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.
Answer
 The questioner was satisfied and accepted the answer, or
 The answer was disputed, but the judge evaluated it as 100% correct.

I'm just going through a book of mathematical proof. I'll be in an intro proofs course next Spring, though.
 answered
 146 views
 $5.00
Related Questions
 Operational Research probabilistic models

Math Proofs: "An alternative notation is sometimes used for the union or intersection of an indexed family of sets."
 Fix any errors in my proof (beginner)
 Discrete Structures  Proving a statement true
 Fix any errors in my proof (beginnner)
 Proof through inclusion (A∆B) ∪ A = A ∪ B
 Topic: Large deviations, in particular: Sanov's theorem
 Prove that: x + y ≤ x + y + x − y.