Is A x B equal to B x A if and only if A equals B?

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Willy_Will
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Hi all...

Homework Statement



Let A, B be non-empty sets, proof that A x B = B x A iff A = B


Homework Equations



A x B = Cartesian Product
iff = if and only if
^ = and


The Attempt at a Solution



Let (x,y) є A x B = B x A
iff (x,y) є (A X B) ^ (x,y) є (B x A)
iff (x є A ^ y є B) ^ (x є B ^ y є A)
iff (x є A ^ y є A) ^ (x є B ^ y є B)
iff (x,y) є A ^ (x,y) є B
iff (x,y) є A = B

Its that right?

Also, if one of the sets if empty, will the statement hold?

Thanks guys!
 
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Your proof is correct in essence, but hugely confusing and ungrammatical. (x,y) for x in A and y in A is not an element of A. It's an element of AxA. First prove if A=B then AxB=BxA. That's pretty easy, right? Now prove if AxB=BxA then A=B. It's actually easiest (and much more clear) to prove this by contradiction. And if one of the sets is empty then the cartesian product is empty. Does that make that case easy?
 
Hi, I do not understand the proof and I see that my proof is inadequate. How would you do this by contradiction? And does AxA need to be in the proof? This is what I did:

assume AxB=BxA
let x be an element of A,B ^ y be an element of A,B
(x,y) is an element of A ^ (x,y) is an element of B
so A=B
 
To prove two sets are equal, you want to prove every element of one is an element of the other. Start with your assumption AxB=BxA. Pick any x in A and any y in B. Then (x,y) is an element of AxB. But since AxB=BxA that mean (x,y) is also an element of BxA. Hence?
 
Thanks for responding and helping me, but I'm not sure if I'm following, here is what I get: I should show

assume AxB=BxA
let x be an element of A ^ y be an element of B
(x,y) is an element of AxB
if yes then (x,y) is an element of BxA
so AxB=BxA
so A=B
 
No, no. You assumed AxB=BxA. You don't conclude it. If (x,y) is an element of BxA then x is an element of B and y is an element of A. But remember x was ANY element of A and y was ANY element of B. So A=B BECAUSE any element of A is an element of B and vice-versa.