Proof or Counterexample:Is every open connected set $D \subset \mathbb C$ is a domain of holomorphy? Def: An open set  $D \subset \mathbb C^n$  is called a domain of Holomorphy if there exists a holomorphic function $f$ on $D$ such that $f$ cannot be extended to a bigger set.
Is every non empty open set $D \subset \mathbb C$ is a domain of holomorphy?
I personally believe that this result is true but I'm unable to find a proof.Any ideas?
 A: Yes, any nonempty open domain $ D\subset \mathbb C$ is a domain of holomorphy.
The proof is in two steps:  
a) One proves that a holomorphically convex domain  $D\subset \mathbb C^n$ is a domain of holomorphy .
This is not difficult: see for example Grauert-Fritzsche, Theorem 6.5, page 81     
b) One proves that for $n=1$ any domain $D\subset \mathbb C$ is holomorphically convex.
This too is rather  easy: given a compact subset $K\subset\subset D$ and a boundary point $a\in \partial D$, consideration of $\frac{1}{z-a}$ shows that the holomorphically convex hull $\hat {K}=\hat {K}_D$ cannot approach $\partial D$ and consideration of the holomorphic function $z$ shows that $\hat {K}$ is bounded.
Since $\hat {K}$ is closed in $D$ these considerations prove that  $\hat {K}$ is compact.  
NB Actually a) above is an equivalence: a domain $D\subset \mathbb C^n$ is  holomorphically convex iff it is  a domain of holomorphy.
This result was proved by Cartan-Thullen  in 1932. Here are the authors reunited 55 years later. 
A: A nonrigorous but perhaps more accessible answer: the way one constructs a function holomorphic on $D$ but not any larger open set is to construct a meromorphic function on $\Bbb C$ that has poles surrounding the boundary of $D$, so closely that they cut off any analytic continuation. Such a meromorphic function can be chosen to be the reciprocal of a holomorphic function with prescribed zeros; the existence of such functions is, as Daniel Fischer suggested, guaranteed by the Weierstrass product theorem.
