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\newtheorem{thm}{Theorem}

\title{Cartesian Products}
\author{Doug Baldwin}
\date{Spring 2021}


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\maketitle

\begin{thm}
If $A$, $B$, and $C$ are sets, then $(A \cap B) \times C = (A \times C) \cap (B \times C)$.
\end{thm}

\begin{proof}
We will prove that if $A$, $B$, and $C$ are sets, then $(A \cap B) \times C = (A \times C) \cap (B \times C)$ by using element chasing to show that each set is a subset of the other.

First, let $(x,y)$ be an element of $(A \cap B) \times C)$. So $x \in A \cap B$ and $y \in C$, i.e., $x \in A$ and $x \in B$. Since $x \in A$ and $y \in C$, $(x,y) \in A \times C$. Furthermore, since $x \in B$ and $y \in C$, $(x,y)$ is also in $B \times C$. Since $(x,y) \in A \times C$ and $(x,y) \in B \times C$, we conclude that $(x,y) \in (A \times C) \cap (B \times C)$.

Now, suppose $(x,y) \in (A \times C) \cap (B \times C)$. Then $(x,y) \in A \times C$ and $(x,y) \in B \times C$. For this to be true, $x$ must be a member of $A$ and $B$, and $y$ must be in $C$. Since $x \in A$ and $x \in B$, $x \in A \cap B$. Finally, since $x \in A \cap B$ and $y \in C$, the pair $(x,y) \in (A \cap B) \times C$.

Having shown that each product is a subset of the other, we have shown that if $A$, $B$, and $C$ are sets, then $(A \cap B) \times C = (A \times C) \cap (B \times C)$.
\end{proof}

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