Holomorphic extension in 3 complex variables The problem is the following: Suppose $U = \{ z \in \mathbb{D}^3 \ | \ \frac{1}{2} < |z_1| \text{ or } \frac{1}{2} < |z_2| \}$. Prove that every $f \in \mathcal{O}(U)$ extends to $\mathbb{D}^3$. 
Here $\mathbb{D}^3$ is the unit polydisc in three complex dimensions.
I've thought about this for a while and the problem is elusive because the set $\mathbb{D}^3 \setminus U$ is not compact in $\mathbb{D}^3$ (it touches the boundary of $\mathbb{D}^3$ - take for example the point (0,0,1)) nor can we fit a Hartogs' figure. 
Any ideas? Thanks in advance. 

I figured it out!
Fix $z_3 = \zeta \in \mathbb{D}$ and let $g : \mathbb{D}^2 \to \mathbb{C}$, where $g(z_1,z_2) := f(z_1,z_2,\zeta)$. Then $g$ is holomorphic on the set $U' := \{ z \in \mathbb{D}^2 \ | \ \frac{1}{2} < |z_1| \text{ or } \frac{1}{2} < |z_2| \}$. However, $g$ extends over the complement of $U'$ in $\mathbb{D}^2$ (by Hartog’s theorem since $U’$ is connected and the complement is compact). It follows that $f$ also extends to this larger set. 
