Question on construction of entire functions Suppose that $x_i$ and $y_i$ are sequences in $\mathbb{C}$. Can you construct a non constant entire function such that $f(x_i)=y_i$?
What happens if $x_i$ have an accumulation point? or what happens if $x_i$ and $y_i$ is an uncountable collection of points?
I have studied the book by Churchill and Brown on complex analysis (up to residues and their applications) but I seem to have never encountered a result that seems to address the following question.
 A: If the sequence $(x_k)$ has no finite accumulation points, i.e., if $x_k\to\infty$ as $k\to\infty$, then for any $(y_k)$ there exists an entire function such that $f(x_k)=y_k$ for all $k$. This is a theorem of Weierstrass. Idea of proof: 


*

*There exists an entire function $g$ with zeros at every $x_k$, also proved by Weierstrass.  

*For each $k$, consider the Taylor series of $g$ at $x_k$: it begins with some term $c_{n_k}(z-x_k)^{n_k}$, $c_n\ne 0$. Use the Mittag-Leffler's theorem to produce a meromorphic function $h$ with pole of order $n_k$ at each $x_k$, and with leading Laurent coefficient such that $g(z)h(z)$ evaluates to $y_k$ at $x_k$ after the singularity is removed. 


If $x_k$ have an accumulation point, the values $y_k$ can no longer be arbitrary. I don't know of any sensible way of expressing a consistency condition on $y_k$ that would be required for $f$ to exist. 
Uncountable case is even worse: such a set will have uncountably many points of accumulation. I'm pretty sure there is nothing constructive to be said. Unless the set has a nice structure: a line or an analytic curve, in which case we are discussing the extensibility of a function defined on such a set. This depends on the radius of convergence of its Taylor series. If the radius is infinite, there is an extension to an entire function; otherwise no.
