\documentclass[12pt]{amsart}
\usepackage{amssymb,amsmath,amsthm, hyperref, verbatim,graphicx}
\oddsidemargin = 0cm \evensidemargin = 0cm \textwidth = 6.2in
\textheight8.0in
\newtheorem*{thm*}{Theorem}
\newtheorem{thm}{Theorem}[section]
\newtheorem*{lemma}{Lemma}
\newtheorem{prop}{Proposition}
\newtheorem{conj}{Conjecture}
\newtheorem{exercise}{Exercise}
\newtheorem*{fact}{Fact}
\newtheorem*{rmk}{Remark}
\theoremstyle{definition}
\newtheorem{definition}[thm]{Definition}
\newtheorem*{dfn}{Definition}
\newtheorem{example}{Example}
\newtheorem{corollary}[thm]{Corollary}
\newcommand{\ord}{\mathrm{ord}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
\begin{document}
\title{Worksheet 1: Practice on set theory}
\author{Math 2106-D}
\maketitle

\begin{enumerate}
\vfill
\item Let $X=\{1,5,8\}$ and $Y=\{1,a,b\}$. Write down the following sets:
\begin{enumerate}
\item $X\cap Y$.
\item $X\cup Y$.
\item $\mathcal P(X\cap Y)$.
\item $X\times Y$
\end{enumerate}
\vfill
\item If $\mathcal I$ is a set, called an {\it index set}, and for each $\alpha\in\mathcal I$, we have a set $A_\alpha$, then we can consider the union and intersection 
\[
\cup_{\alpha\in\mathcal I}A_{\alpha}=\left\{x | x\in A_{\alpha} \ \text{ for at least one } A_{\alpha} \ \mathrm{with} \  \alpha\in\mathcal I  \right\},
\]
\[
\cap_{\alpha\in\mathcal I}A_{\alpha}=\left\{x | x\in A_{\alpha} \ \text{ for every } A_{\alpha} \ \mathrm{with} \  \alpha\in\mathcal I  \right\}.
\]
Determine what the following sets are, and explain your answer.
\begin{enumerate}
\item $\cap_{n\in\mathbb N}\left(-\frac 1n,\frac 1n\right)$
\item $\cup_{n\in\mathbb N}\left(-\frac 1n,\frac 1n\right)$
\item $\cup_{a\in\R}\{(x,y,z) | x^2+y^2=a^2, z=a\}$
\end{enumerate}
\vfill
\item Explain why $|A\cup B|=|A|+|B|-|A\cap B|$. Can you find a similar formula for $|A\cup B\cup C|$ in terms of cardinalities of $A,B,C$ and intersections between these sets?
\vfill
\item[(***)] Let $X$ be the set of all sets that are not elements of themselves, that is, 
\[
X=\{A | A \ \text{ is a set}, \ A\not\in A\}
.
\]
Explain why the existence of this ``set'' doesn't make logical sense.  (Hint: Is $X\in X$?) This conundrum, known as {\bf Russell's Paradox}, shows that in order to be $100\%$ rigorous with set theory, one must be very careful when describing which sets are ``allowed''; the standard resolution to this problem is to base set theory on a precise set of {\bf axioms}, such as {\bf ZFC}.

\end{enumerate}


\pagebreak
\center
{\bf Handy notation cheat sheet:}


\begin{itemize}
\vfill
\item
$a\in A$, read ``$a$ is in $A$,'' means that $a$ is an element of $A$
\vfill
\item
$A\subseteq B$, read ``$A$ is a subset of $B$,'' means that every element of $A$ is also an element of $B$
\item $A=B$, read ``$A$ equals $B$,'' if $A$ and $B$ have the same elements
\vfill
\item
$|A|$ is the cardinality, or size of $A$, namely the number of its elements (if $A$ is finite; if $A$ is infinite, one often writes $|A|=\infty$ as a shorthand, although this isn't very precise)
\vfill
\item
$\emptyset$ denotes the {\bf empty set}, which is the only set with no elements
\vfill
\item
$\mathbb N=\{1,2,3,\dots\}$ is the set of {\bf natural numbers}
\vfill
\item
$\mathbb Z=\{\ldots,-3,-2,-3,0,1,2,3,\dots\}$ is the set of {\bf integers} (whole numbers)
\vfill
\item
$\mathbb Q=\{\frac{a}{b}\, | \, a,b\in\mathbb Z, \text{ and } b\neq0\}$ is the set of {\bf rational numbers} (fractions)
\vfill
\item
$\mathbb R$ is the set of {\bf real numbers}
\vfill
\item
$A\times B$ denotes the {\bf Cartesian product} $A\times B=\left\{(a,b)\, |\, a\in A\ , b\in B\right\}$ consisting of {\bf ordered pairs} of elements from $A$ and $B$. The $n$-fold Cartesian product $A\times\ldots\times A$ is denoted by $A^n$.
\vfill
\item
The notations $\mathcal P(A)$ and $2^A$ both denote the {\bf power set of $A$}, namely the set of all subsets of $A$. That is, $\mathcal P(A)=\{X\, |\, X\subseteq A\}$.
\vfill
\item 
The {\bf union} of $A$ and $B$ is $A\cup B=\{x | x\in A \ \mathrm{ or } \ x\in B\}$ (where {\it or} in math always means that one or the other or both conditions hold)
\vfill
\item 
The {\bf intersection} of $A$ and $B$ is $A\cap B=\{x | x\in A \ \mathrm{ and } \ x\in B\}$ 
\vfill
\item The {\bf set difference} of $A$ and $B$ is $A\setminus B=A-B=\{x| x\in A \ \mathrm{ and } \ x\not\in B\}$
\vfill
\item If $U$ is the universal set, then the {\bf complement of $A$} is $A^{\mathrm c}=\overline A=U\setminus A$ (i.e., the set of everything not inside of $A$, in whatever universal set you have given by the context)
\end{itemize}







\end{document}