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\begin{document}


\noindent
{\bf Solutions to HW 1.}

\bigskip
  
\begin{enumerate}
\item[1.]
Define
    $V_0=\emptyset$, $V_1={\mathcal P}(V_0)$, 
$V_2={\mathcal P}(V_1)$, 
$V_3={\mathcal P}(V_2)$, and so on.
\begin{enumerate}
\item  List the elements of $V_0, V_1, V_2$ and $V_3$.
\vfill
\color{blue}
\begin{enumerate}
\item $V_0 = \emptyset$,
\item $V_1 = {\mathcal P}(\emptyset) = \{\emptyset\}$,
\item $V_2 = {\mathcal P}(V_1) = \{\emptyset,\{\emptyset\}\}$,
\item $V_3 = {\mathcal P}(V_2) =
  \{\emptyset,\{\emptyset\},\{\{\emptyset\}\},\{\emptyset,\{\emptyset\}\}\}$,
  \end{enumerate}
\color{black}

\vfill
  
\item  Draw a directed graph whose ``dots'' are the sets in $V_3$
    and where $x\to y$ means $x\in y$. (Hint: your graph
    should have four ``dots'' and four edges.)
\vfill
\color{blue}
  \begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(60,30)
\thicklines
\put(0,0){\circle*{2}}
\put(10,10){\circle*{2}}
\put(0,20){\circle*{2}}
\put(20,20){\circle*{2}}

\put(2,2){\vector(1,1){6}}
\put(12,12){\vector(1,1){6}}
\put(8,12){\vector(-1,1){6}}
\put(0,2){\vector(0,1){16}}

\put(-1,-6){$\emptyset$}
\put(9,4){$\{\emptyset\}$}
\put(19,14){$\{\{\emptyset\}\}$}
\put(-16,19){$\{\emptyset,\{\emptyset\}\}$}
\end{picture}                             
\end{center}
\color{black}
\bigskip

\vfill
  
\end{enumerate}
\end{enumerate}

\bigskip

\begin{enumerate}
\item[2.]
Find sets $A$ and $B$ satisfying the given conditions.
\begin{enumerate}
\item[(a)] $A\in B$ and $A\not\subseteq B$.
\vfill
\color{blue}
There are many answers,
such as $A=\{\emptyset\}$ and $B=\{\{\emptyset\}\}$.
\color{black}

\vfill
  
\item[(b)] $A\in B$ and $A\subseteq B$.
\vfill
\color{blue}
You could take $A=\emptyset$ and $B=\{\emptyset\}$.  
\color{black}

\vfill
  
\item[(c)] $A\notin B$ and $A\subseteq B$.
\vfill
\color{blue}
  You could take $A=B=\emptyset$.
\color{black}

\vfill
  
\end{enumerate}
\end{enumerate}

  
\bigskip
\bigskip

\begin{enumerate}
\item[3.]
Show that $\bigcup {\mathcal P}(x) = x$.
\end{enumerate}

\vfill
\color{blue}
According to the Axiom of Extensionality,
to show that the sets $\bigcup {\mathcal P}(x)$ and $x$ are equal,
we must show that they have the same elements.

\smallskip
\noindent
    {\bf Claim 1:}
    We show that any element $y\in \bigcup {\mathcal P}(x)$
    is an element of $x$.
\smallskip

If $y\in \bigcup {\mathcal P}(x)$, then
$y$ is an element of an element of ${\mathcal P}(x)$
according to the definition of union. This means
that there is some set $z$ such that
$y\in z\in {\mathcal P}(x)$.  Since $z\in {\mathcal P}(x)$,
we know that $z\subseteq x$ according to
the definition of ${\mathcal P}(x)$.
But now we know that $y\in z$ and $z\subseteq x$,
so we derive $y\in x$ from the definition of $\subseteq$.
\smallskip

\noindent
    {\bf Claim 2:} We show that any element $y\in x$ is an element of
    $\bigcup {\mathcal P}(x)$.
\smallskip

Choose any $y\in x$.
Let $z=\{y\}$ (which \underline{is} a set according to the Pairing Axiom).
Observe $z\subseteq x$
by the definition of subset.
Now we have
$y\in z$ and $z\subseteq x$, hence $y\in z\in {\mathcal P}(x)$.
This yields $y\in \bigcup {\mathcal P}(x)$.

\end{document}
