Fig. 41. (0) is not in ran g. Exercises 18. Prove that the following statement i
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Fig. 41. (0) is not in ran g. Exercises 18. Prove that the following statement is equivalent to the axiom of choice: For any set whose members are nonempty sets, there is a function f with domain such that f (X)e X for all X in 19. Assume that H is a function with finite domain 1 and that H(1) is nonempty for each i e I. Without using the axiom of choice, show that there is a function fwith domain I such that f (i) e Hi) for each ie I. [Suggestion: Use induction on card 1] 20. Assume that A is a nonempty set and R is a relation such that (Yx EA)(3y E A) yRx. Show that there is a functionf: A with/(n+ R(n) for all n in o. 21. (Teichmüller-Tukey lemma) Assume that o is a nonempty set such that for every set B, BEdevery finite subset of B is a member of Show that has a maximal element, i.e, an element that is not a subset of any other element of st. 22. Show that the following statement is another equivalent version of the axiom of choice: For any set A there is a function F with dom F UA and such that xe F(x)E A for all x e UA 23. Show that in the proof to Theorem 6N, we have g[n]-h(n). 24. How would you define the sum of infinitely many cardinal numbers? Infinite products? 25. Assume that s is a function with domain such that S(n)s S(n*) for each ne o. (Thus S is an increasing sequence of sets.) Assume that B is a subset of the union s S(n) such that for every infinite subset B of BExplanation / Answer
For any set A, there is a function F with dom F=A such that xF(x)A for all xA. Call this statement S.
To prove that S is equivalent to the axiom of choice, we show that S holds iff version 4 of the axiom of choice holds: Let A be a set such that (a) each member of A is a nonempty set, and (b) any two distinct members of A are disjoint. Then there exists a set C containing exactly one element from each member of A (i.e., for each BA, the set CB is a singleton {x} for some x. Call this version S.
First, we show that S S. Assume that S holds and that A is a set such that each member is a nonempty set and any two members of A are disjoint. Then, there is a function f such that dom f=A and xA (xf(x)A). We now have dom f ran f A, and so ran f by the definition of A. Further, we have ran f B={f(x)}, for some xA and for any BA. So ran f=C, and S holds where CB, for any BA is a singleton.
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