# Thread: Product Sets and relations

1. ## Product Sets and relations

So i've been stuck on this for a while now and im just not sure what to do. I'm giving this definition of a product set:

A relation R on a set U is said to be a product set if there are subsets A,B of U such that R = A × B.

How do I use that to prove the following theorem? Let U be a set and let R a relation on U. If there exists four elements a,b,c,d of U such that (a,b) ∈ R, (c,d) ∈ R and (a,d) is not in R, then R is not a product set.

Also, is the set {(x, y) ∈ R2 : xy not equal 0} a product set, is the set {(x, y) ∈ R2 : xy not equal 1} a product set? and how would I prove them? I'm completely blank.

2. Originally Posted by tn11631
So i've been stuck on this for a while now and im just not sure what to do. I'm giving this definition of a product set:

A relation R on a set U is said to be a product set if there are subsets A,B of U such that R = A × B.

How do I use that to prove the following theorem? Let U be a set and let R a relation on U. If there exists four elements a,b,c,d of U such that (a,b) ∈ R, (c,d) ∈ R and (a,d) is not in R, then R is not a product set.

Suppose the relation is a product set $\Longrightarrow R=A\times B\,,\,\,A,B\subset U$ , but then as $(a,b)\,,\,(c,d)\in R=A\times B$ , then

we'd get that $a,c\in A\,,\,\,b,d \in B\Longrightarrow (a,d)\in R$ , by definition of cartesian product, in contradiction with

the given $(a,d)\notin R$

Check carefully the above and then try, with the new understanding, to answer the following two questions

Tonio

Also, is the set {(x, y) ∈ R2 : xy not equal 0} a product set, is the set {(x, y) ∈ R2 : xy not equal 1} a product set? and how would I prove them? I'm completely blank.
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3. Originally Posted by tonio
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Thanks! I am 100% understanding the proof now, I feel kinda lame for that passing me. However for some reason I just can't grasp my brain around the second part of the problem. I'm staring at it and I'm setting it up the same way as in the proof like XxY and stuff but i'm getting no where. Aye, must be the lack of practice.