To prove simple set containment

Question

$ A \times (B \cup C) = (A \times B) \cup (A \times C) $

Let

$p=(x,y) \in A \times (B \cup C)$. So $x \in A$ and $y \in (B \cup C)$. Three cases arise from here:

1. $y \in B$

2.$y \in C$

3.$y \in B \cap C$

Now proceeding from each case os trivial. But my question is that my textbook does not make third case which i do not understand why so.

Thanks

Answer 1

Since your third case is included in each of the two previous ones, it is already solved by your preceding reasonning.

Assuming you have solved case 1, then $y \in B \cap C \implies y \in B \implies p \in ( A \times B) \cup (A \times C)$

Written logically, $y∈(B∪C) \Leftrightarrow (y∈B$ or $y∈C)$ and there is no third case to consider.

Answer 2

The case $y\in B\cap C$ have been already considered, as $B\cap C$ is contained in both $B$ and $C$. On the other hand, it is easier to prove the statement as follows:

\begin{align*} p=(x,y)\in A\times (B\cup C) &\Leftrightarrow (x\in A) \text{ and }(y\in B\cup C)\\ &\Leftrightarrow (x\in A) \text{ and }(y\in B \text{ or }y\in C)\\ &\Leftrightarrow (x\in A \text{ and }y\in B)\text{ or }(x\in A \text{ and }y\in C) \\ &\Leftrightarrow p\in (A\times B)\text{ or }p\in(A\times C)\\ &\Leftrightarrow p\in (A\times B)\cup (A\times C) \end{align*}

Hence, $A\times (B\cup C)=(A\times B)\cup (A\times C)$.

Answer 3

We have $(x,y)\in A\times (B\cup C)$ iff $x\in A\wedge y\in B\cup C$ iff $x\in A\wedge (y\in B\vee y\in C)$ iff $(x\in A\wedge y\in B)\vee(x\in A\wedge y\in B)$ iff $(x,y)\in A\times B \vee (x,y)\in A\times C$ iff $(x,y)\in (A\times B)\cup (A\times C)$.