%EXERCISE FOUR
          
\documentclass{amsart}
\usepackage{amssymb}
\thispagestyle{empty}
\newtheorem{thm}{Theorem}
\begin{document}
             
\begin{center}
	\textbf{Exercise Four: Set Notation, Greek and Hebrew Letters}
\end{center}  

If $A=\{\heartsuit, \spadesuit\}$ and $B=\{\epsilon, \theta\}$, then 

$A \cup B = \{\heartsuit, \spadesuit, \epsilon, \theta\}$

$A \cap B = \emptyset$ 
 
$\mathcal{P}(A\cap B)=\{\emptyset\}$ 

\begin{thm}
If $A\subseteq B$ and $B\subseteq C$, then $A\subseteq C$.
\end{thm}

\begin{proof}
We assume $A\subseteq B$ and $B\subseteq C$, and we want to show $A\subseteq C$.

Suppose $x\in A$. We must show that $x\in C$ also.

Since $A\subseteq B$, we know $x\in B$. 

Similarly, since $B\subseteq C$, we also know $x\in C$.

But this means $A\subseteq C$, as desired.
\end{proof}

\begin{thm}
If $A\subseteq B$, then $\mathcal{P}(A)\subseteq \mathcal{P}(B)$.
\end{thm}

\begin{proof}
We assume $A\subseteq B$, and we want to show 
$\mathcal{P}(A)\subseteq \mathcal{P}(B)$.

Suppose $X\in\mathcal{P}(A)$. We must show $X\in\mathcal{P}(B)$ also.

By the definition of power set, $X\in\mathcal{P}(A)$ means that 
$X$ is a subset of $A$.

Since $X\subseteq A$ and $A\subseteq B$, the previous theorem tells 
us that $X\subseteq B$ also.

Then, using the definition of power set again, this means 
$X\in\mathcal{P}(B)$.

Therefore, $\mathcal{P}(A)\subseteq \mathcal{P}(B)$, as desired.
\end{proof}

\noindent\textbf{Puzzling Power Sets}
% The command \noindent ... well, I think you can guess what it does.

$\mathcal{P}(\emptyset)=\{\emptyset\}$

$\mathcal{P}(\mathcal{P}(\emptyset))=\{\emptyset, \{\emptyset\} \}$

$\mathcal{P}(\mathcal{P}(\mathcal{P}(\emptyset)))=
\{\emptyset, \{\emptyset\}, 
  \{\emptyset, \{\emptyset\} \}, 
  \{\{\emptyset\}\} \}$ 

\end{document} 
