I was working through these questions:
(c) Prove that f(C1 ∩ C2) ⊆ f(C1) ∩ f(C2).
(d) Give a counterexample to the opposite inclusion in (c).
The opposite of c. (that is, the RHS is not a subset of the LHS) is false. This is their answer which makes sense to me:
(d) Let f : R → R be given by f(x) = x^2.
Let C1 = {−2}, and let C2 = {2}. Then f(C1 ∩ C2) = f(∅) = ∅, but f(C1) ∩ f(C2) = {4} ̸⊆ ∅.
However, if it is false, I was trying to prove that utilizing methods I have learned (utilizing the pre-image of a function).
I proved c. using essentially the same logic but reversed. In my proof of the false statement, this logic seems true to me but I know it should be a false statement... any help would be appreciated.