# Critique my proof of: Suppose $A$ and $B$ are sets. Then $A \times B = B \times A \iff A = \emptyset, B = \emptyset,$ or $A = B$

Critique my proof on correctness, structure, etc.

Proof.

$$(\rightarrow)$$ Suppose $$A \times B = B \times A$$ and let $$P = (x, y)$$ be an arbitrary element of $$A \times B$$. Assume for the sake of contradiction that $$A \neq \emptyset$$, $$B \neq \emptyset$$, and $$A \neq B$$. By definition of cartesian product, $$x \in A$$ and $$y \in B$$, and because $$A \times B = B \times A$$, $$\ x \in B$$ and $$y \in A$$. Because $$x$$ and $$y$$ are arbitrary elements and the definition of subset, it follows that $$A \subseteq B$$ and $$B \subseteq A$$, so $$A = B$$. This is a contradiction, so we can conclude that if $$A \times B = B \times A$$, then either $$A = \emptyset$$, $$B = \emptyset$$, 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 $$A \times B = B \times A$$.

I feel like this proof is too long and that there may be some objection to the claim that "$$x$$ and $$y$$ are arbitrary elements".

• I think that the forward direction can just be proven directly without the need of a contradiction Dec 27, 2021 at 20:08
• @wjmccann I agree, but I was concerned about extending my proof by having to account for the $A = \emptyset$ and $B = \emptyset$ cases. I figured if I did it by contradiction, I wouldn't have to since I'm assuming $A \neq \emptyset$ and $B \neq \emptyset$. Dec 27, 2021 at 20:11

For the left-to-right direction, you cannot start by letting $$P$$ be an arbitrary element of $$A \times B$$, because you are not proving a statement of the form "for all $$P$$ in $$A \times B,\ \ldots$$." Your strategy was to prove $$A \subseteq B$$ and $$B \subseteq A$$, so that means your proof should have looked like this: "Let $$x$$ be an arbitrary element of $$A$$. (Proof of $$x \in B$$ goes here.) Therefore $$A \subseteq B$$. Now let $$y$$ be an arbitrary element of $$B$$. (Proof of $$y \in A$$ goes here.) Therefore $$B \subseteq A$$."
You can tell that there is something wrong with your proof because your proof never used the assumption that $$A \ne \varnothing$$ and $$B \ne \varnothing$$, but those assumptions are necessary. If you fill in the proof outline above, you will find that you need to use those assumptions to complete the proof.
• I knew that was odd, but I just accepted that I didn't have to use $A \neq \emptyset$ and $B \neq \emptyset$. I see where I went wrong now. Dec 27, 2021 at 20:47