Symmetry in set theory

Question

Let $A$ and $B$ be non-empty sets. Prove that $A \times B = B \times A$ if and only if $A = B$. Where do you use that $A$ and $B$ are non-empty?

$A \times B$ is empty if $A$ or $B$ is empty. See this question: Cartesian Product of Empty Set — FullofDill, Sep 13 '17 at 23:19
@FullofDill typesetting notes: A\times B produces $A\times B$. Also [link name](http://linklocation) for links. — JMoravitz, Sep 13 '17 at 23:20
One direction is immediate. For the other, consider an element in B and not in A or vice versa and use a pair containing that element. Such a pair cannot be in both products. — Peter, Sep 13 '17 at 23:21
At some point you are going to want to say something like let $a$ be an element in $A$ that is not in $B$... and you can only make that statement (or you would need to qualify it) if you know that $A$ is non-empty. — Doug M, Sep 13 '17 at 23:21
Every element of $A \times B$ is an element of $B \times A$ and vice versa. — FullofDill, Sep 13 '17 at 23:25
Cartesian product. $A \times B = {(a,b) | a \in A, b \in B}$. I recommend reading the section this question comes from before attempting a solution. Not knowing the notation is a sign you're in trouble. — FullofDill, Sep 13 '17 at 23:30

score 1 · Answer 1 · answered Sep 13 '17 at 23:30

At some point, you'll want to use these two facts:

$\forall a\in A,\exists p\in A\times B,\exists b\in B, p=(a,b)$

$\forall b\in B,\exists p\in A\times B,\exists a\in A, p=(a,b)$

However, the first statement is false if $A\ne\emptyset$ and $B=\emptyset$, while the second one is false if $A=\emptyset$ and $B\ne\emptyset$.

The thesis is trivially true if $A=B=\emptyset$.

score 1 · Answer 2 · answered Sep 14 '17 at 02:17

1

If a in A, there is some b in B with (a,b) in AxB.
Thus (a,b) in BxA and a in B. Whence A subset B.
As B subset A by symmetry, A = B.

answered Sep 14 '17 at 02:17

William Elliot

2 Answers2