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\usepackage[utf8]{inputenc}
\title{Math 220.201 Class 14 Solutions}
\begin{document}
\maketitle
\begin{enumerate}
\item For each of the following relations $f$ from $A$ and $B$, is it a function? If it is, write an expression for $f(a)$ in terms of $a$. If not, explain why.
\begin{enumerate}
\item $A=B=\{1, 2, 3, 4\}$, $f=\{(1, 2), (2, 3), (1, 3), (4, 4)\}$.
\vskip 0.15in
{\bf Solution:} It is not a function, because $(1, 2)$ and $(1, 3)$ are both in $f$.
\vskip 0.15in
\item $A=\mathbb{R}-\{1\}$, $B=\mathbb{R}$, $f=\{(a, b)|\frac{1}{a-1}=b\}$.
\vskip 0.15in
{\bf Solution:} It is a function, and $f(a)=\frac{1}{a-1}$.
\vskip 0.15in
\item $A=B=\mathbb{N}$, $f=\{(2n-1, n)|n\in\mathbb{N}\}\cup \{(2n, n)|n\in\mathbb{N}\}$.
\vskip 0.15in
{\bf Solution:} It is a function, and
$$f(a)=\begin{cases} \frac{a}{2} \hskip 0.1in \text{ if } a \text{ is even}\\
\frac{a+1}{2}\hskip 0.1in \text{ if } a \text{ is odd}mbc
\end{cases}$$
This can also be written as $f(a)=\lceil a/2\rceil$ (called the \emph{ceiling} function of $a/2$).
\vskip 0.15in
\end{enumerate}
\item Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be the function defined by $f(x)=x^2$. Determine the following sets.
\begin{enumerate}
\begin{multicols}{2}
\item $f([0,4])$
\vskip 0.2in
{\bf Solution:} $[0,16]$
\vskip 0.2in
\item $f([-1,2])$
\vskip 0.2in
{\bf Solution:} $[0,4]$
\vskip 0.2in

\item $f^{-1}([0,9])$
\vskip 0.2in
{\bf Solution:} $[-3,3]$
\vskip 0.2in
\item $f^{-1}([1,4])$
\vskip 0.2in
{\bf Solution:} $[0,16]$
\vskip 0.2in
\end{multicols}
\end{enumerate}

\item Suppose that $A, B$ are sets and $f:A \rightarrow B$ is a function.
\begin{enumerate}
\item If $C \subseteq A$, is it necessarily true that $f^{-1}(f(C))=C$?
\vskip 0.2in
{\bf Solution:} No, not necessarily. Consider the function $f:\{1, 2\} \rightarrow \{3, 4\}$ defined by $f(1)=4$ and $f(2)=4$. Then let $C=\{1\}$. Then $f(C)=\{4\}$, and $f^{-1}(f(C))=\{1, 2\}$. In general, $f^{-1}(f(C))\supseteq C$.
\vskip 0.2in
\item If $D \subseteq B$, is it necessarily true that $f(f^{-1}(D))=D$?
\vskip 0.2in
{\bf Solution:} No, not necessarily. Consider the same function as above and let $D=\{3\}$. Then $f^{-1}(D)=\emptyset$, and so $f(f^{-1}(D))=\emptyset$. In general, $f(f^{-1}(D))\subseteq D$.
\end{enumerate}
\end{enumerate}
\end{document}