20180611

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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.

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.

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