Show that f−1(A∪B)=f−1(A)∪f−1(B) but not necessarily
f−1(A∩B)=f−1(A)∩f−1(B).
Let S=A∪B
I know that f−1(S)={x:f(x)∈S} assuming that that f is one to one.
Is this true {x:f(x)∈S}={x:f(x)∈A}∪{x:f(x)∈B}?
Why doesn't the intersection work?
Sources : ♦ 2nd Ed, P219 9.60(d), Mathematical Proofs by Gary Chartrand,
♦ P214, Theorem 12.4.#4, Book of Proof by Richard Hammack,
♦ P257-258, Theorem 5.4.2.#2(b), How to Prove It by D Velleman.
Answer
Your exercise is incorrect.
f−1[A∩B]:={x∈dom(f):f(x)∈A∩B}={x∈dom(f):f(x)∈A and f(x)∈B}={x∈dom(f):f(x)∈A}∩{x∈dom(f):f(x)∈B}=:f−1[A]∩f−1[B].
You'll proceed similarly to show that f−1[A∪B]=f−1[A]∪f−1[B], trading "and" for "or".
On the other hand, while we have f[A∪B]=f[A]∪f[B] and f[A∩B]⊆f[A]∩f[B], we don't generally have equality in the last case, unless f is one-to-one. Pick any constant function on your personal favorite set of two or more elements, then choose two disjoint subsets A and B for an example where the inclusion is strict.
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