I'm building off of this question
if-a-and-b-are-nonempty-sets-prove ...
but instead $A$ and $B$ are not equal:
Prove that if $A$ and $B$ are nonempty sets where $A \neq B$, then $A \times B \neq B \times A$
So far I have
where $x \in A$ and $y \in B$ so $(x, y) \in A \times B$
If $A \neq B,$ there is an elements in $A$ not in $B$ (or, in $B$ and not in $A$). How are the elements of $A\times B$ and $B \times A$? They are ordered pairs.
Observe that
$$A\neq B\iff A\not\subset B\;\;\text{or}\;\;B\not\subset A$$
Suppose WLOG $\;A\not\subset B\;\implies\;\exists\,a\in A\setminus B\;$ , and since $\;B\neq\emptyset\;$, there exists $\;b\in B\;$ , so
$$(a,b)\in A\times B\;,\;\;\text{yet}\;\;(a,b)\notin B\times A\;\;\text{ since}\;\;a\notin B$$
You can prove by Contrapositive, namely: $(A \times B)=(B \times A) \to (A= \emptyset \vee B =\emptyset \vee A=B)$
In fact:
Theorem:$$(A \times B)=(B \times A) \leftrightarrow (A= \emptyset \vee B =\emptyset \vee A=B)$$ Proof: $$"\to" \, \text{ by Contradiction}$$ Supppose that $\emptyset \neq A \neq B \neq \emptyset$, we have $$ A \nsubseteq B \vee B \nsubseteq A \to \exists x \in A:(x \notin B) \vee \exists s \in B:(s \notin A) \to $$ $$ \to (x,y) \in (A \times B) \, [\text{with } y \in B\neq \emptyset] \, \vee \, (r,s) \in (A \times B) \,[\text{with } r \in A \neq \emptyset] \to$$ $$ \to (x,y) \in (B\times A) \, \vee \, (r,s) \in(B\times A)\,[\text{because} (A\times B)=(B \times A)] \to $$$$\to x \in B \, \vee \, s \in A \,\text{ (which contradicts our assumptions } x \notin B, s\notin A) $$ $$"\leftarrow"$$ Suppose $A=B$, we have $$(A \times B)=(A \times A)=(B\times B)=(B\times A)$$ Suppose $A =\emptyset \vee B = \emptyset$, we have $$(A \times B)=\emptyset=(B \times A)$$
Observe that, if $A\times B\ne\emptyset,$ then $$A=\{x:\exists y\ (x,y)\in A\times B\}$$ and $$B=\{y:\exists x\ (x,y)\in A\times B\}.$$ Hence, if $A\times B=C\times D\ne\emptyset,$ then $A=C$ and $B=D.$
In particular, if $A\times B=B\times A\ne\emptyset,$ then $A=B.$
