Thread: Proof involving functions View Single Post
 P: 239 Well, one of our earlier assignments was to disprove the case when we took unions. So I know that you don't get a function when you take f union g. I'm still not convinced that the intersection claim is correct though. Earlier, on another forum, someone came up with the counterexample: $$A = \left\{ {1,2,4} \right\}\,,\,B = \left\{ {p.q,r} \right\}\,,\,C = \left\{ {2,4,6} \right\}\,\& \,D = \left\{ {r,s,t} \right\}$$ $$f:A \mapsto B,\quad f = \left\{ {(1,p),(2,r),(4,q)} \right\}$$ $$g:C \mapsto D,\quad g = \left\{ {(2,r),(4,t),(6,s)} \right\}$$ But $$f \cap g = \left\{ {(2,r)} \right\}$$ while $$A \cap C = \left\{ {2,4} \right\}$$ clearly $$f \cap g:A \cap C \not{\mapsto} B \cap D$$ There is no mapping for the term 4.