Hartog's Theorem proof clarification. From Huybrechts:
Let $\epsilon=(\epsilon_1,...,\epsilon_n)$, $\epsilon'=(\epsilon_1',...,\epsilon_n')$be given such that for each $1 \leq j \leq n$, ($n \geq 2$) we have
$$\epsilon_j'< \epsilon_j.$$ If $n >1$, then any holomorphic map
$$f: B_\epsilon(0) \setminus \overline{B_{\epsilon'(0)}} \rightarrow \mathbb{C}$$
can uniquely be extended to a holomorphic map
$$f: B_\epsilon(0) \rightarrow \mathbb{C}.$$
The proof starts by considering the unit poly disk in $\mathbb{C}^n$. I can see why it is enough to take $\epsilon_j=1$ for each $1 \leq j \leq n$. Then they say there is a $\delta>0$ such that the open set
$$\{z \in \mathbb{C}^n : 1 - \delta < \vert z_1 \vert < 1, \vert z_{j \neq 1}\vert <1\} \cup \{z \in \mathbb{C}^n : 1 - \delta < \vert z_2 \vert < 1, \vert z_{j \neq 2}\vert <1\}$$
is contained inside the complement of $B_{\epsilon'}(0)$ (is the complement with respect to $B_\epsilon(0)$ that is, $B_\epsilon(0) \setminus B_{\epsilon'}(0))$.
My questions are, for the existence of this $\delta$ are we relying on the fact that the $\epsilon_j' < \epsilon_j$? Also what is the significance of stopping at the union of the first two components of $z$ having a "small" modulus. Is it for the existence of the annulus they define? in which one needs two "radii"? Just wondering why only the union of modulus of $z_1,z_2$ components being small will suffice. Typsetting this I think it IS due to this annulus that has a convergent Laurent Series. Moreover, for what map in $\mathbb{C}$ does this result fail?
 A: The existence of such a $\delta$ follows from the fact that $e_j' < e_j$. If we had equality then we might not be able to choose the $\delta$ parameter.
Edited: As you say, the choice of $\delta$ guarantees that the set lies in $B_{\epsilon}(0) \setminus B_{\epsilon '}(0)$. The reason we choose this set to restrict only the first two coordinates is that these coordinates are all we need to establish the holomorphicity of the Laurent series expansion of $f$ in the entire $B_{\epsilon}(0)$. We use $z_1$ to write the Laurent expansion as a polynomial in $z_1$ with holomorphic coefficients $\alpha_i(w)$. The restriction on $z_2$ and the preceding lemma allows us to analyze the coefficients $\alpha_i(w)$ and conclude that they must vanish identically on $B_{\epsilon}(0)$ for $i < 0$. Thus the Laurent expansion for $f$ actually yields a function which is holomorphic on all of $B_{\epsilon}(0)$ which agrees with $f$ on $B_{\epsilon}(0) \setminus B_{\epsilon'}(0)$.
As for a domain in $\mathbb{C}$ for which this fails, the classic example is the unit disk minus the origin $D =\mathbb{D} - \{0\}$ and the function $f(z) = \frac{1}{z}$; $f$ cannot be extended to the origin but is holomorphic in $D$.
