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We'll look at Levi's typed solution to 27. I forgot to do this Friday - oops. Anyone, please remind me.

We'll have the following present:

  • 41.1 - Jason
  • 41.5 - Emilee
  • 42 - Shawn
  • 43 - Junseong
  • 44 - James/Jai (two boards)
  • 45 - Brent
  • 46 - Levi
  • 47

The following students were absent:

  • Connor
From Ben: We did the second half of this on Friday. The first half is still available.

Problem 41 (More Practice With Universal Quantifiers)

For each $n\in \mathbb{N}$, define $\ds a_n = \frac{n-1}{n}$. Only one of the statements below is true. Rewrite each statement using the quantifiers $\forall$ and $\exists$. Then determine which statement is true, prove that statement is true, and then prove the other statement is false.

  1. For each real number $\varepsilon>0$, there exists $N\in \mathbb{N}$ such that for every $n\in \mathbb{N}$, if $n>N$ then $\ds \left|a_n-0\right|<\varepsilon$.
  2. For each real number $\varepsilon>0$, there exists $N\in \mathbb{N}$ such that for every $n\in \mathbb{N}$, if $n>N$ then $\ds \left|a_n-1\right|<\varepsilon$.
From Ben: The following problem came because of a conjecture we created in class (thanks Jason). If you can prove parts 1 and 5 are true, then every other fact follows immediately.

Conjecture 41.5

Let $S$ and $T$ be nonempty bounded sets of real numbers, and let $x\in\mathbb{R}$. Define the following sets:

  • $-S=\{-s\mid s\in S\}$,
  • $x+S=\{x+s\mid s\in S\}$, and
  • $S+T=\{s+t\mid s\in S,t\in T\}$.

Prove or disprove each of the following:

  1. $\inf(-S)=-\sup S$
  2. $\sup(-S)=-\inf S$
  3. $\inf(x+S)=x+\inf S$
  4. $\sup(x+S)=x+\sup S$
  5. $\inf(S+T)=\inf S+\inf T$
  6. $\sup(S+T)=\sup S+\sup T$

We use Cartesian products all the time without realizing it. When we look for relationships between two groups, we are looking at a subset of a Cartesian product. Family history involves looking at a Cartesian product. For example, we could let $A$ and $B$ both be the set of all humans that have lived on the planet. In our family tree, we often care to know when person $x$ descends from person $y$. One goal of family history is to determine all the pairs $(x,y)$ where person $x$ descends from person $y$. This is a subset of $A\times B$. Similarly, any time we start grouping objects together from two sets, we're formally stating a relationship between the two sets and looking for a subset of the Cartesian product. We call these subsets relations.

Definition (Relation Between Two Sets)

Let $A$ and $B$ be sets. A relation $\mathrm{R}$ between $A$ and $B$ is a subset of $A\times B$, so $\mathrm{R}\subseteq A\times B$. Given $a\in A$ and $b\in B$, we write $(a,b)\in\mathrm{R}$ precisely when $a$ is related to $b$.

A function is one important kind of relation that we've been studying since middle school. Here's a formal definition.

Definition (Function)

Let $A$ and $B$ be sets. A function $f$ from $A$ into $B$, written $f:A\to B$, is a relation between $A$ and $B$ (so $f\subseteq A\times B$) such that for every $x\in A$, there exists a unique $y\in B$ such that $(x,y)\in f$. When $f$ is a function from $A$ into $B$, we'll use the notation $y=f(x)$ or $f(x)=y$ rather than the more cumbersome notation $(x,y)\in f$ used for sets.

Problem 42: (Which Relations Are Functions)

In each number below, you are given a relation $\mathrm{R}$ between a set $A$ and a set $B$. Prove that the relation is a function from $A$ into $B$ or give a counter example to show that the relation is not a function from $A$ into $B$.

  1. Let $\mathrm{R}$ be the relation between $\mathbb{N}$ and $\mathbb{Z}$ given by $(n,m)\in \mathrm{R}$ if and only if $n^2=m$.
  2. Let $\mathrm{R}$ be the relation between $\mathbb{N}$ and $\mathbb{Z}$ given by $(n,m)\in \mathrm{R}$ if and only if $n=m^2$.
  3. Let $\mathrm{R}$ be the relation between $\mathbb{R}\times\mathbb{R}$ and $\mathbb{R}$ given by $((x,y),z)\in \mathrm{R}$ if and only if $x+y=z$.
  4. Let $\mathrm{R}$ be the relation between $\mathbb{R}\times\mathbb{R}$ and $\mathbb{R}$ given by $((x,y),z)\in \mathrm{R}$ if and only if $x/y=z$.
Definition (Domain And Codomain)

Let $A$ and $B$ be sets and let $f$ be a function from $A$ into $B$, so we have $f:A\to B$.

  • We call the set $A$ the domain of $f$.
  • We call the set $B$ the codomain of $f$.
Definition (Injective, Surjective, And Bijective)

Let $D$ and $R$ be sets and let $f$ be a function from $D$ into $R$, so we have $f:D\to R$.

  • We say that $f$ is injective (or one-to-one) if and only if for every $a,b\in D$ we have $f(a)=f(b)$ implies $a=b$.
  • We say that $f$ is surjective (or onto) if and only if for every $y\in R$ there exists an $x\in D$ such that $y=f(x)$.
  • We say that $f$ is bijective if and only if the function $f$ is both injective and surjective.

Problem 43: (Practice With Injective And Surjective)

For each function below, state the domain and codomain, determine if the function is injective, and then determine if the function is surjective.

  1. Let $f:\mathbb{R}\to\mathbb{R}$ be defined by $f(x)=x^2$.
  2. Let $f:[0,\infty)\to\mathbb{R}$ be defined by $f(x)=x^2$.
  3. Let $f:\mathbb{R}\to [0,\infty)$ be defined by $f(x)=x^2$.
  4. Let $f:[0,\infty)\to[0,\infty)$ be defined by $f(x)=x^2$.

As always, remember to justify each claim you make.

Exercise (Induction With The Sum Of Cubes)

Prove that for every $n\in \mathbb{N}$, we have $$1^3+2^3+\cdots+n^3=\left(\frac{n(n+1)}{2}\right)^2.$$

Click to see a solution.

Consider the set $$S=\left\{n\in\mathbb{N}\mid 1^3+2^3+\cdots+n^3=\left(\frac{n(n+1)}{2}\right)^2\right\}.$$ We need to prove that $S=\mathbb{N}$. We proceed by induction. We know $1\in S$ because $1^3 =1$ and $\left(\frac{1(1+1)}{2}\right)^2 = (1)^2=1$. Assume for some $k\in \mathbb{N}$ that $k\in S$, which means we've assumed that $$1^3+2^3+\cdots+k^3=\left(\frac{k(k+1)}{2}\right)^2.$$ We need to prove that $k+1\in S$, which means we need to prove that $$1^3+2^3+\cdots+k^3+(k+1)^3=\left(\frac{(k+1)((k+1)+1)}{2}\right)^2.$$ We'll start with the expression $1^3+2^3+\cdots+k^3+(k+1)^3$ and modify it to obtain the right hand side of the equation above. We now compute $$\begin{align} 1^3+2^3+\cdots+k^3+(k+1)^3 &= \left(1^3+2^3+\cdots+k^3\right)+(k+1)^3 && (\text{group the first $k$ terms})\\ &= \left(\frac{k(k+1)}{2}\right)^2+(k+1)^3 && (\text{substitute using our assumption})\\ &= \frac{k^2(k+1)^2}{4}+\frac{4(k+1)^3}{4} && (\text{prepare to combine fractions})\\ &= \frac{(k+1)^2(k^2+4(k+1))}{4}&& (\text{combine and factor})\\ &= \frac{(k+1)^2(k^2+4k+4)}{4}&& (\text{expand})\\ &= \frac{(k+1)^2(k+2)^2}{4}&& (\text{factor})\\ &= \left(\frac{(k+1)((k+1)+1)}{2}\right)^2&& (\text{note $k+2=(k+1)+1$}) .\end{align}$$ This shows that if $k\in S$, then $k+1\in S$. By mathematical induction, we know that $S=\mathbb{N}$. This prove that for every $n\in \mathbb{N}$, we have $$1^3+2^3+\cdots+n^3=\left(\frac{n(n+1)}{2}\right)^2.$$

Problem 44: (Induction With Sum Of Odds)

Notice that $$ \begin{align} 1&=1 \\ 1+3&=4 \\ 1+3+5&=9 \\ 1+3+5+7&=16 \\ 1+3+5+7+9&=25. \end{align} $$ Make a conjecture about the sum of odd numbers by giving a general formula for what you see above. Then prove that your formula is valid by using mathematical induction.

Problem 45: (Relationships Between Subsets, Infimums, And Supremums)

Suppose that $S$ and $T$ are nonempty bounded subsets of $\mathbb{R}$ and that $S\subseteq T$. Prove that $$\inf T \leq \inf S \leq \sup S\leq \sup T.$$

From Ben: Based on Brent's proof of the theorem above, we'll need two additional facts to complete his proof.

Conjecture 45.A: Infimums are limit points when not in the set

Suppose $S$ is a nonempty set that is bounded below. Then either $\inf S\in S$ or $\inf S$ is a limit point of $S$. A similar result holds for the supremum of $S$.

Conjecture 45.B: Infimums are a lower bound on the set of limit points

Suppose $S$ is a nonempty set that is bounded below. If $p$ is a limit point of $S$, then we must have $\inf S\leq p$. A similar result holds for the supremum of $S$.

Definition (Epsilon Neighborhoods And Deleted Neighborhoods)

Given $\varepsilon>0$, an $\varepsilon$-neighborhood of the real number $x$ is the interval $$N_{\varepsilon}(x) = (x-\varepsilon,x+\varepsilon) = \{y\in \mathbb{R}\colon |x-y|<\varepsilon\}.$$ A deleted $\varepsilon$-neighborhood of $x$ is the same interval minus the point $x$, which we'll write as $$N^*_{\varepsilon}(x) = N_{\varepsilon}(x)\setminus\{x\} = (x-\varepsilon,x)\cup(x,x+\varepsilon) = \{y\in \mathbb{R}\colon 0<|x-y|<\varepsilon\}.$$

Problem 46: (Practice With Neighborhoods)

Consider the interval $S=(1,8)$.

  1. Find real numbers $x$ and $\varepsilon$ so that $S=N_\varepsilon(x)$.
  2. Find real numbers $x$ and $\varepsilon$ so that $N_\varepsilon(x)$ is a proper subset of $S$.
  3. For every $x\in S$, give a formula for $\varepsilon$ so that $N_\varepsilon(x)\subseteq S$. Your choice of $\varepsilon$ will depend on $x$.
Definition (Interior Point, Open Set, Closed Set)

Let $S\subseteq \mathbb{R}$.

  • We say that $x$ is an interior point of $S$ if and only if there exists an $\varepsilon>0$ such that $N_\varepsilon(x)\subseteq S$.
  • The interior of $S$ is the collection of interior points of $S$.
  • We say that $S$ is an open set if and only if for every $x\in S$ there exists an $\varepsilon>0$ such that $N_\varepsilon(x)\subseteq S$ (so every point in $S$ is an interior point, or equivalently $S$ equals the interior of $S$).
  • We say that $S$ is a closed set if and only if the complement $\mathbb{R}\setminus S$ is open.

Problem 47: (Open Intervals Are Open Sets)

Prove that if $a$ and $b$ are real numbers such that $a<b$, then the interval $S=(a,b)$ is an open set.

Problem 48: (Closed Intervals Are Closed Sets)

Prove that if $a$ and $b$ are real numbers such that $a<b$, then the interval $ S=[a,b] $ is a closed set.

For more problems, see AllProblems