Show that $f^{-1}(A\cup B) = f^{-1}(A)\cup f^{-1}(B)$ but not necessarily
$f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$.
Let $S=A\cup B$
I know that $f^{-1}(S)=\{x:f(x)\in S\}$ assuming that that $f$ is one to one.
Is this true $\{x:f(x)\in S\}=\{x:f(x) \in A\}\cup\{x:f(x)\in B\}$?
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.
$$\begin{align}f^{-1}[A\cap B] &:= \{x\in\text{dom}(f):f(x)\in A\cap B\}\\ &= \{x\in\text{dom}(f):f(x)\in A\text{ and }f(x)\in B\}\\ &= \{x\in\text{dom}(f):f(x)\in A\}\cap\{x\in\text{dom}(f):f(x)\in B\}\\ &=: f^{-1}[A]\cap f^{-1}[B].\end{align}$$
You'll proceed similarly to show that $f^{-1}[A\cup B] = f^{-1}[A]\cup f^{-1}[B],$ trading "and" for "or".
On the other hand, while we have $f[A\cup B]=f[A]\cup f[B]$ and $f[A\cap B]\subseteq f[A]\cap f[B],$ 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.
No comments:
Post a Comment