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Jacobpm64
#5
Oct27-07, 04:20 PM
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:
[tex]A = \left\{ {1,2,4} \right\}\,,\,B = \left\{ {p.q,r} \right\}\,,\,C = \left\{ {2,4,6} \right\}\,\& \,D = \left\{ {r,s,t} \right\}[/tex]
[tex]f:A \mapsto B,\quad f = \left\{ {(1,p),(2,r),(4,q)} \right\}[/tex]
[tex]g:C \mapsto D,\quad g = \left\{ {(2,r),(4,t),(6,s)} \right\}[/tex]

But [tex]f \cap g = \left\{ {(2,r)} \right\}[/tex] while [tex]A \cap C = \left\{ {2,4} \right\}[/tex] clearly [tex]f \cap g:A \cap C \not{\mapsto} B \cap D[/tex]
There is no mapping for the term 4.