Quick proof

So two posts ago I posited that ‘for any given open set U ⊆ ℂ, (∀z∈U: ζ(z) = ζ(z+k)) ⇒ (k = 0)’. This is almost true; the correct statement begins "for any given nonempty open set U ⊆ ℂ", and follows almost immediately from the identity theorem.

First, note that both ζ(z) and ζ_k(z) = ζ(z+k) are holomorphic everywhere in ℂ, except at their respective singularities, 1 and 1-k. Note also that the set S = ℂ\{1, 1-k} is open (since its complement, the set {1, 1-k}, has an empty interior is equal to its boundary and is therefore closed) and connected (since, e.g., it is path-connected).

Assume the antecedent holds for some open set U. It then also holds for the open set U' = U ∩ S. Select any point z in this set, and any open ball V containing z, but lying within U'. (One such must exist; if it did not, z would be on the boundary of U', and since U' is open it excludes its boundary.) The preconditions for the identity theorem are thereby met: ζ = ζ_k on a neighborhood of a point, z, which lies within a connected open set on which both ζ and ζ_k are holomorphic — namely, S.

Thus, ζ = ζ_k on S.

Now, assume for the sake of contradiction that k ≠ 0. Then ζ(1-k) is some complex number, since ζ is defined everywhere except at 1. Consider some sequence z_n which approaches 1-k in the limit, but contains neither 1-k nor 1. (Such sequences must exist: consider the disjoint sequences (1-k)+2^-n and (1-k)-2^-n, only one of which might contain 1.). ζ is complex-differentiable and therefore Heine-continuous, so the sequence ζ(z_n) approaches ζ(1-k). Moreover, since ζ_k's singularity at 1-k is a pole, ζ_k(z_n) must approach infinity. However, this sequence is contained in S, and so ζ(z_n) = ζ_k(z_n) for all n. Since the sequences are identical, so too must be their limits; but this means that ζ(1-k) = ∞, which is a contradiction.

Therefore, k = 0.

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Edit 2008-12-21: fixed a silly topology error — an empty interior does not imply closedness. (Specific counterexample: the set of all Gaussian rationals has an empty interior, but is not a closed set.)