# If a set is arbitrary, can you prove something about it by constructing an example of it?

Let $$R$$ be a relation from $$A$$ to $$B.$$ Prove that $$\operatorname{Domain}\left(R\right)\times \operatorname{Range}\left(R\right)\not\subset R.$$

I need to show that there is some couple $$(a,b)\in \operatorname{Domain}(R)\times \operatorname{Range}(R)$$ such that $$(a,b)\notin R.$$ It is easy to show by example that the theorem is true:

A={1,2}
B={3,4}
R={(1,3),(2,4)}
Domain(R)={1,2}
Range(R)={3,4}
Domain(R) x Range(R)={(1,3),(1,4),(2,3),(2,4)}

In this case, the theorem is true. But the $$R$$ is arbitrary and has an implicit "forall" quantifier, right? So I need to show that the theorem is true for all $$R,$$ so I can't assume anything about the contents of $$R,$$ right? Or is it valid to use my above example ?

• By definition a relation R is any subset (proper or not) of $dom(R) \times ran(R).$ In particular R might coincide with the latter. So the "theorem" is false, or else needs some more hypotheses. May 27 at 15:21
• yes, i think it should have been of the form: "its not true that for any R, R is a subset of dom(R) x ran(R)". And then i could have proved existence of such a set with the example above. Is this correct? May 27 at 15:36
• Why don't you use contradiction? Assume $Dom(R) \times Ran(R) \subset R$. Thus there exists some $(a,b) \in R$ such that $(a,b) \notin Dom(R) \times Ran(R)$. And from here derive a contradiction to establish $Dom(R) \times Ran(R) \not\subset R$. May 27 at 15:49
• @lightyourassonfire Yes if that was the wording and the symbol $\subset$ is interpreted as "is a proper subset of". [I have seen that symbol used to include the case of equality] May 27 at 16:05