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\item[2.3]\begin{center}
\section*{Chapter 2}
\end{center}
\begin{enumerate}
\item[2.1] This problem is split into $5$ parts.
\begin{enumerate}
\item Since every element of $A$ is in $A$, we have $A\subseteq A$. However, $A=A$, which means $A\not\subset A$.
\item Since every element of $B$ is in $A$, we have $B\subseteq A$. Also, $B\neq A$, which means $B\subset A$ as well.
\item We have an element $\{4,5\}\in C$ which is not in $A$, so we have $C\not\subseteq A$ and $C\not\subset A$.
\item We have the elements of $B$ are $2$,$3$, and $4$, so $4\in B$. However, the elements of $C$ are $3$ and $\{4,5\}$, so $4\not\in C$.
\item The subsets of $B$ are $\{\}$, $\{4\}$, $\{3\}$, $\{2\}$, $\{2,3\}$, $\{2,4\}$, $\{3,4\}$, and $\{2,3,4\}$. There are $8$ subsets in all.
\end{enumerate}
\item[2.2]$\emptyset$ has no elements, every element of
$\emptyset$ is in any set, as there is no element to check
it for. So, the empty set is a subset of any set, as desired.
\item[2.3]$B$ is the set of all subsets of $A$
such that $1$ is not in said subset. From here, it is easy
to list out the elements of $B$:
$$B=\{\{\},\{2\},\{3\},\{6\},\{2,3\},\{2,6\},\{3,6\},\{2,3,6\}\}.$$
\item[2.2.1]$\{\emptyset\}$ is the set with
$\emptyset$ as its only element. Since
$\emptyset$ is the set with no elements, this means
$\emptyset\neq\{\emptyset\}$.
\item[2.2.2]$7$ parts.
\begin{enumerate}
\item Since $\{1,3,5,9\}$ has elements $1$, $3$, $5$, and $9$, we have that our statement is true.
\item Since we can remove the $2$ in the right hand side (redundancy), and order does not matter, we have that our statement is true.
\item Since the element so of $\{3,5,9\}$ are $3$, $5$, and $9$, and $\{5\}$ is not among these numbers, we have that our statement is false.
\item The only element of $\{\{4\}\}$ is $\{4\}$, which is not the same as $4$. So, our statement is false.
\item Since the empty set is a subset of any set by 2.2, and $\emptyset\neq\{1,2,9\}$, we have that our statement is true.
\item We have the elements of $\{\emptyset,\{1\},82\}$ are $\emptyset$, $\{1\}$, and $82$. Since $\emptyset$ is included in these elements, we have our statement is true.
\item Since every even integer is an integer, and there are integers which are not integers, we have $\{x\mid x\textrm{ is an even integer}\}\subseteq\mathbb{Z}$ and $\{x\mid x\textrm{ is an even integer}\}\neq\mathbb{Z}$. Combining these, we get that our statement is true.
\end{enumerate}
\item[2.2.3]$A$ is in $A$, by the
definition of $\subseteq$ we have
$A\subseteq A$.
\item[2.2.4]$A\subseteq B$ means every element of $A$ is
in $B$. $B\subseteq A$ means every element of
$B$ is in $A$. Suppose, for the sake of
contradiction, that we can take an element $x$ such that
$x\in A$ but $x\not\in B$. (The choice of
$A$ and $B$ can be switched easily.) This is
the same as $A\neq B$. However, all elements in
$A$ must be in $B$, so this $x$
cannot exist. (The choice of $x\in B$ and
$x\not\in A$ can be disproved similarly.) Therefore, we
have a contradiciton, so $A=B$, as desired.
\item[2.2.5]Suppose we have two sets such that $A\subseteq B$ and
$B\subseteq C$. This means that every element in
$A$ is in $B$, and every element in
$B$ is in $C$. Since every element in
$A$ is in $B$, we have to have every element
in $A$ is in $C$, therefore
$A\subseteq C$, as desired.
For proper subsets, suppose we have two sets such that $A\subset B$ and $B\subset C$. We already have that $A\subseteq C$, by the previous argument. Since $B\subset C$, we have that there must be an element in $C$ not in $B$. That element would therefore not be able to be part of $A$, so we must have $A\neq C$. So, we have $A\subset C$, as desired.
\item[2.2.6] We would have every element in $B$ be in $\emptyset$. Since $\emptyset$ has no elements, we have $B$ cannot have any elements, so $B=\emptyset$.
\item[2.2.7] First, for the sake of contradiction, assume that we have two sets $A$ and $B$ such that $A\subseteq B$ but $\#(A)>\#(B)$. This would mean we have more elements in $A$ than in $B$, so we have to have some element in $A$ not in $B$ (by the pigeonhole principle). This violiates $A\subseteq B$, so this is a contradiction. Therefore, we must have $\#(A)\leq \#(B)$, as desired.
Now, assume we have two sets such that $A\subseteq B$ and $\#(A)=\#(B)$. Then, we have all elements of $A$ are in $B$, and $A$ has the same number of elements as $B$. This has to mean that $A$ and $B$ have the same elements, so $A=B$.
Finally, let $A$ and $B$ be two sets such that $A\subset B$. We have that $\#(A)\leq \#(B)$, as $A\subset B$ means $A\subseteq B$ as well. However, we cannot have $\#(A)=\#(B)$ as that would mean $A=B$, violating $A\subset B$. So, we have $\#(A)<\#(B)$.
\item[2.2.8] Assume, for the sake of contradiction, that we have two sets $A$ and $B$ such that $A\subset B$ and $B\subset A$. This means that $A\subseteq B$ and $B\subseteq A$, which, by 2.2.4, we have that it means $A=B$. However, $A\subset B$ (and $B\subset A$) means that $A\neq B$, so we have a contradiction. So, we cannot have two sets $A$ and $B$ such that $A\subset B$ and $B\subset A$, as desired.
\item[2.2.9] We have the only subset of $\emptyset$ is $\{\}$. So, we have $\mathcal{P}(\emptyset)=\{\{\}\}.$ So, the only subsets of $\mathcal{P}(\emptyset)$ are the empty set and the set containing the empty set. Therefore, we have
$$\mathcal{P}(\mathcal{P}(\emptyset))=\{\{\},\{\{\}\}\},$$
which is our answer.
\item[2.2.10] We have $\#(\mathcal{P}(S))$ is the number of subsets of $S$. We can think of each subset as follows: for each element in $S$, we choose whether or not that element is in our desired subset. This means that for each element in $S$, we have $2$ possibilities. Multiplying it out, we get $2^n$ subsets, so $\#(\mathcal{P}(S))=2^n$, as desired.
\item[2.4] This problem is split into two parts.
\begin{enumerate}
\item We get that $A\cup B=B$. To show this, first we show that every element of $A\cup B$ is in $B$. Since $A\subseteq B$, we have every element of $A$ is in $B$. Since $A\cup B$ consists of all elements in either $A$ or $B$, but every element in $A$ is already in $B$, we have that every element in $A\cup B$ is in $B$, so $A\cup B\subseteq B$, as desired.
Now, we show that every element in $B$ is in $A\cup B$. This is true by definition, as $A\cup B$ is the set of all elements either one of $A$ or $B$. So, $B\subseteq A\cup B$.
Therefore, we have $A\cup B=B$, as desired.
\item We get that $A\cap B=A$. To show this, first show that every element of $A\cap B$ is in $A$. Since $A\cap B$ is the set of all elements in both $A$ and $B$, we have that $A\cap B\subseteq A$, by definition.
Now, we show that every element in $A$ is in $A\cap B$. Since $A\subseteq B$, we have every element in $A$ is in $B$. Also, $A\cap B$ is the set of all elements in both $A$ and $B$, which means that $A\cap B$ contains every element in $A$, as each element in $A$ is in $B$. So, we have $A\subseteq A\cap B$.
Therefore, we have $A\cap B=A$, as desired.
\end{enumerate}
\item[2.5] To prove that
$$A\cap(B\cup C)=(A\cap B)\cup(A\cap C),$$
we have to show that
$$A\cap(B\cup C)\subseteq (A\cap B)\cup(A\cap C)\textrm{ and }(A\cap B)\cup(A\cap C)\subseteq A\cap(B\cup C).$$
To show the first statement, let $x\in A\cap (B\cup C)$. This means that $x\in A$ and $x\in B\cup C$. $x\in B\cup C$ implies that either $x\in B$ or $x\in C$ (or both). If $x\in B$, then $x\in A\cap B$, so $x\in (A\cap B)\cup(A\cap C)$. Similarly, if $x\in C$, then $x\in A\cap C$, so $x\in (A\cap B)\cup(A\cap C)$. (Note that $x$ being in both $B$ and $C$ would fit either case, so we are okay.) This means that
$$A\cap (B\cup C)\subseteq(A\cap B)\cup(A\cap C),$$
as desired.
Now, we show the second statement. Let $x\in (A\cap B)\cup(A\cap C)$. This means either $x\in A\cap B$ or $x\in A\cap C$ (or both). If $x\in A\cap B$, then we have $x\in A$ and $x\in B$, so $x\in B\cup C$ and therefore is also in $A\cap(B\cup C)$. Similarly, if $x\in A\cap C$, then we have $x\in A$ and $x\in C$. This means $x\in B\cup C$, and therefore $x\in A\cap(B\cup C)$. (Again, note that $x$ being in both $A\cap B$ and $A\cap C$ would fit either case, so we are okay.) We therefore have
$$(A\cap B)\cup(A\cap C)\subseteq A\cap (B\cup C),$$
so we have
$$A\cap(B\cup C)=(A\cap B)\cup(A\cap C),$$
as desired.
\item[2.3.1] We get that $A\cup \emptyset=A$. Note that we have, by definition, $A\subseteq A\cup\emptyset$. Also, since $A\cup \emptyset$ consists of all elements in $A$ or in $\emptyset$. Since there are no elements in $\emptyset$, we have $A\cup\emptyset$ consists of the elements in $A$, so $A\cup\emptyset\subseteq A$. Therefore, we have $A\cup\emptyset =A$, as desired.
Also, we get that $A\cap \emptyset=\emptyset$. Note that we have, by definition, that $A\cap\emptyset\subseteq \emptyset$. Since $\emptyset$ is a subset of any set (by 2.2), we have $\emptyset\subseteq A\cap \emptyset$. So, we have $A\cap \emptyset=\emptyset$. So, we are done.
\item[2.3.2] First, we show that $A\cup A=A$. We have, by definition, that $A\subseteq A\cup A$. Also, since $A\cup A$ consists only of elements both in $A$ and $A$, we have $A\cup A$ consists of elements in $A$, as $A$ and $A$ are the same set. So, we have $A\cup A\subseteq A$. Therefore, we have $A\cup A=A$, as desired.
Now, we show that $A\cap A=A$. By defintion, we have $A\cap A\subseteq A$. Now, since $A\cap A$ is the set of all elements in both $A$ and $A$, which is the same as the set of all elements in $A$, we have $A\subseteq A\cap A,$ so $A\cap A=A$, as desired.
Since $A\cup A=A$ and $A\cap A=A$, we have $A\cup A=A\cap A$, so we get that
$$A\cup A=A\cap A=A,$$
as desired.
\item[2.3.3] We get that $S\cup \{x\}=S$. First, note that, by definition, $S\subseteq S\cup \{x\}$. Also, we have $S\cup \{x\}$ is the set of all elements in $S$ or in $\{x\}$. Since $x\in S$, we have that $S\cup \{x\}$ is the set of all elements in $S$. Therefore, $S\cup\{x\}\subseteq S$, so $S\cup\{x\}=S$, as desired.
Also, we get that $S\cap\{x\}=S$. To show this, first, note that, by definition, $S\cap\{x\}\subseteq S$. Since $S\cap\{x\}$ is the set of all elements in both $S$ and $\{x\}$, but $x\in S$, we have $S\cap\{x\}$ is the set of all elements in $S$. So, we have $S\subseteq S\cap\{x\}$, which means $S\cap\{x\}=S$, as desired. So, we are done.
\item[2.3.4] To show this, we show each of the following statements:
$$A\cup(B\cap C)\subseteq (A\cup B)\cap(A\cup C)\textrm{ and }(A\cup B)\cap(A\cup C)\subseteq A\cup(B\cap C).$$
To show the first one, let $x\in A\cup(B\cap C)$. We have either $x\in A$ or $x\in B\cap C$ (or both). If $x\in A$, we have $x\in A\cup B$ and $x\in A\cup C$, so we have $x\in(A\cup B)\cap(A\cup C)$. If we have $x\in B\cap C$, we have $x\in B$ and $x\in C$. This means $x\in A\cup B$ and $x\in A\cup C$, so we have $x\in(A\cup B)\cap(A\cup C)$. Therefore, we have $A\cup(B\cap C)\subseteq (A\cup B)\cap(A\cup C)$, as desired.
Now, we show the second statement. Let $x\in(A\cup B)\cap (A\cup C)$. This means $x\in A\cup B$ and $x\in A\cup C$. So, we have the following cases:
\textit{Case 1: $x\in A$ and $x\in A$}\newline
In this case, we have $x\in A$. So, we have $x\in A\cup(B\cap C)$, as desired.
\textit{Case 2: $x\in A$ and $x\in C$}\newline
In this case, we have $x\in A$. So, we have $x\in A\cup(B\cap C)$, as desired.
\textit{Case 3: $x\in B$ and $x\in A$}\newline
In this case, we have $x\in A$, so we have $x\in A\cup(B\cap C)$, as desired.
\textit{Case 4: $x\in B$ and $x\in C$}\newline
In this case, we have $x\in B\cap C$, so we have $x\in A\cup(B\cap C)$, as desired.
Therefore, we get that $(A\cup B)\cap(A\cup C)\subseteq A\cup(B\cap C),$ which means
$$(A\cup B)\cap(A\cup C)=A\cup(B\cap C),$$
as desired.
\item[2.3.5] This problem is split into two parts.
\begin{enumerate}
\item We have that $A\cup B$ is the set of all elements in either $A$ and $B$. Since this is the same as $A$, we must have every element in $B$ in $A$, so we have $B\subseteq A$.
\item We have that $A\cap B$ is the set of all elements in both $A$ and $B$. Since this is the same as $A$, we must have every element of $A$ in $B$, so we have $A\subseteq B$.
\end{enumerate}
\item[2.3.6] We have $\mathcal{P}(S\cap T)$ is the set of all subsets of $S\cap T$. This is the same as all sets in both $\mathcal{P}(S)$ and $\mathcal{P}(T)$. (This is true as every subset in both cannot use elements in only one of $S$ or $T$.) So, we have $\mathcal{P}(S\cap T)$ is the same as $\mathcal{P}(S)\cap\mathcal{P}(T)$.
However, we still need to prove this! So, first, let $x\in \mathcal{P}(S\cap T)$. This means that $x$ is a subset of $S\cap T$, so $x$ consists of elements in both $S$ and $T$. So, we have $x\in\mathcal{P}(S)$ and $x\in\mathcal{P}(T)$, and so $\mathcal{P}(S\cap T)\subseteq\mathcal{P}(S)\cap\mathcal{P}(T).$
Now, let $x\in\mathcal{P}(S)\cap\mathcal{P}(T)$. This means $x$ must be a subset of both $S$ and $T$. The only way for this to be possible is for $x$ to consist only of elements in $S$ and $T$; that is, for $x$ to consist only of elements in $S\cap T$. This is the same as saying that $x$ is a subset of $S\cap T$, so $x\in\mathcal{P}(S\cap T),$ so we have $\mathcal{P}(S)\cap\mathcal{P}(T)\subseteq\mathcal{P}(S\cap T).$ Therefore, we have that $\mathcal{P}(S\cap T)=\mathcal{P}(S)\cap\mathcal{P}(T),$ as desired.
Note that $\mathcal{P}(S\cup T)$ involves subsets using elements in both $S$ and $T$. Since $\mathcal{P}(S)$ and $\mathcal{P}(T)$ only involve elements in one of $S$ or $T$, we have $\mathcal{P}(S\cup T)$ contains elements not in $\mathcal{P}(S)$ and $\mathcal{P}(T)$, so we cannot describe $\mathcal{P}(S\cup T)$ in a similar way.
\item[2.6] We construct the following truth table:
$$\begin{tabular}{c|c|c}
$p$ & $\neg p$ & $p\wedge (\neg p)$\\ \hline
T & F & F\\
F & T & F
\end{tabular}$$
This shows that '$p$ and (not $p$)' is always false for any statement $p$, as desired.
\item[2.7] We construct the following truth table:
$$\begin{tabular}{c|c|c|c|c|c|c}
$p$ & $q$ & $p\vee q$ & $\neg p$ & $\neg q$ & $(\neg p)\wedge(\neg q)$ & $\neg((\neg p)\wedge(\neg q))$\\ \hline
T & T & \textbf{T} & F & F & F & \textbf{T}\\
T & F & \textbf{T} & F & T & F & \textbf{T}\\
F & T & \textbf{T} & T & F & F & \textbf{T}\\
F & F & \textbf{F} & T & T & T & \textbf{F}
\end{tabular}$$
We bolded each final value to make things easier for us. As we can see, this means that either $p\vee q$ and $\neg((\neg p)\wedge(\neg q))$ are both true or both false, as desired.
\item[2.4.1] This problem is split into $6$ parts.
\begin{enumerate}
\item This is a statement.
\item This is a statement.
\item This is not a statement, because infinity could be really cool to one person, but not cool to someone else; it's an opinion.
\item This is a statement.
\item This is not a statement, as this asks someone something; it's a question.
\item This is not a statement; it's an expression.
\end{enumerate}\end{enumerate}