Where A and B are both non-empty sets, Prove that AxB does not equal BxA unless A equals B.?

Question

I've worked on this a bit and this is what I came up with:

If A does not equal B, then

There's some element a that exists in A, but does not exist in B

OR

There's some element b that exists in B, but does not exist in A

A = (a, ......)
B = (b, ......)

AxB = ( (a, b), .....)
BxA = ( (b, a), .....)

Since (a, b) and (b, a) are both ordered pairs:

(a, b) does not equal (b, a) unless a = b.

However, if a = b, then a (or b) exists in both A and B which cannot be true unless A = B

Therefore, AxB does not equal BxA unless A=B.

Is this a sound proof? If not, what did I do wrong and how should I correct it?

Answer 1

what do you mean by AXB?



are you multiplying sets? or you're multiplying the elements of A to its corresponding value in B?



i don't get why you came up with these:



A = (a, ......)

B = (b, ......)



AxB = ( (a, b), .....)

BxA = ( (b, a), .....)





cuz they look like ordered pairs to me.

Answer 2

"There's some element a that exists in A, but does not exist in B



OR



There's some element b that exists in B, but does not exist in A"

Good



"A = (a, ......)

B = (b, ......)



AxB = ( (a, b), .....)

BxA = ( (b, a), .....)"





The proper way is to write

A={a,...}, B={b,...}

AxB={(a,b),...}

BxA={(b,a),...}

because this is notation for sets, though i don't like the "..." part.

Better to write: since aεA and bεB then (a,b)εAxB and

(b,a)εBxA.



"Since (a, b) and (b, a) are both ordered pairs:



(a, b) does not equal (b, a) unless a = b.



However, if a = b, then a (or b) exists in both A and B which cannot be true unless A = B"



True, you proved that (a,b) cannot be (b,a). But this is not all!

You have to prove that (a,b) cannot be any element of BxA.



So we do like this: suppose (a,b)εBxA, so it must me

(a,b)=(x_1,x_2), where x_1 εB and x_2 ε A.

That means a=x_1 and b=x_2. In particular, both a and b are in A and B, contradiction.

Note that this proves already that (a,b) cannot be (b,a)

Answer 3

That works. Though I might break it down into two parts, where you first explicitly assume there is at least one element "a" in A that is not in B, then show why AxB has an element that's not in BxA, and then repeat the same steps with an element that is in B but not A. You don't need any ordering. You can show that if "a" can be a first element in at least one pair in AxB, then it can't be a first element in any pairs in BxA.



(And please folks, if you don't know what the cartesian product (AxB) means in set theory, then don't answer)