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% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[
solutions,
singlenumbering
]{../../resources/ormc_handout}
\uptitlel{Advanced 1}
\uptitler{Summer 2023}
\title{The Size of Sets, Part 1}
\subtitle{Prepared by Mark on \today{}}
\begin{document}
\maketitle
\input{parts/0 sets.tex}
\input{parts/1 really big.tex}
\input{parts/2 cartesian.tex}
\input{parts/3 functions.tex}
\input{parts/4 dense.tex}
\vfill
\pagebreak
\section{Bonus Problems}
\problem{}
Using only sets, how can we build an ordered pair $(a, b)$? \par
$(a, b)$ should be equal to $(c, d)$ if and only if $a = b$ and $c = d$. \par
Of course, $(a, b) \neq (b, a)$.
\begin{solution}
$(a, b) = \{ \{a\}, \{a, b\}\}$
\end{solution}
\vfill
\problem{}
Let $R$ be the set of all sets that do not contain themselves. \par
Does $R$ exist? \par
\hint{If $R$ exists, do we get a contradiction?}
\vfill
\problem{}
Suppose $f: A \to B$ and $g: B \to C$ are both one-to-one. Must $h(x) = g(f(x))$ be one-to-one? \par
Provide a proof or a counterexample.
\vfill
\problem{}
Suppose $f: A \to B$ and $g: B \to C$ are both onto. Must $h(x) = g(f(x))$ be onto? \par
Provide a proof or a counterexample.
\vfill
\pagebreak
\end{document}

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Advanced/Size of Sets/parts/0 sets.tex Normal file
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\section{Set Basics}
\definition{}
A \textit{set} is a collection of objects. \par
If $a$ is an element of set $S$, we write $a \in S$. This is pronounced \say{$a$ in $S$.} \par
The position of each element in a set or the number of times it is repeated doesn't matter. \par
All that matters is \textit{which} elements are in the set.
\vspace{2mm}
We say two sets $A$ and $B$ are equal if every element of $A$ is in $B$, and every element of $B$ is in $A$. This is known as the \textit{principle of extensionality.}
\problem{}
Convince yourself that $\{a, b\} = \{b, a\} = \{a, b, a, b, b\}$.
\definition{}
A set $A$ is a \textit{subset} of a set $B$ if every element of $A$ is in $B$. \par
For example, $\{a, b\}$ is a subset of $\{a, b, c\}$. This is written $\{a, b\} \subseteq \{a, b, c\}$. \par
Note that the \say{subset} symbol resembles the \say{less than or equal to} symbol.
\vspace{2mm}
We can also write $\{a, b\} \subset \{a, b, c\}$, which denotes a \textit{strict subset.} \par
The relationship between $\subseteq$ and $\subset$ is the same as the relationship between $\leq$ and $<$. \par
For example, $\{a, b, c\} \subseteq \{a, b, c\}$ is true, but $\{a, b, c\} \subset \{a, b, c\}$ is false.
\definition{}
The \textit{empty set}, usually written $\varnothing$, is the unique set containing no elements. \par
\note[Note]{The $\varnothing$ symbol is called \say{varnothing.} If you'd like to know why, ask an instructor.}
\problem{}
Which of the following are true?
\begin{itemize}
\item $\{1, 3\} = \{3, 3, 1\}$
\item $\{1, 2\} \subset \{2\}$
\item $\{1, 2\} \subset \{1, 2\}$
\item $\{1, 2\} \subseteq \{1, 2\}$
\item $\{2\} \subseteq \{1, 2\}$
\item $\varnothing \subseteq \{1, 2\}$
\end{itemize}
\vfill
\pagebreak
\problem{}
Let $A$ and $B$ be sets. Convince yourself that $A \subseteq B$ and $B \subseteq A$ implies $A = B$.
\vspace{2mm}
\hint{Whenever you start a proof, you should first look at definitions. \\
As stated on the previous page, $A = B$ if every element in $A$ is in $B$ and every element of $B$ is in $A$.}
\vspace{2mm}
As we saw before, the $\subseteq$ relation behaves a lot like the $\leq$ relation. \par
The statement above is very similar to the statement \say{$x \leq y$ and $y \geq x$ implies $x = y$}.
\definition{}
Let $A$ be a set. The \textit{power set} of $A$, written $\mathcal{P}(A)$, is the set of all subsets of $A$.
\problem{}
What is the power set of $\{1, 2, 3\}$? \par
\hint{It has eight elements.}
\vfill
\problem{}
Let $A$ be a set with $n$ elements. \par
How many elements does $\mathcal{P}(A)$ have? \par
\hint{Binary may help.}
\vfill
\pagebreak
\definition{Set Operations}
$A \cap B$ is the \textit{intersection} of $A$ and $B$. \par
It is the set of objects that are in both $A$ and $B$.
\vspace{3mm}
$A \cup B$ is the \textit{union} of $A$ and $B$. \par
It is the set of objects that are in either $A$ or $B$.
\vspace{3mm}
$A - B$ is the \textit{difference} of $A$ and $B$. \par
It is the set of objects that are in $A$ but are not in $B$.
\problem{}
What is $\{a, b, c\} \cap \{b, c, d\}$?
\vfill
\problem{}
What is $\{a, b, c\} \cup \{b, c, d\}$?
\vfill
\problem{}
What is $\{a, b, c\} - \{b, c, d\}$?
\vfill
\pagebreak

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\section{Really Big Sets}
\definition{}<infiniteset>
%We say a set $S$ is \textit{finite} if there exists a bijection from $S$ to $\{1, 2, 3, ..., n\}$ for some integer $n$. \par
%In other words
We say set is \textit{finite} if its elements can be consecutively numbered from 1 to some maximum index $n$. \par
Informally, we could say that a set is finite if it \say{ends.} \par
For example, the set $\{\star, \diamond, \heartsuit\}$ is (obviously) finite. We can number its elements 1, 2, and 3.
\vspace{2mm}
If a set is not finite, we say it is \textit{infinite}.
\vspace{2mm}
\problem{}
Which of the following sets are finite?
\begin{itemize}
\item $\{\texttt{A}, \texttt{B}, ..., \texttt{Z}\}$
\item $\{ \text{all rats in Europe} \}$
\item $\{ \text{all positive numbers} \}$
\item $\{ \mathbb{ \text{all rational numbers} } \}$
\end{itemize}
\vfill
\generic{Remark:}
Note that our definition of \say{infinite-ness} is based on a property of the set. Saying \say{a set is infinite} is much like saying \say{a cat is black} or \say{a number is even}. There are many different kinds of black cats, and there are many different even numbers --- some large, some small. \par
\vspace{2mm}
In general, \textbf{$\infty$ is not a well-defined mathematical object\footnotemark{}}. Infinity is not a number. There isn't a single \say{infinity.} Infinity is the the general concept of endlessness, used in many different contexts.
\vspace{2mm}
%The Russian language (as well as many others, no doubt) captures this well: \say{infinity} in Russian is \say{бес-конеч-ность}, which can be literally translated as \say{without-end-ness}.
\footnotetext{
In most cases. There are exceptions, but you need not worry about them for now. If you're curious, you may ask an instructor to explain. There's also a chance we'll see a well-defined \say{infinity} in a handout later this quarter.
}
\vfill
\pagebreak
%Say we have two finite sets $A$ and $B$. Comparing the sizes of these is fairly easy: all we need to do is count the elements %in each. It is not difficult to see that $\{1, 2, 3\}$ is bigger than $\{1, 2\}$.
%
%\vspace{2mm}
%
%We could extend this notion of \say{size} to infinite sets. \par
%For example, consider $\mathbb{R}$ and $\mathbb{Z}$. Intuitively, we'd expect $\mathbb{R}$ to be larger, \par
%since there are many elements in $\mathbb{R}$ between every two elements in $\mathbb{Z}$.
%
%\vspace{1mm}
%
%We could also try to compare the sizes of $\mathbb{Q}$ and $\mathbb{Z}$. There are bIntuitively, we'd expect $\mathbb{R}$ to %be larger, \par
%since there are many elements in $\mathbb{R}$ between every two elements in $\mathbb{Z}$.
%
%
%\vfill
%\pagebreak

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\section{Common Sets and Cartesian Products}
\definition{}
There are a few sets we use often. They have special names:
\begin{itemize}
\item $\mathbb{N} = \{0, 1, 2, 3, ...\}$ is the set of \textit{natural numbers}.
\item $\mathbb{Z} = \{..., -2, -1, 0, 1, 2, ...\}$ is the set of \textit{integers}.
\item $\mathbb{Q}$ is the set of \textit{rational numbers}.
\item $\mathbb{R}$ is the set of \textit{real numbers}.
\end{itemize}
\note[Note]{$\mathbb{Z}$ is called \say{blackboard zee} or \say{big zee.} Naturally, $\mathbb{N}$, $\mathbb{Q}$, and $\mathbb{R}$ have similar names. \\ This, of course, depends on context. Sometimes \say{zee} is all you need.}
\problem{}
Which of the following sets contain 100? \par
\hint{There may be more than one answer in all the problems below.}
\begin{tcolorbox}[
colback=white,
colframe=black,
width=0.5\textwidth,
toprule=0.3mm,
bottomrule=0.3mm,
leftrule=0.3mm,
rightrule=0.3mm,
]
\hfill $\mathbb{N}$ \hfill $\mathbb{Z}$ \hfill $\mathbb{Q}$ \hfill $\mathbb{R}$ \hfill\null
\end{tcolorbox}
\vfill
\problem{}
Which of the following sets contain {\large $\frac{1}{2}$}? \par
\begin{tcolorbox}[
colback=white,
colframe=black,
width=0.5\textwidth,
toprule=0.3mm,
bottomrule=0.3mm,
leftrule=0.3mm,
rightrule=0.3mm,
]
\hfill $\mathbb{N}$ \hfill $\mathbb{Z}$ \hfill $\mathbb{Q}$ \hfill $\mathbb{R}$ \hfill\null
\end{tcolorbox}
\vfill
\problem{}
Which of the following sets contain $\pi$? \par
\begin{tcolorbox}[
colback=white,
colframe=black,
width=0.5\textwidth,
toprule=0.3mm,
bottomrule=0.3mm,
leftrule=0.3mm,
rightrule=0.3mm,
]
\hfill $\mathbb{N}$ \hfill $\mathbb{Z}$ \hfill $\mathbb{Q}$ \hfill $\mathbb{R}$ \hfill\null
\end{tcolorbox}
\vfill
\problem{}
Which of the following sets contain $\sqrt{-1}$? \par
\begin{tcolorbox}[
colback=white,
colframe=black,
width=0.5\textwidth,
toprule=0.3mm,
bottomrule=0.3mm,
leftrule=0.3mm,
rightrule=0.3mm,
]
\hfill $\mathbb{N}$ \hfill $\mathbb{Z}$ \hfill $\mathbb{Q}$ \hfill $\mathbb{R}$ \hfill\null
\end{tcolorbox}
\vfill
\pagebreak
\definition{}
Consider the sets $A$ and $B$. The set $A \times B$ consists of all ordered\footnotemark{} pairs $(a, b)$ where $a \in A$ and $b \in B$. \par
This is called the \textit{cartesian product}, and is usually pronounced \say{$A$ cross $B$}.
\footnotetext{This means that order matters. $(a, b) \neq (b, a)$.}
\vspace{2mm}
For example, $\{1, 2, 3\} \times \{\heartsuit, \star\} = \{(1,\heartsuit),~ (1, \star),~ (2,\heartsuit),~ (2, \star),~ (3,\heartsuit),~ (3, \star)\}$ \par
You can think of this as placing the two sets \say{perpendicular} to one another:
\begin{center}
\begin{tikzpicture}[
scale=1,
bullet/.style={circle,inner sep=1.5pt,fill}
]
\draw[->] (-0.2,0) -- (4,0) node[right]{$A$};
\draw[->] (0,-0.2) -- (0,3) node[above]{$B$};
\draw (1,0.1) -- ++ (0,-0.2) node[below]{$1$};
\draw (2,0.1) -- ++ (0,-0.2) node[below]{$2$};
\draw (3,0.1) -- ++ (0,-0.2) node[below]{$3$};
\draw (0.1, 1) -- ++ (-0.2, 0) node[left]{$\heartsuit$};
\draw (0.1, 2) -- ++ (-0.2, 0) node[left]{$\star$};
\node[bullet] at (1, 1){};
\node[bullet] at (2, 1) {};
\node[bullet] at (3, 1) {};
\node[bullet] at (1, 2) {};
\node[bullet] at (2, 2) {};
\node[bullet] at (3, 2) {};
\draw[rounded corners] (0.5, 0.5) rectangle (3.5, 2.5) {};
\node[above] at (2, 2.5) {$A \times B$};
\end{tikzpicture}
\end{center}
\problem{}
Let $A = \{0, 1\} \times \{0, 1\}$ \par
Let $B = \{ a, b\}$ \par
What is $A \times B$?
\vfill
\problem{}
What is $\mathbb{R} \times \mathbb{R}$? \par
\hint{Use the \say{perpendicular} analogy}
\vfill
\pagebreak
\definition{}
$\mathbb{R}^n$ is the set of $n$-tuples of real numbers. \par
In English, this means that an element of $\mathbb{R}^n$ is a list of $n$ real numbers: \par
\vspace{4mm}
Elements of $\mathbb{R}^2$ look like $(a, b)$, where $a, b \in \mathbb{R}$. \hfill \note{\textit{Note:} $\mathbb{R}^2$ is pronounced \say{arrgh-two.}}
Elements of $\mathbb{R}^5$ look like $(a_1, a_2, a_3, a_4, a_5)$, where $a_n \in \mathbb{R}$. \par
$\mathbb{R}^1$ and $\mathbb{R}$ are identical.
\vspace{4mm}
Intuitively, $\mathbb{R}^2$ forms a two-dimensional plane, and $\mathbb{R}^3$ forms a three-dimensional space. \par
$\mathbb{R}^n$ is hard to visualize when $n \geq 4$, but you are welcome to try.
\problem{}
Convince yourself that $\mathbb{R} \times \mathbb{R}$ is $\mathbb{R}^2$. \par
What is $\mathbb{R}^2 \times \mathbb{R}$?
\vfill
\problem{}
What is $\mathbb{N}^2$?
\vfill
\problem{}
What is $\mathbb{Z}^3$?
\vfill
\pagebreak

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\section{Functions and Maps}
\definition{}<deffun>
A \textit{function} or \textit{map} $f$ from a set $A$ to a set $B$ is a rule that assigns an element of $B$ to each element of $A$. We write this as $f: A \to B$.
\vspace{1mm}
Let $L = \{\texttt{a}, \texttt{b}, \texttt{c}, \texttt{d}, ..., \texttt{z}\}$ be the set of lowercase english letters. \par
Let $C = \{\texttt{A}, \texttt{B}, \texttt{C}, \texttt{D}, ..., \texttt{Z}\}$ be the set of uppercase english letters. \par
\vspace{1mm}
Say we have a function $g: L \to C$ that capitalizes english letters. \par
We can think of this function as a \textit{map} from $A$ to $B$, shown below using arrows:
\begin{center}
\begin{tikzpicture}[scale=0.5]
\node[anchor=south] at (0, 1) {Set $A$};
\draw[line width = 0.25mm, rounded corners=2mm] (-1, 1) rectangle (1, -5);
\node at (0, 0) {\texttt{a}};
\node at (0, -1) {\texttt{b}};
\node at (0, -2) {\texttt{c}};
\node at (0, -3) {\texttt{d}};
\node at (0, -4) {\texttt{...}};
\draw[line width = 0.4mm, ->, gray] (0.5, 0) -- (3.5, 0);
\draw[line width = 0.4mm, ->, gray] (0.5, -1) -- (3.5, -1);
\draw[line width = 0.4mm, ->, gray] (0.5, -2) -- (3.5, -2);
\draw[line width = 0.4mm, ->, gray] (0.5, -3) -- (3.5, -3);
\node[fill=white, text=gray] at (2, 0) {$g$};
\node[anchor=south] at (4, 1) {Set $B$};
\draw[line width = 0.25mm, rounded corners=2mm] (3, 1) rectangle (5, -5);
\node at (4, 0) {\texttt{A}};
\node at (4, -1) {\texttt{B}};
\node at (4, -2) {\texttt{C}};
\node at (4, -3) {\texttt{D}};
\node at (4, -4) {\texttt{...}};
\end{tikzpicture}
\end{center}
\definition{}
We say a map $f$ is \textit{one-to-one} if $a = b$ implies $f(a) = f(b)$ for all $a, b \in A$. \par
In other words, this means that no two elements of $A$ are mapped to the same $b$:
\null\hfill
\begin{minipage}{0.48\textwidth}
\begin{center}
\textbf{A one-to-one function:} \par
\vspace{2mm}
\begin{tikzpicture}[scale=0.5]
\node[anchor=south] at (0, 1) {Set $A$};
\draw[line width = 0.25mm, rounded corners=2mm] (-1, 1) rectangle (1, -4);
\node at (0, 0) {\texttt{1}};
\node at (0, -1) {\texttt{2}};
\node at (0, -2) {\texttt{3}};
\node at (0, -3) {\texttt{4}};
\draw[line width = 0.4mm, ->, gray] (0.5, 0) -- (3.5, 0);
\draw[line width = 0.4mm, ->, gray] (0.5, -2) -- (3.5, -1);
\draw[line width = 0.4mm, ->, gray] (0.5, -1) -- (3.5, -3);
\draw[line width = 0.4mm, ->, gray] (0.5, -3) -- (3.5, -2);
\node[fill=white, text=gray] at (2, 0) {$f$};
\node[anchor=south] at (4, 1) {Set $B$};
\draw[line width = 0.25mm, rounded corners=2mm] (3, 1) rectangle (5, -5);
\node at (4, 0) {\texttt{a}};
\node at (4, -1) {\texttt{b}};
\node at (4, -2) {\texttt{c}};
\node at (4, -3) {\texttt{d}};
\node at (4, -4) {\texttt{e}};
\end{tikzpicture}
\end{center}
\end{minipage}
\hfill
\begin{minipage}{0.48\textwidth}
\begin{center}
\textbf{NOT a one-to-one function:} \par
\vspace{2mm}
\begin{tikzpicture}[scale=0.5]
\node[anchor=south] at (0, 1) {Set $A$};
\draw[line width = 0.25mm, rounded corners=2mm] (-1, 1) rectangle (1, -4);
\node at (0, 0) {\texttt{1}};
\node at (0, -1) {\texttt{2}};
\node at (0, -2) {\texttt{3}};
\node at (0, -3) {\texttt{4}};
\draw[line width = 0.4mm, ->, gray] (0.5, 0) -- (3.5, 0);
\draw[line width = 0.4mm, ->, gray] (0.5, -2) -- (3.5, -1);
\draw[line width = 0.4mm, ->, gray] (0.5, -1) -- (3.4, -1.8);
\draw[line width = 0.4mm, ->, gray] (0.5, -3) -- (3.4, -2.2);
\node[fill=white, text=gray] at (2, 0) {$f$};
\node[anchor=south] at (4, 1) {Set $B$};
\draw[line width = 0.25mm, rounded corners=2mm] (3, 1) rectangle (5, -5);
\node at (4, 0) {\texttt{a}};
\node at (4, -1) {\texttt{b}};
\node at (4, -2) {\texttt{c}};
\node at (4, -3) {\texttt{d}};
\node at (4, -4) {\texttt{e}};
\draw[line width = 0.4mm, ->] (5.5, -2) -- (4.5, -2);
\node[anchor=west] at (5.5, -2) {!!!};
\end{tikzpicture}
\end{center}
\end{minipage}
\hfill\null
\vfill
\definition{}
We say a map $f$ is \textit{onto} if for every $b \in B$, there is an $a \in A$ so that $b = f(a)$. \par
In other words, this means that every element of $B$ has some element of $A$ mapped to it:
\null\hfill
\begin{minipage}{0.48\textwidth}
\begin{center}
\textbf{An onto function:} \par
\vspace{2mm}
\begin{tikzpicture}[scale=0.5]
\node[anchor=south] at (0, 1) {Set $A$};
\draw[line width = 0.25mm, rounded corners=2mm] (-1, 1) rectangle (1, -5);
\node at (0, 0) {\texttt{1}};
\node at (0, -1) {\texttt{2}};
\node at (0, -2) {\texttt{3}};
\node at (0, -3) {\texttt{4}};
\node at (0, -4) {\texttt{5}};
\draw[line width = 0.4mm, ->, gray] (0.5, 0) -- (3.5, 0);
\draw[line width = 0.4mm, ->, gray] (0.5, -1) -- (3.5, -2);
\draw[line width = 0.4mm, ->, gray] (0.5, -2) -- (3.5, -1);
\draw[line width = 0.4mm, ->, gray] (0.5, -3) -- (3.4, -3);
\draw[line width = 0.4mm, ->, gray] (0.5, -4) -- (3.4, -3.4);
\node[fill=white, text=gray] at (2, 0) {$f$};
\node[anchor=south] at (4, 1) {Set $B$};
\draw[line width = 0.25mm, rounded corners=2mm] (3, 1) rectangle (5, -4);
\node at (4, 0) {\texttt{a}};
\node at (4, -1) {\texttt{b}};
\node at (4, -2) {\texttt{c}};
\node at (4, -3.1) {\texttt{d}};
\end{tikzpicture}
\end{center}
\end{minipage}
\hfill
\begin{minipage}{0.48\textwidth}
\begin{center}
\textbf{NOT an onto function:} \par
\vspace{2mm}
\begin{tikzpicture}[scale=0.5]
\node[anchor=south] at (0, 1) {Set $A$};
\draw[line width = 0.25mm, rounded corners=2mm] (-1, 1) rectangle (1, -5);
\node at (0, 0) {\texttt{1}};
\node at (0, -1) {\texttt{2}};
\node at (0, -2) {\texttt{3}};
\node at (0, -3) {\texttt{4}};
\node at (0, -4) {\texttt{5}};
\draw[line width = 0.4mm, ->, gray] (0.5, 0) -- (3.5, -0.8);
\draw[line width = 0.4mm, ->, gray] (0.5, -1) -- (3.5, -2);
\draw[line width = 0.4mm, ->, gray] (0.5, -2) -- (3.5, -1.2);
\draw[line width = 0.4mm, ->, gray] (0.5, -3) -- (3.4, -3);
\draw[line width = 0.4mm, ->, gray] (0.5, -4) -- (3.4, -3.4);
\node[fill=white, text=gray] at (2, -0.4) {$f$};
\node[anchor=south] at (4, 1) {Set $B$};
\draw[line width = 0.25mm, rounded corners=2mm] (3, 1) rectangle (5, -4);
\node at (4, 0) {\texttt{a}};
\node at (4, -1) {\texttt{b}};
\node at (4, -2) {\texttt{c}};
\node at (4, -3.1) {\texttt{d}};
\draw[line width = 0.4mm, ->] (5.5, 0) -- (4.5, 0);
\node[anchor=west] at (5.5, 0) {!!!};
\end{tikzpicture}
\end{center}
\end{minipage}
\hfill\null
\vfill
\pagebreak
\generic{Remark:}
The words \say{function} and \say{map} are two views of the same mathematical object. We usually think of functions as \say{machines} that take an input, change it, and produce an output. We think of maps as \say{rules} that match each element of a set $A$ to an element of a set $B$.
\vspace{2mm}
Again, functions and maps are \textit{identical}. They do the same thing. The only difference between \say{functions} and \say{maps} is how we think about them.
% one-to-one = injective
% onto = surjective
\problem{}
Is the \say{capitalize} function in \ref{deffun} one-to-one? Is it onto?
\vfill
\problem{}
Consider the function $f: \mathbb{Z} \to \mathbb{Z}$ defined by $f(x) = x^2$. \par
Is this function one-to-one? Is it onto?
\vfill
\problem{}
Consider the function $f: \mathbb{N} \to \mathbb{N}$ defined by $f(x) = x^2$. \par
Is this function one-to-one? Is it onto?
\vfill
\problem{}
Consider the function $f: \mathbb{Z} \to \mathbb{Z}$ defined below. \par
Is this function one-to-one? Is it onto?
\[
f(x) = \begin{cases}
0 & \text{if } x = 0 \\
x + 1 & \text{otherwise}
\end{cases}
\]
% TODO:
% bijections, same size if exists bijection
\vfill
\pagebreak
\definition{Invertible Functions}
A function $g$ is an \textit{inverse} of a function $f$ if $g(f(x)) = x$ for any $x$. \par
In other words, the function $g$ \say{undoes} $f$. Usually, the inverse of a function $f$ is written $f^{-1}$. \par
We say a function is \textit{invertible} if it has an inverse.
\vspace{2mm}
Intuitively, we could say that the inverse of $f$ reverses the \say{arrows} of $f$.
\problem{}
Is the following function invertible? \par
Draw the inverse, or explain why you can't.
\begin{center}
\begin{tikzpicture}[scale=0.5]
\node[anchor=south] at (0, 1) {$A$};
\draw[line width = 0.25mm, rounded corners=2mm] (-1, 1) rectangle (1, -4);
\node at (0, 0) {\texttt{1}};
\node at (0, -1) {\texttt{2}};
\node at (0, -2) {\texttt{3}};
\node at (0, -3) {\texttt{4}};
\draw[line width = 0.4mm, ->, gray] (0.5, 0) -- (3.5, 0);
\draw[line width = 0.4mm, ->, gray] (0.5, -2) -- (3.5, -1);
\draw[line width = 0.4mm, ->, gray] (0.5, -1) -- (3.5, -3);
\draw[line width = 0.4mm, ->, gray] (0.5, -3) -- (3.5, -2);
\node[fill=white, text=gray] at (2, 0) {$f$};
\node[anchor=south] at (4, 1) {$B$};
\draw[line width = 0.25mm, rounded corners=2mm] (3, 1) rectangle (5, -4);
\node at (4, 0) {\texttt{a}};
\node at (4, -1) {\texttt{b}};
\node at (4, -2) {\texttt{c}};
\node at (4, -3) {\texttt{d}};
\end{tikzpicture}
\end{center}
\vfill
\problem{}
Is the following function invertible? \par
Draw the inverse, or explain why you can't.
\begin{center}
\begin{tikzpicture}[scale=0.5]
\node[anchor=south] at (0, 1) {Set $A$};
\draw[line width = 0.25mm, rounded corners=2mm] (-1, 1) rectangle (1, -4);
\node at (0, 0) {\texttt{1}};
\node at (0, -1) {\texttt{2}};
\node at (0, -2) {\texttt{3}};
\node at (0, -3) {\texttt{4}};
\draw[line width = 0.4mm, ->, gray] (0.5, 0) -- (3.5, 0);
\draw[line width = 0.4mm, ->, gray] (0.5, -1) -- (3.5, -1);
\draw[line width = 0.4mm, ->, gray] (0.5, -2) -- (3.5, -3);
\draw[line width = 0.4mm, ->, gray] (0.5, -3) -- (3.5, -2);
\node[fill=white, text=gray] at (2, 0) {$f$};
\node[anchor=south] at (4, 1) {Set $B$};
\draw[line width = 0.25mm, rounded corners=2mm] (3, 1) rectangle (5, -5);
\node at (4, 0) {\texttt{a}};
\node at (4, -1) {\texttt{b}};
\node at (4, -2) {\texttt{c}};
\node at (4, -3) {\texttt{d}};
\node at (4, -4) {\texttt{e}};
\end{tikzpicture}
\end{center}
\vfill
\problem{}
Is the following function invertible? \par
Draw the inverse, or explain why you can't.
\begin{center}
\begin{tikzpicture}[scale=0.5]
\node[anchor=south] at (0, 1) {Set $A$};
\draw[line width = 0.25mm, rounded corners=2mm] (-1, 1) rectangle (1, -5);
\node at (0, 0) {\texttt{1}};
\node at (0, -1) {\texttt{2}};
\node at (0, -2) {\texttt{3}};
\node at (0, -3) {\texttt{4}};
\node at (0, -4) {\texttt{5}};
\draw[line width = 0.4mm, ->, gray] (0.5, 0) -- (3.5, 0);
\draw[line width = 0.4mm, ->, gray] (0.5, -1) -- (3.5, -2);
\draw[line width = 0.4mm, ->, gray] (0.5, -2) -- (3.5, -1);
\draw[line width = 0.4mm, ->, gray] (0.5, -3) -- (3.4, -3);
\draw[line width = 0.4mm, ->, gray] (0.5, -4) -- (3.4, -3.4);
\node[fill=white, text=gray] at (2, 0) {$f$};
\node[anchor=south] at (4, 1) {Set $B$};
\draw[line width = 0.25mm, rounded corners=2mm] (3, 1) rectangle (5, -4);
\node at (4, 0) {\texttt{a}};
\node at (4, -1) {\texttt{b}};
\node at (4, -2) {\texttt{c}};
\node at (4, -3.1) {\texttt{d}};
\end{tikzpicture}
\end{center}
\vfill
\pagebreak
\definition{Bijections}
One-to-one maps are also called \textit{injective} maps. \par
Onto maps are also called \textit{surjective} maps.
\vspace{2mm}
If a function is both one-to-one and onto, we say it is a \textit{bijection}.
\vspace{4mm}
\theorem{}
All bijective functions are invertible. All invertible functions are bijections. \par
You should review the problems on the previous page and convince yourself that this is true.
\problem{}
We say a set $S$ is \textit{finite} if there exists a bijection from $S$ to $\{1, 2, 3, ..., n\}$ for some integer $n$.\par
Convince yourself that this definition of \say{finite-ness} is the same as the one in \ref{infiniteset}.
\problem{}
Is there a bijection between the sets $\{1, 2, 3\}$ and $\{\texttt{A}, \texttt{B}, \texttt{C}\}$? \par
If a bijection exists, find one; if one doesn't, prove it. \par
\vfill
\problem{}
Is there a bijection between the sets $\{1, 2, 3, 4\}$ and $\{\texttt{A}, \texttt{B}, \texttt{C}\}$? \par
If a bijection exists, find one; if one doesn't, prove it. \par
\vfill
\problem{}<samesize>
Let $A$ and $B$ be two sets of different sizes. \par
Show that no bijection between $A$ and $B$ exists.
\vfill
\ref{samesize} reveals a very important fact: if we can find a bijection between two sets $A$ and $B$, these sets must have the same number of elements. Similarly, if we know that a bijection doesn't exist, we know that $A$ and $B$ must have a different number of elements.
\vspace{2mm}
Intuitively, you can think of a bijection as a \say{matching} between elements of $A$ and $B$. If we were to draw a bijection, we'd see an arrow connecting every element in $A$ to every element in $B$. If a bijection exists, every element of $A$ directly corresponds to an element of $B$, therefore $A$ and $B$ must have the same number of elements.
\pagebreak

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\section{Dense Orderings}
\note[Note]{This section is fairly difficult. If you get stuck here, try the bonus problems on the next page.}
\definition{}
An \textit{ordered set} is a set with an \say{order} attached to it. \par
A few examples are below:
\begin{itemize}
\item $\mathbb{Z}$ is an ordered set under $<$.
\item $\{\texttt{A}, \texttt{B}, ..., \texttt{Z}\}$ is an ordered set under $\diamond$,\par
Where $\alpha \diamond \beta$ holds iff the letter $\alpha$ comes before letter $\beta$ in the alphabet.
\end{itemize}
\definition{}
We say an ordered set $A$ is \textit{dense} if for any $a, b \in A$ there is a $c \in A$ so that $a < c < b$.\par
Intuitively, this means that there is an element of $A$ between any two elements of $A$.
\problem{}
Show that the ordered set $(\mathbb{Q}, <)$ is dense.
\vfill
\problem{}
Show that the ordered set $(\mathbb{R}, <)$ is dense.
\vfill
\problem{}
Show that there is a real number between every two rationals.
\vfill
\problem{}
Show that there is a rational number between every two reals.
\vfill
\pagebreak