Is this proof correct? To prove F(A)∩F(B)⊆F(A∩B) for all functions F.
Let any number y∈F(A)∩F(B). We want to show y∈F(A∩B).
Therefore, y∈F(A) and y∈F(B), by definition of intersection.
By definition of inverse, y=F(x) for some x∈A and x∈B
And so, y=F(x) for some x∈A∩B
And therefore, y∈F(A∩B)
I have a gut feeling deep down that something is wrong. Can anyone help me pinpoint the mistake? I am not sure why am I having so much problems with functions. Am I not thinking in the right direction?
Sources : 2nd Ed, P219 9.60 = 3rd Ed, P235 9.12, 9.29 - Mathematical Proofs, by Gary Chartrand,
P214 Theorem 12.4 - Book of Proof, by Richard Hammack,
P257-258 - How to Prove It, by D Velleman.
Answer
The third line is mistaken. You only know that there exists an x in A such that F(x)=y, and you know there is a z∈B such that F(z)=y.
It is extremely easy to find a counterexample: just draw two sets A, B that are disjoint, and map an a∈A and a b∈B to a single point. Then you have that y∈F(A)∩F(B), but F(A∩B)=∅.
No comments:
Post a Comment