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3 modified argument to take care of all real values

Regarding 3), this "Big Picard" stuff is serious overkill.

Think like an undergraduate real analysis student:

The p-series zeta(p) converges for real p > 1, whereas zeta(1) = sum of the harmonic series = oo.

An easy argument using (e.g.) the integral test shows that

lim_{p -> oo} zeta(p) = 1.

The function zeta(p) is continuous in p [the convergence is uniform on right half-planes, hence on compact subsets], so by the intermediate value theorem it takes on every positive integer value n >= 2 at least once -- and, since it is a decreasing function of p, exactly once -- on the real line.

Thus A_n is nonempty for all n >= 2.

EDIT: Let me show that zeta(s) takes on all real values infinitely many times on the negative real axis.

For this, note that for all n >=1,

zeta(-(2n-1)) = - B_{2n}/(2n),

where B_{2n} is the (2n)th Bernoulli number. It is known that the B_{2n}'s alternate in sign and grow rapidly in absolute value:

|B_{2n}| \sim 4 \sqrt{\pi n} (n/(pi e))^{2n}.

The claim follows from this and the Intermediate Value Theorem.

2 added 7 characters in body

Regarding 3), this "Big Picard" stuff is serious overkill.

Think like an undergraduate real analysis student:

The p-series zeta(p) converges for real p > 1, whereas zeta(1) = sum of the harmonic series = oo.

An easy argument using (e.g.) the integral test shows that

lim_{p -> oo} zeta(p) = 01.

The function zeta(p) is continuous in p [the convergence is uniform on right half-planes, hence on compact subsets], so by the intermediate value theorem it takes on every positive integer value n >= 2 at least once -- and, since it is a decreasing function of p, exactly once -- on the real line.

Thus A_n is nonempty for all n >= 12.

1

Regarding 3), this "Big Picard" stuff is serious overkill.

Think like an undergraduate real analysis student:

The p-series zeta(p) converges for real p > 1, whereas zeta(1) = sum of the harmonic series = oo.

An easy argument using (e.g.) the integral test shows that

lim_{p -> oo} zeta(p) = 0.

The function zeta(p) is continuous in p [the convergence is uniform on right half-planes, hence on compact subsets], so by the intermediate value theorem it takes on every positive integer value at least once -- and, since it is a decreasing function of p, exactly once -- on the real line.

Thus A_n is nonempty for all n >= 1.