I am studying the book From Holomorphic Functions to Complex Manifolds by Grauert and Fritzsche. In proposition $2.7$ the authors showed that domain of convergence of a power series in $\mathbb C^n$ is a complete Reinhardt domain and the power series converges compactly in its domain of convergence. But the proof seems a bit unclear to me which is the following $:$
Except for the connectedness of $B^{\circ}$ I can follow other line of arguments in the proof. My understanding of the proof (except for the connectedness argument) goes along the following lines $:$
Choose $w \in B^{\circ}.$ Since $B^{\circ}$ is open there exists $\varepsilon \gt 0$ such that the polydisk $P^n (w, \varepsilon) \subseteq B^{\circ}.$ Let $S : = \{i \in \{1,\cdots,n\}\ |\ w_i \neq 0 \},$ where $w = (w_1, \cdots, w_n).$ Let $d = \min\limits_{i \in S} |w_i|.$ Choose any $0 \lt \varepsilon' \lt \min \left \{\varepsilon, d \right \}.$ Then for any $v = (v_1, \cdots, v_n) \in \mathbb C^n$ with $|v_i - w_i| \lt \varepsilon'$ we have $v_i \neq 0.$ This shows that $T^n (w,\varepsilon') \subseteq \left (\mathbb C^{\ast} \right )^n$ for any $0 \lt \varepsilon' \lt \min \left \{\varepsilon, d \right \}$ where $T^n(w, \varepsilon)$ denotes the distinguished boundary of the polydisk $P^n (w, \varepsilon').$ This, in turn, implies that the punctured polydisk $P^n \left (w, \min \left \{d, \varepsilon \right \} \right ) \setminus \{w\} \subseteq \left (\mathbb C^{\ast} \right )^n.$ Now take $v = \lambda w$ where $\lambda = 1 + \frac {\varepsilon'} {1 + |w|} \gt 1.$ Then it is easy to see that $v \in P^n \left (w, \min \left \{d, \varepsilon \right \} \right ) \setminus \{w\} \subseteq P^n(w,\varepsilon) \cap \left (\mathbb C^{\ast} \right )^n$ and $w \in P_v (0)$ where $P_v(0) = P^n (0, r)$ for $r = (|v_1|, \cdots, |v_n|)$ (Note that $P_v (0)$ is a polydisk in $\mathbb C^n$ as $v \in \left (\mathbb C^{\ast} \right )^n$). Since $v \in T_v (0) = T^n (0,r)$ (for $r$ as above) and $v \in P^n (w, \varepsilon) \subseteq B^{\circ}$ it follows that the given power series converges at a point $v$ in the distinguished boundary $T_v (0)$ of $P_v (0).$ So it converges compactly on $P_v (0).$ Since $w \in P_v (0)$ it follows that $T_w (0) \subseteq P_v (0)$ and hence $T_w (0) \subseteq B^{\circ}$ and moreover if $w \in \left (\mathbb C^{\ast} \right )^n$ then the given series is convergent at the point $w$ in the distinguished boundary $T_w(0)$ of the polydisk $P_w(0)$ and consequently the given series converges compactly on $P_w (0).$ Therefore $P_w (0) \subseteq B^{\circ}.$
Also it is clear that $B^{\circ} = \bigcup\limits_{w \in B^{\circ}} P^n \left (w, \frac {\varepsilon (w)} {2} \right )$ and each of the polydisks in the union is relatively compact in $B^{\circ}.$ Hence given any $K \subseteq B^{\circ}$ there exist $N \in \mathbb N$ and $w_1, \cdots, w_N \in B^{\circ}$ such that $K \subseteq \bigcup\limits_{i = 1}^{N} P^n \left (w_i, \frac {\varepsilon (w_i)} {2} \right ) \subseteq \bigcup\limits_{i = 1}^{N} \overline {P^n \left (w_i, \frac {\varepsilon (w_i)} {2} \right )} : = K'.$ Now let $K_i : = \overline {P^n \left (w_i, \frac {\varepsilon (w_i)} {2} \right )}$ for $i = 1, \cdots, N.$ Then $K = \bigcup\limits_{i=1}^{N} K_i$ and we know that the given series converges normally on each $K_i$ and hence it converges normally on $K'$ as $\sup\limits_{z \in K'} \left \lvert a_{\alpha} z^{\alpha} \right \rvert = \max\limits_{1 \leq i \leq N} \sup\limits_{z \in K_i} \left \lvert a_{\alpha} z^{\alpha} \right \rvert \leq \sum\limits_{i=1}^{N} \sup\limits_{z \in K_i} \left \lvert a_{\alpha} z^{\alpha} \right \rvert.$ for any $\alpha \geq 0$ and therefore we have $$\sum\limits_{\alpha \geq 0} \sup\limits_{z \in K'} \left \lvert a_{\alpha} z^{\alpha} \right \rvert \leq \sum\limits_{i = 1}^{N} \sum\limits_{\alpha \geq 0} \sup\limits_{z \in K_i} \left \lvert a_{\alpha} z^{\alpha} \right \rvert \lt \infty.$$ So the given power series converges normally on $K'$ and hence converges normally on $K,$ proving that the power series converges compactly on $B^{\circ}.$
Could anyone clarify the connectedness argument in the proof? Thanks for your time.
EDIT $:$ I think that I got it now. For connectedness of $B^{\circ}$ we can argue as follows $:$ Consider the compact set $C_v : = P_v (0) \cup T_v (0)$ (obtained by taking the union of the polydisk $P_v (0)$ and its distinguished boundary $T_v (0)$). It is easy to see that $C_v$ is convex. Since the given power series converges at $v \in T_v (0)$ it follows from Abel's theorem that the given power series converges compactly on $P_v (0).$ In particular, $0 \in B^{\circ}.$ From the above considerations it follows that given $w \in B^{\circ}$ there exists $v(w) \in B^{\circ} \cap \left (\mathbb C^{\ast} \right )^n$ such that $w \in P_{v(w)} (0) \subseteq C_{v(w)}.$ Now since $v \in T_{v(w)} (0) \subseteq C_{v(w)}$ by convexity of $C_{v(w)}$ there exist lines $L_1(w)$ connecting $w$ and $v(w)$ and $L_2(w)$ connecting $v(w)$ and $0.$ So for any pair of points $w_1, w_2 \in B^{\circ}$ there exists a polygonal path joining $w_1$ and $w_2$ namely $$L_1(w_1) \longrightarrow L_2(w_1) \longrightarrow \text{Reverse}\ L_2(w_2) \longrightarrow \text{Reverse}\ L_1(w_2).$$ So $B^{\circ}$ is polygonally connected and hence connected in particular. $\square$
Could anyone kindly have a look at my argument on connectedness of $B^{\circ}\ $? Thanks again for your time.