5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $f^{-1}[S]=\{x:f(x)\in S\}$ by definition; this has nothing to do with whether $f$ is one-to-one. $\endgroup$ – Brian M. Scott Apr 10 '13 at 23:52
  • $\begingroup$ then f^-1 isnt a function right if f isnt one to one $\endgroup$ – sarah Apr 10 '13 at 23:53
  • 2
    $\begingroup$ If $f:A\to B$, then $f^{-1}:P(B)\to P(A)$ is a function on the power sets. $\endgroup$ – Berci Apr 10 '13 at 23:55
  • $\begingroup$ That’s right, but it has no bearing on this problem. The statement that $f^{-1}[A\cup B]=f^{-1}[A]\cup f^{-1}[B]$ is true for all $A,B$, and $f$. $\endgroup$ – Brian M. Scott Apr 10 '13 at 23:55
  • $\begingroup$ I think of $f^{-1}(S)$ informally as "stuff that lands inside $S$ (when I hit it with $f$)". You're being asked to show that "x lands inside $A\cup B$" iff "x lands inside $A$ or $B$", which is obviously true. You're also being asked to disprove the claim that "x lands inside $A\cap B$" iff "x lands inside both $A$ and $B$", but this claim is obviously true too, as Cameron Buie says. $\endgroup$ – Billy Apr 11 '13 at 0:25
9
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Am I suppose to disprove * ? $\endgroup$ – sarah Apr 11 '13 at 0:15
  • $\begingroup$ No, actually. You can and should prove it. Your exercise is incorrect. $\endgroup$ – Cameron Buie Apr 11 '13 at 0:17
  • $\begingroup$ I think there's a typo (twice) on the line *, where x should say f(x). $\endgroup$ – Billy Apr 11 '13 at 0:18
  • $\begingroup$ @Billy: Good catch, thanks. $\endgroup$ – Cameron Buie Apr 11 '13 at 0:19
  • $\begingroup$ why does the * need justification? $\endgroup$ – sarah Apr 11 '13 at 0:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.