Monday, September 26, 2016

elementary set theory - Show that f1(AcupB)=f1(A)cupf1(B)



Show that f1(AB)=f1(A)f1(B) but not necessarily




f1(AB)=f1(A)f1(B).



Let S=AB



I know that f1(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.



f1[AB]:={xdom(f):f(x)AB}={xdom(f):f(x)A and f(x)B}={xdom(f):f(x)A}{xdom(f):f(x)B}=:f1[A]f1[B].



You'll proceed similarly to show that f1[AB]=f1[A]f1[B], trading "and" for "or".






On the other hand, while we have f[AB]=f[A]f[B] and f[AB]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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