# Holomorphic function defined on a nonopen set?

Every definition I've seen for what it means for a function to be complex differentiable is stated for an open subset of $\mathbb{C}$. Some textbooks use the term "domain" to refer specifically to open subsets, and some use the term "region" in this way, but one way or another, it seems that one always assumes that the domain of holomorphicity is open. Is this necessarily the case, or does it make sense to say that a function is holomorphic on a set which is not open. I understand that it is possible to make up a definition for what this could mean, e.g. a function is holomorphic on a nonopen set if its holomorphic on an open set containing the nonopen set, but is this standard?

No, it does not make sense a priori for a function to be holomorphic on a non-open subset of $\mathbb C$.
a) It does make sense, given a compact subset $K\subset \mathbb C$, to consider the set $A(K)\subset \mathcal C(K)$ consisting of those continuous functions on $K$ which are holomorphic on the interior $\mathring {K}$ of $K$.
The set $A(K)$ is then a Banach algebra (for the sup norm), which is even better than in the usual case of the set $\mathcal O(U)$ of holomorphic functions on an open subset $U\subset \mathbb C$ since $\mathcal O(U)$ is only a Fréchet algebra.
b) Another possibility, given a subset $S\subset \mathbb C$, is to consider the subalgebra $\mathcal H(S)\subset \mathcal C(S)$ consisting of those continuous functions on $S$ which are the restriction of a holomorphic function defined on some open neighbourhood of $S$.
It is reasonable to call the functions in $\mathcal H(S)$ "holomorphic functions on $S$".