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\centerline{\bf Math 365 -- Wednesday 1/30/19}

\noindent Course website: https://zdaugherty.ccnysites.cuny.edu/teaching/m365s19/





\begin{warmup} Recall that the power set of a set $A$ is 
$$\cP = \{ X ~|~ X \subseteq A\}.$$
\begin{enumerate}
\item What is $|\emptyset|$? What is $|\{\emptyset\}|$?
\item Let $A = \{x\}$. Calculate $\cP(A)$ and $\cP(\cP(A))$. 
\item Let $A = \emptyset$. Calculate $\cP(A)$ and $\cP(\cP(A))$. 
\item Give an example of a set $A$ such that $A \cap \cP(A) = \emptyset$.
\item Give an example of a set $A$ such that $A \cap \cP(A) \neq \emptyset$. 
\item True or false and why: For any set $A$, $\{\emptyset\} \subseteq \cP(\cP(A))$. 
\item Explain why $\cP(A) \cap \cP(\cP(A)) \neq \emptyset$. 
\end{enumerate}
\end{warmup}

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\begin{try}
Fix two sets $A$ and $B$ and a universal set $U$. 
Draw and shade in diagrams to decide whether the following identities are true or false. For those that are false, give a concrete example illustrating the failure of the identity. (Your argument against should start with something like \emph{``Let $A=???$ and $B = ???$. Then\dots''}.) 
\begin{enumerate}[(a)]
\item $\overline{(A \cap B)} = \overline{A} \cup \overline{B}$
\item $\overline{(A \cup B)} = \overline{A} \cup \overline{B}$
\item $A \cap (A \cup B) = (U-(B-A)) - \overline{(A \cup B)}$
\end{enumerate}
\end{try}

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\begin{try}
Let $A_i = \{1,2,\dots,i\}$ for $i=1,2,3,\dots$. \\
Calculate
$$\bigcup_{i=1}^n A_i  \qquad \text{ and } \qquad \bigcap_{i=1}^n A_i$$
for $n=2,4,5$ (meaning compute $\bigcup_{i=1}^2 A_i$, $\bigcup_{i=1}^3 A_i$, etc.). Make a hypothesis about what $\bigcup_{i=1}^n A_i$ and $\bigcap_{i=1}^n A_i$
are for general $n$. Then make a hypothesis about what $\bigcup_{i=1}^\infty A_i$ and $\bigcap_{i=1}^\infty A_i$ are. Explain in words why. (More formal proof comes next.)
\end{try}

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\begin{try}\label{ex:set-proof}
Show that if  $A_i = \{1,2,\dots,i\}$ for $i=1,2,3,\dots$, then 
$\bigcap_{i=1}^\infty A_i = \{1\}.$\\
{[}Hint: Call $\cA = \bigcap_{i=1}^\infty A_i$ for brevity. First argue that $\{1\} \subseteq \cA$. For the reverse, explain why  $\cA = A_1 \cap \bigcap_{i=2}^\infty A_i$, and use that expression to show that $\cA \subseteq \{1\}$. You will want to use the fact that the $X \cap Y$ of two sets $X$ and $Y$ is a subset of $X$ (also it's a subset of $Y$){]}
\end{try}


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\begin{try}
Argue formally that $\overline{A \cup B}= \overline{A} \cap \overline{B}$.
\end{try}


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\begin{try}Find 
$$\bigcup_{i = 1}^\infty A_i \qquad \text{ and } \qquad \bigcap_{i = 1}^\infty A_i$$ for the following. In each case, justify your answer with a formal argument (like in Exercise \ref{ex:set-proof}).
\begin{enumerate}[(a)]
\item $A_i = \{i, i+1, i+2, \dots\}$
\item $A_i = \{0,i\}$
\item $A_i = (i,\infty) = \{x \in \RR ~|~ i < x \}$
\end{enumerate}
\end{try}
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