Exercise 3.1.7

Question

Let $A$, $B$, $C$ be sets. Show that $A \cap B \subseteq A$ and $A \cap B \subseteq B$.
Furthermore, show that $C \subseteq A$ and $C \subseteq B$ if and only if $C \subseteq A \cap B$. In a similar spirit, show that $A \subseteq A \cup B$ and $B \subseteq A \cup B$, and furthermore that $A \subseteq C$ and $B \subseteq C$ if and only if $A \cup B \subseteq C$.

Khagan Gulusoy · Accepted Answer

\begin{proof}
\begin{enumerate}
    \item ($A\cap B\subseteq A$) Let $x\in A\cap B$ be arbitrary. By Definition 3.1.23, we have $x\in A$ and $x\in B.$ In particular, $x \in A$, as desired. We prove that $A\cap B\subseteq B$ similarly.
    %Thus, we can conclude that $x\in A\cap B\implies x\in A$ is true since $x\in A\cap B$ is true and $x\in A$ is true, as desired.
    
    %\item ($A\cap B\subseteq B$) Follows similarly. 
    %Let $x\in A\cap B$ be arbitrary. We have $x\in A$ and $x\in B$ by Definition 3.1.23. In particular, $x\in B$ is also true. Thus, we have $A\cap B\subseteq B$, as desired.
    
    \item ($C\subseteq A\ 	ext{and} \ C\subseteq B\iff C\subseteq A\cap B$) It suffices to show that $C\subseteq A\ 	ext{and} \ C\subseteq B\implies C\subseteq A\cap B$ and vice versa.
    \begin{itemize}
        \item ($C\subseteq A\ 	ext{and} \ C\subseteq B\implies C\subseteq A\cap B$) Suppose that $C$ is such that $C \subseteq A$ and $C \subseteq B$ simultaneously. Let $x\in C$ be arbitrary. By Definition 3.1.15, since $C$ is a subset of $A$, we have $x\in A$, and since $C$ is a subset of $B$, we have $x\in B$. By Definition 3.1.23, we have $x\in A\cap B$ (because $x\in A$ and $x\in B$). %Since $x\in C$, by Definition 3.1.15 we have $C\subseteq A\cap B$.
        %Let $x\in C$ such that $C\subseteq A$ and $C\subseteq B$ be arbitrary. Since $x\in C,$ we have $x\in A$ and $x\in B$ by Definition 3.1.15 (because $C\subseteq A$ and $C\subseteq B$). Therefore, by Definition 3.1.23 we have $x\in A\cap B.$ Thus, we have $x\in C\implies x\in A\cap B$, which means that $C\subseteq A\cap B$. 
        %Therefore, it is true that $C\subseteq A\ 	ext{AND} \ C\subseteq B\implies C\subseteq A\cap B$.
        \item ($C\subseteq A\cap B\implies C\subseteq A\ 	ext{and} \ C\subseteq B$) Suppose that $C$ is such that $C\subseteq A\cap B$. Let $x\in C$ be arbitrary. Since $C\subseteq A\cap B$, we have $x\in A\cap B$ by Definition 3.1.15, which by Definition 3.1.23 means that $x\in A$ and $x\in B.$ Since by assumption $x\in C$, by Definition 3.1.15 we have $C\subseteq A$ (because $x\in A$) and $C\subseteq B$ (because $x\in B$).
        %Therefore, the statements $x\in C\implies x\in A$ and $x\in C\implies x\in B$ are also true since $x\in C, x\in A$ and $x\in B$ all hold. Therefore, by Definition 3.1.15 we have $C\subseteq A$ and $C\subseteq B.$ 
        %and thus, we get $C\subseteq A\cap B\implies C\subseteq A\ 	ext{and} \ C\subseteq B$.
    \end{itemize}
    %Since both implications hold, the statement we had to prove is true by the definition of equivalence.
    
    \item ($A\subseteq A\cup B$ and $B\subseteq A\cup B$) Let $x\in A$ be arbitrary. Since $x\in A,$ by Axiom 3.4 we can conclude that $x\in A\cup B.$ We prove that $B\subseteq A\cup B$ similarly. 
    %Since $x\in A$ by assumption, $A\subseteq A\cup B$ hold, as desired.
    %Therefore, $x\in A\implies x\in A\cup B$ is true. But the previous statement by the definition of subsets means that $A\subseteq A\cup B,$ as desired.
    
    %\item ($B\subseteq A\cup B$) Follows similarly.
    %Let $x\in B$ be arbitrary. Since $x\in B,$ by Axiom 3.4 we can conclude that $x\in A\cup B.$ Since $x\in B$ by assumption, $B\subseteq A\cup B$ hold, as desired.
    %Therefore, $x\in B\implies x\in A\cup B$ is also true. But the previous statement by the definition of subsets means that $B\subseteq A\cup B,$ as desired.
    
    \item ($A\subseteq C\ 	ext{and}\ B\subseteq C\iff A\cup B\subseteq C$) It suffices to show that $A\subseteq B\ 	ext{and} \ B\subseteq C\implies A\cup B\subseteq C$ and vice versa. To prove that $A\subseteq C\ 	ext{and}\ B\subseteq C\implies A\cup B\subseteq C$ we suppose that $A \subseteq C$ and $B \subseteq C$. Let $x \in A\cup B$ be arbitrary. By Axiom 3.4 this means that $x\in A$ or $x\in B$. In the case when $x\in A$ we have $x\in C$ since $A\subseteq C$ by assumption. In the case when $x\in B$ we have $x\in C$ since $B\subseteq C$ by assumption. As you can see, in both cases we have $x \in C$. Since our choice of $x$ was arbitrary, this follows for all $x\in A \cup B$. The other part $A\cup B\subseteq C\implies A\subseteq C\ 	ext{and}\ B\subseteq C$ follows similarly.

%\begin{itemize}

%\item ($A\subseteq С\ 	ext{AND} \ B\subseteq C\implies A\cup B\subseteq C$) Let $x\in A$ be arbitrary. By assumption we have $x\in A$
        %By assumption we have $x\in C.$ Since $x\in A$, $x\in A\cup B$ is also a true statement. Since $x\in C,$ $x\in A\cup B\implies x\in C$ is also true. But this by the definition of subsets means that $A\cup B\subseteq C,$ as desired.
        
        %\item ($A\cup B\subseteq C\implies A\subseteq B\ 	ext{AND} \ B\subseteq C$) Let $x\in A\cup B$ be arbitrary. By the definition of subsets we have $x\in A\cup B\implies x\in C$. By Axiom 3.4, $x\in A\cup B$ means that $x\in A$ or $x\in B$. We now look at these cases.
        %\begin{enumerate}
         %   \item ($x\in A$) 
         %   \item 
        %\end{enumerate}
        % By assumption we have $x\in B,$ and therefore, $x\in C.$ Since $x\in A$, $x\in A\cup B$ is also a true statement. Since $x\in C,$ $x\in A\cup B\implies x\in C$ is also true. But this by the definition of subsets means that $A\cup B\subseteq C,$ as desired.

%\end{itemize}

\end{enumerate}
\end{proof}

Exercise 3.1.7

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Exercise 3.1.7

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