Critique my proof of: Suppose $A$, $B$, $C$, and $D$ are sets. Prove that $(A \times B) \cup (C \times D) \subseteq (A \cup C) \times (B \cup D)$.

Critique my proof on correctness, structure, etc.

Proof. Let $p = (x,y) \in (A \times B) \cup (C \times D)$. Thus, $(x, y) \in (A \times B)$ or $(x, y) \in (C \times D)$.

Case #1

Suppose $(x, y) \in (A \times B)$. By definition of cartesian product, $x \in A$ and $y \in B$. It follows that $x \in (A \cup C)$ and $y \in (B \cup D)$ by definition of union.

Case #2

Suppose $(x, y) \in (C \times D)$. By definition of cartesian product, $x \in C$ and $y \in D$. It follows that $x \in A \cup C$ and $y \in C \cup D$ by definition of union.

$\therefore$ Because $(x, y)$ is an arbitrary element of $(A \times B) \cup (C \times D)$, $(A \times B) \cup (C \times D) \subseteq (A \cup C) \times (B \cup D)$.

Solution 1:

This proof is very tight and if I was grading it I'd give it full credit.

If you want to be more concise you could say that since the union operator is commutative, that proving only case one is necessary since case two is the same argument just relabeling the sets.

Something along the lines of

Without loss of generality, assume that $(x,y)\in A\times B$. This means that $x\in A$ and $y\in B$ which, by definition of the union means that $x\in A\cup C $ and $y\in B\cup D \implies (x,y)\in(A\cup C)\times (B\cup D)$. The same argument can be made if $(x,y)\in C\times D$.

$\therefore$ since all $p\in (A\times B)\cup(C\times D)\implies p\in (A\cup C)\times(B\cup D)$ we can say $(A\times B)\cup(C\times D)\subseteq (A\cup C)\times(B\cup D)$. Q.E.D.

However, perhaps I should not be encouraging people to be lazy in proofs like me