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[SOLVED] Sets so that the cartesian product is commutative

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,028
Hey!! :eek:

Let $A,B$ be sets, such that $A\times B=B\times A$. I want to show that one of the following statements hold:
  • $A=B$
  • $\emptyset \in \{A,B\}$


I have done the following:

Let $A$ and $B$ be non-empty set.

Let $a\in A$. For each $x\in B$ we have that $(a,x)\in A\times B$. Since $A\times B=B\times A$, it follows that $(a,x)\in B\times A$. So $a\in B$.

That means that $A\subseteq B$.


Let $b\in B$. For each $y\in A$ we have that $(y,b)\in A\times B$. Since $A\times B=B\times A$, it follows that $(y,b)\in B\times A$. So $b\in A$.

That means that $B\subseteq A$.


From these two relations we have that $A=B$.



If one of $A$ and $B$ is the emptyset, then it holds that $A\times B=B\times A=\emptyset$.

It also holds that the cartesian product is the empty set, then one of the setsmust be the empty set.

So it holds that $A\times B=\emptyset \iff A=\emptyset \ \text{ or } \ B=\emptyset$.




I am not really sure if the strusture of my proof is correct. At the first case I consider that both $A$ and $B$ are non-empty and I show that it must hold that $A=B$. Then at the other case I just say that if at least one of $A$ and $B$ is empty, then it holds that $A\times B=B\times A$ which is the empty set. But shouldn't I start by $A\times B=B\times A$ and conclude that one of the set must be empty? I am confused now. (Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,713
Hey mathmari !!

I think your proof is fine.

We start with $A\times B=B\times A$ and we consider 2 cases.
Either $A\times B$ is empty, or it is not empty.
Your proof follows naturally. (Happy)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,028
I think your proof is fine.

We start with $A\times B=B\times A$ and we consider 2 cases.
Either $A\times B$ is empty, or it is not empty.
Your proof follows naturally. (Happy)
Ok!! Thanks a lot!! (Sun)