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:

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 ?

  • $\begingroup$ 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. $\endgroup$
    – coffeemath
    May 27 at 15:21
  • $\begingroup$ 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? $\endgroup$ May 27 at 15:36
  • $\begingroup$ 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 $Dom(R) \times Ran(R) \not\subset R$. $\endgroup$
    – Juan Jose
    May 27 at 15:49
  • $\begingroup$ @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] $\endgroup$
    – coffeemath
    May 27 at 16:05


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