I'm trying to prove the following statement and I thought that it is a special case of Hartogs' extension theorem but I wasn't able to show that $U \setminus \mathbb{C}^{n-2}$ is connected. Any suggestions?
"Let $U$ be an open set in $\mathbb{C}^n$ and let $\varphi: U \setminus \mathbb{C}^{n-2} \to \mathbb{C}$ be a holomorphic function. Then there is a unique holomorphic extension $\tilde{\varphi} : U \to \mathbb C$."
Best Answer
Assuming that $U$ is connected, this is purely topological. You can use the following Lemma applied to $m = 2n$ and $k = 4$ to cover $U \setminus \mathbb C^{n-2}$ by path-connected sets.
Lemma. The set $\mathbb R^m \setminus \mathbb R^{m-k}$ is path-connected for $2 \leq k \leq m$.
Proof Sketch. Note that $\mathbb R^m \setminus \mathbb R^k$ retracts to $\mathbb R^{m-k} \setminus \{0\}$, which is homotopy equivalent to $S^{m-k}$, which is path-connected. Q.E.D.
The way to find a path-connected cover of $U \setminus \mathbb C^{n-2}$ is now to cover $U$ it by random balls (which works because $U$ is open) and then noting that all these balls are homeomorphic to $\mathbb R^{2n} \setminus \mathbb R^{2n-4}$.