction
- We defined the set of natural numbers and explored its successor property:
if $n$ $\epsilon$ $\mathbb{N}$ then $n+1$ also in $\mathbb{N}$.
- We defined Induction, a method which is used to prove an infinite number of successive propositions. Induction always begins by defining a base case. If some proposition holds for this base case then one may assume the proposition hold for all $n$ and then prove that it must then hold true for $n+1$.
- We also defined the set of rational numbers $\mathbb{Q}$ and the set of integers $\mathbb{Z}$ both of which have the property of being ordered meaning that there is a notion of some greater than and less than between all of their elements.
- The Rational Zeros Theorem states that the only rational candidates for solutions of the polynomial:
$c_nx^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0=0$
have the form $r=\frac{c}{d}$ where $c$ divides $c_0$ and $d$ divides $c_n$
Example: $\sqrt{2}$ is not a rational number
The only rational solutions of $x^2-2=0$ are $\pm1$, $\pm2$. $\sqrt{2}$ is a solution to $x^2-2=0$ thus it is not a rational number.
It can be seen that the Rational Zeros Theorem provides a convenient way of identifying potential rational roots of very complex polynomials.
- Next we discussed the set of Real Numbers $\mathbb{R}$ which can be represented as the real number line.
- $\mathbb{R}$ is an ordered field. It is the mathematical system on which we will do our analysis.
- maximum: Let $S\subset\mathbb{R}$, $\alpha$ $\epsilon$ $S$ $\alpha$ is a maximum if $\forall$ $\beta$ $\epsilon$ $S$, $\alpha$ $\geq$ $\beta$
- minimum: Let $S\subset\mathbb{R}$, $\alpha$ $\epsilon$ $S$ $\alpha$ is a minimum if $\forall$ $\beta$ $\epsilon$ $S$, $\alpha\leq\beta$
- {$r$ $\epsilon$ $\mathbb{Q}$ : $0 \leq r \leq \sqrt{2}$} has a minimum of 0 but does not have a maximum because $\sqrt{2}$ does not belong to the set but there are rationals in the set arbitrarily close to $\sqrt{2}$
- supremum If S is bounded above and S has a least upper bound then we call this the supremeum of S
- infimum If S is bounded below and has a greatest lower bound then we call this the infimum of S
Example: inf{$n^{(-1)^n}$ : $n \epsilon \mathbb{N}$} = 0
The Completeness Axiom: Every nonempty subset S of $\mathbb{R}$ that is bounded above has a least upper bound i.e. $\sup$S exists. Same goes for lower bounds and the greatest lower bound.
- The Completeness Axiom does not hold for $\mathbb{Q}$
- Archimedean Property: if a > 0 and b > 0 then $\exists n \epsilon \mathbb{Z}^+$ : na > b
- Denseness of $\mathbb{Q}$: If a,b $\epsilon \mathbb{R}$ and a < b, then $\exists r \epsilon \mathbb{Q}$: a < r < b
- $\infty$ and $-\infty$ are not real numbers but are very useful to include. $[a, \infty) = \{x \epsilon \mathbb{R} : a \leq x \}$ $(a, \infty) = \{x \epsilon \mathbb{R} : a < x \}$ $( -\infty, b] = \{ x \epsilon \mathbb{R} : x \leq b \}$ $( -\infty, b) = \{x \epsilon \mathbb{R} : x < b \}$
Ross Ch. 2 Sequences
- definition of sequence convergent: $(s_n)$ converges to $s$ if ${\forall \epsilon > 0, \exists N}$ such that ${n > N \Rightarrow |s_n - s| < \epsilon.}$
example: show that $\lim_{n \rightarrow \infty}{\frac{1}{n^2}} = 0$.
Let $\epsilon > 0$, ${N = \frac{1}{\sqrt{\epsilon}}}$. Then ${n > N \Rightarrow n > \frac{1}{\sqrt{\epsilon}} \Rightarrow n^2 > \frac{1}{\epsilon}}$ and hence ${\epsilon > \frac{1}{n^2}}$. Thus $n > N \Rightarrow |\frac{1}{n^2} - 0| < \epsilon$.
example: Let $\lim_{n \rightarrow \infty}{s_n} = s \neq 0$ Prove $\inf\{|s_n| : n \epsilon \mathbb{N}\} > 0$
Let $\epsilon = \frac{1}{2}|s| > 0$ ${\exists N \epsilon \mathbb{N}}$ so that ${n > N \Rightarrow |S_n - s| < \frac{|s|}{2}}$. $n > N \Rightarrow |s_n| \geq \frac{|s|}{2}$ otherwise the triangle inequality would yield $|s| < |s|$. If we set $m = min\{\frac{|s|}{2}, |s_1|, …, |s_N|\}$ then we have ${m > 0, |s_n| \geq m \forall n \epsilon \mathbb{N}}$ thus $\inf\{|s_n| : n \epsilon \mathbb{N}\} \geq m > 0$
- 9.1 Theorem: Convergent sequences are bounded
- Additivity and homogeneity are properties of convergence. i.e. if $\lim{s_n} = s$ and $k \epsilon \mathbb{R}$ then $\lim{ks_n} = ks$ and if $\lim{s_n} = s$, $\lim{t_n} = t$ then $\lim{s_n + t_n} = s + t$
- $\lim{\frac{1}{s_n}} = \frac{1}{s}$, it follows that $\lim{\frac{t_n}{s_n}} = \frac{t}{s}$
- $\lim{s_n} = \infty \Leftrightarrow \lim{\frac{1}{s_n}} = 0$
- 10.2 Theorem: All bounded monotone sequences converge
example: Prove $(s_n)$ is monotone bounded.
${s_1 = 5}$ ${s_n = \frac{s_{n-1}^2 + 5}{2s_{n-1}}}$ for ${n \geq 2}$
soln:
${s_n > \sqrt{5}}$ for $n \leq 2$. Assume this holds for some $n \geq 2$
Need to show ${\frac{s_{n-1}^2 + 5}{2s_{n-1}} < s_{n+1} \Leftrightarrow s_{n+1}^2 > 5}$
this holds because $s_{n+1} > \sqrt{5}$ by the inductive hypothesis. To show ${s_{n+2} > \sqrt{5}}$ we need ${\frac{s_{n-1}^2 + 5}{2s_{n-1}} > \sqrt{5} \Leftrightarrow {s_{n+1}- \sqrt{5}}^2 > 0}$ Thus ${n+1 \leq n \forall n \epsilon \mathbb{N}}$ by induction.
definition: ${\lim \sup s_n = \lim_{N \rightarrow \infty} \sup{s_n : n > N}}$
definition: ${\lim \inf s_n = \lim_{N \rightarrow \infty} \inf{s_n : n > N}}$
It follows that ${\lim \sup s_n \leq \sup{s_n : n \epsilon \mathbb{N}}$ and ${\lim \inf s_n \geq \inf{s_n : n \epsilon \mathbb{N}}$
definition: Cauchy Sequence $s_n$ is Cauchy if for each ${\epsilon > 0 \exists N}$ such that ${m,n > N \Rightarrow |s_n-s_m| < \epsilon}$
- 10.11 Theorem: Convergent $\Leftrightarrow$ Cauchy
Subsequences
-11.2 Theorem Let $s_n$ be a sequence
(i) If $t$ is in $\mathbb{R}$ then there is a subsequence of $(s_n)$ converging to t iff $\n \epsilon \mathbb{N} : |s_n - t| < \epsilon \}$ is infinite for all $\epsilon > 0$
-11.3 Theorem
If $(s_n)$ converges, then every subsequence converges to same limit.
-11.4 Theorem
Every sequence $(s_n)$ has a monotonic subsequence
-11.5 Bolzano-Weierstrauss Theorem
Every bounded sequence has a convergent subsequence.
lim sup and lim inf
Rudin ch. 2 Basic Topology
First we define the set definitions of function, image, inverse image, onto, and one-to-one. I offer the definition of one-to-one below:
Let A and B be two sets and let $f$ be a mapping of A into B. If $\forall y \epsilon B$, $f^{-1}(y)$ consists of at most one element of A then $f$ is a one-to-one mapping of A into B.
Then we introduce the notions of size such as finite, infinite, countable, uncountable, and at most countable
- 2.8 Theorem
Every infinite subset of a countable set is countable.
union and intersection
- the infinite union of countable sets is countable.
Metric Spaces
A metric is a distance function which must satisfy the following properties:
for points p and q in X
1. $d(p, q) > 0$ if $p \neq q$, $d(p, p) = 0$
2. $d(p, q) - d(q, p)$
3. $d(p, q) \leq d(p, r) + d(r, q) \forall r \epsilon X$
Distance in $\mathbb{R}^k$:
$d(x, y) = |x - y| for x, y \epsilon \mathbb{R}^k$
open and closed balls
An open ball with center $x$ and radius $r$ is defined as $\{y \epsilon \mathbb{R}^k : |y - x| < r\}$
A closed ball with center $x$ and radius $r$ is defined as $\{y \epsilon \mathbb{R}^k : |y - x| \leq r\}$
We are given several definitions as follows:
- neighborhood of $p$ of radius $r$ := the set of all $q$ such that $d(p, q) < r$
- $p$ is a limit point of set E if every neighborhood of $p$ contains some point $q \neq p$ such that $q \epsilon E$.
- E is closed if every limit point of E is point in E.
- $p$ is an interior point of E if there is a neighborhood N of $p$ such that N $\subset$ E.
- E is open if every point of E is an interior point of E.
- The complement of E is the set of all points $p \epsilon X$ such that $p \notin E$ where $E \subset X$.
- E is bounded if $\exists M \epsilon \mathbb{R}$ and $q \epsilon X$ such that $d(q, p) < M \forall p \epsilon E$
- E is dense in X if every point of X is a limit point and\or a point of E.
Example: The set of all integers
Open: No Closed: Yes Perfect: No Bounded: No
2.19 Theorem
Every neighborhood is an open set.
2.20 Theorem
If $p$ is a limit point of E then every neighborhood of $p$ contains infinitely many points in E.
2. Questions
1. I do not understand the solution to HW10 question 4:
Prove $f(x) = \sum_{n=1}^{\infty} 4^{-n} \varphi (4^n x)$ is everywhere continuous but nowhere differentiable, $\varphi$(x)=min{$|x-n||n \epsilon \mathbb{Z}$}.
The goal of this problem is to essentially create a function which has edges everywhere so that it is not differentiable at any point. The graph of $\varphi$ as given in the problem is nondifferentiable at $x=n+\frac{1}{2}$ where $n$ is any integer.We are then gping to define $h_n(x)$ to be a multiple of $\frac{1}{4}$ or $\frac{-1}{4}$ function that is to be honest I still have no idea what this proof is saying lol.
2. Proof of Theorem 19.2 from Ross: If $f$ is continuous on closed interval [a, b] then $f$ is uniformly continuous on [a, b]. Proof states that a subsequence $x_{n_k}$ of $x_n$ converges to $x_0 = \lim_k{x_{n_k}} = \lim_k{y_{n_k}}$ since $f$ is continuous at $x_0$. I am unsure how they achieve $\lim_k{x_{n_k}} = \lim_k{y_{n_k}}$ also I am unsure how this theorem is true given we can have very steep slopes on closed intervals which make it difficult or impossible to assign a delta to every $\epsilon > 0$ such that uniform continuity is achieved.
3. Theorem 9.1 of Ross: Convergent sequences are bounded. What if $s_n = \frac{1}{n-1}$ and $n$ starts at 1?
4. In proof of Ross Theorem 33.4 (ii): ${(c, d) \subseteq [a, b]}$, how could ${(c, d) = [a, b]}$?
5. In proof of Ross Theorem 33.5, I am unsure how one can claim M($|f|$, S) $-$ ${m}$($|f|$, S) $\leq$ M($|f|$, U) $-$ ${m}$($|f|$, U)
I believe the answer to this question is due to the fact that $\inf |f| \leq \inf f$
6. Does a subsequence have to have an equation (in terms of k) for generating the n-index of $s_n$ or could we technically just pick random n indices?
7. From Rudin ch. 2 pg. 33: Why is the set of all complex $z$ such that $|z| \leq 1$ a perfect set and why is $z$ such that $|z| < 1$ not perfect?
8. Why is a nonempty finite set closed? Answer: It has no interior points. Thus it cannot be open. Furthermore a single point is closed and we have the property that the union of finitely many closed sets is closed. A nonempty finite set is obviously the union of single points and thus is closed.
