Critique my proof on correctness, structure, etc.


$(\rightarrow)$ 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".

  • 1
    $\begingroup$ I think that the forward direction can just be proven directly without the need of a contradiction $\endgroup$
    – wjmccann
    Dec 27, 2021 at 20:08
  • $\begingroup$ @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 $B \neq \emptyset$. $\endgroup$ Dec 27, 2021 at 20:11

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Dec 27, 2021 at 20:47

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