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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.

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  • $\begingroup$ $f^{-1}[S]=\{x:f(x)\in S\}$ by definition; this has nothing to do with whether $f$ is one-to-one. $\endgroup$ Apr 10, 2013 at 23:52
  • $\begingroup$ then f^-1 isnt a function right if f isnt one to one $\endgroup$
    – sarah
    Apr 10, 2013 at 23:53
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    $\begingroup$ If $f:A\to B$, then $f^{-1}:P(B)\to P(A)$ is a function on the power sets. $\endgroup$
    – Berci
    Apr 10, 2013 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 $f$. $\endgroup$ Apr 10, 2013 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, 2013 at 0:25

1 Answer 1

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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 non-empty disjoint subsets $A$ and $B$ for an example where the inclusion is strict.

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  • $\begingroup$ Am I suppose to disprove * ? $\endgroup$
    – sarah
    Apr 11, 2013 at 0:15
  • $\begingroup$ No, actually. You can and should prove it. Your exercise is incorrect. $\endgroup$ Apr 11, 2013 at 0:17
  • $\begingroup$ I think there's a typo (twice) on the line *, where x should say f(x). $\endgroup$
    – Billy
    Apr 11, 2013 at 0:18
  • $\begingroup$ @Billy: Good catch, thanks. $\endgroup$ Apr 11, 2013 at 0:19
  • $\begingroup$ why does the * need justification? $\endgroup$
    – sarah
    Apr 11, 2013 at 0:33

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