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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 $x/y=z$.

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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.

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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 $$N^*_{\varepsilon}(x) = N_{\varepsilon}(x)\setminus\{x\} = (x-\varepsilon,x)\cup(x,x+\varepsilon) = \{y\in \mathbb{R}\colon 0<|x-y|<\varepsilon\}.$$

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.

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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.

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Definition (Sequence)

A sequence of real numbers is a function $a:\mathbb{N}\to\mathbb{R}$.

• We'll often use the notation $a_n$ rather than $a(n)$ to denote the value of the sequence at $n$. We call $a_n$ the $n$th term of the sequence.
• We may refer to the sequence by writing $(a_n)$ or by writing $(a_1,a_2,\ldots)$.
• We may change the domain from $\mathbb{N}$ to $\mathbb{N}\cup \{0\}$ by adding in $0$, or we might remove several of the first entries from the sequence. When we do this, we'll use the notation $(a_n)_{n=m}^{\infty}$ where $m$ is the first term we want to consider.

Definition (Convergent Sequence)

Let $(a_1,a_2,\ldots)$ be a sequence of real numbers.

• We say that $(a_n)$ converges to $L$ and write $(a_n)\to L$ if and only if for every $\varepsilon>0$, there exists a real number $M$ such that for every $n\in \mathbb{N}$ we have $n> M$ implies $|a_n-L|<\varepsilon$.
• When $(a_n)$ converges to $L$, we call $L$ the limit of $(a_n)$.
• We say that $(a_n)$ is a convergent sequence if and only if $(a_n)$ converges to $L$ for some real number $L$.
• We say that $(a_n)$ is a divergent sequence if and only if $(a_n)$ is not a convergent sequence.

Problem 49: (Showing A Sequence Converges)

Consider the sequence $\ds (a_n)=\left(\frac{n+1}{n}\right)$. Prove that $(a_n)$ converges to $L=1$.

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For more problems, see AllProblems

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