Now that I think about it, while on one hand you can patch together/interpolate (not uniquely!) an analytic function by specifying its expansion at a discrete set of points, on the other hand on compact Riemann surfaces like the Riemann sphere there are very strong geometric constraints given by the Riemann–Roch theorem, or even just consider Liouville’s theorem that if a function is holomorphic on the entire Riemann sphere then you already know it must be constant.
I was thinking about this. What about the following situation? Let D be the set \{1, 2, 2 1/2, 3, 3 1/3, ..., n, n+1/n,...\}. Then D is certainly discrete. Let the function f be 1 at integers and 0 at non-integers. Can this be extended to an analytic function? I do not know the answer to this question. On the other hand, say that a set is uniformly discrete if there is a lower bound to the distant between any two points. Then I am certain your conclusion is correct.
Actually, the set \{1, 1/2, 1/3, 1/4, ...\} is also discrete and you can’t even get a continuous function that interpolates the function that alternates between 1 and 0. So you will certainly require not only discrete but also closed. A uniformly discrete set is obviously closed.
I was clearly being sloppy when I typed “discrete”, as your example
\{1,\,\frac{1}{2},\,\frac{1}{3},\,\ldots\}
shows.
Now eg Theorem 26.7 in Forster’s Lectures on Riemann Surfaces shows that if X is a non-compact Riemann surface and a_1,\, a_2,\,a_3, \,\ldots is a sequence of distinct points on X which has no accumulation point, then given arbitrary complex numbers c_n\in\mathbb{C} there exists a holomorphic function on X with f(a_n) = c_n.
In particular, your set \{1,\,2,\,2\frac{1}{2},\,3,\,3\frac{1}{3},\,\ldots\} is not problematic; it does not have to be “uniformly discrete”, merely discrete and closed.
I assume you mean that given an arbitrary sequence \{c_n\in\mathbb C\},…