# Critique my proof (beginner)

Critique my proof on correctness, structure, etc.

Theorem.
Suppose $A$ and $B$ are sets. Then $A \times B = B \times A \iff A = \emptyset, B = \emptyset,$ or $A = B$

Proof.

($\rightarrow$) Assume for the sake of contradiction that $A \neq \emptyset$ and $B \neq \emptyset$ and $A \neq B$. Let $x$ be an arbitrary element such that $x \in A$. Because $A \times B = B \times A$, it follows by definition of cartesian product that $x \in B$ as well. Likewise, let y be an arbitrary element such that $y \in B$. By the same definitions, it follows that $y \in A$. Thus by definition of substet, $A \subseteq B$ and $B \subseteq A$, so $A = B$. This is a contradiction, so we can conclude that if $A \times B = B \times A$, then $A = \emptyset, B = \emptyset$, or $A = B$

$(\leftarrow)$ Suppose $A = \emptyset$, $B = \emptyset$, or $A = B$.
Case #1
Let $A = \emptyset$. Then $\emptyset \times B = \emptyset = B \times \emptyset$, so $A \times B = B \times A$.
Case #2
Let $B = \emptyset$. Then $A \times \emptyset = \emptyset = \emptyset \times A$, so $A \times B = B \times A$.
Case #3
Let $A = B$. $A \times B = A \times A = B \times A$, so $A \times B = B \times A$.
$\therefore$ Because all cases have been exhausted, we can conclude that if $A = \emptyset$, $B = \emptyset$, or $A = B$, then $A \times B = B \times A$.

This is a revision of a proof I did earlier. I was told to incorporate $A \neq \emptyset$ and $B \neq \empty$ into the proof somehow, but I'm stumped on how to do that and why it's even necessary.

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