Find all values of such that there exists a function that satisfies the following properties:
1), if and only if .
Proof: does not fit into the criteria because it requires . Fix .
Bertrand’s Postulate (a.k.a. Tchebyshev’s theorem): , with prime.
Now we aim to find a such that .
For , take . For odd, , apply Tchebyshev’s theorem to , there exists a prime . But now is even and thus is not prime, and thus we have , which shows that is the desired prime number as . For even, apply Tchebyshev’s theorem to to yield a prime which gives . The has been chosen for all .
Let , . Let .
(E.g. let , then .
Define an equivalence relation on : for , if and only if . It is easy to verify that this is indeed an equivalence relation. Partition into equivalence classes , by this relation. Let be indexed by the set .
Axiom of Choice: If is a family of sets indexed by the set , then the Cartesian product is non-empty.
Apply Axiom of Choice to the equivalence classes of , and we get a family of elements , with each corresponding to exactly one .
Let be the -adic evaluation of . (That is, reduce to simplest form , if does not divide , and if .
Let . For , suppose . Define . (E.g. if .Similarly, . This gives .) We set
Indeed, we show first show that the function satisfies the conditions.
First, we see that does not vanish for non-zero .
We claim that if . This is obvious by substituting the definitions of and , and then LHSRHS.
Now, we verify the conditions. Condition 1 is already checked. If , . If , let . For with , we have , and thus . Since , there are such s. .
Thus . We have found the desired function, for each of .
Comment: to be written later.