I was studying some basic matters of several complex variables (here $\Omega\subseteq\Bbb C^n$, open):
After this, before the proof, the author pointed what follows:
So I'm going to tell you what my problem is.
Since the theorem says: "in any polydisc $P(z_0,r)\subset\Omega$", I can choose the polydisc I prefer, provided that it is contained in $\Omega$. Clear.
Thus I can take $P(z_0,r)$, with every choice of $r=(r_1,\dots,r_n)$, $r_j\in]0,R_j]$, where $R_j:=\operatorname{dist}(z_0^j,\partial\Omega_j)$ (here we use the notation $\Omega=\Omega_1\times\dots\times\Omega_n$, $z_0=(z_0^1,\dots,z_0^n)$).
In particular, we can take $r_j=R_j$ and the other $r_k,\;k\neq j$ as small as I want.
And till here, I'm still ok. The problem is that the comment says "$r_j$ approaches to the distance of $z_0$ to $\partial\Omega$ along the $z_j$-plane, at the expenses of the other radii $r_k$ for $k\neq j$, which possibly approach to $0$": reading this, it seems that I can take $r_j=D-(n-1)\varepsilon$ and $r_k=\varepsilon$ for $k\neq j$, where $D:=R_1+\dots+R_n$ is the distance of $z_0$ to $\partial\Omega$. But this is wrong since I'd get $r_j\notin\Omega_j$ for $\varepsilon$ small enough.
In this way, it sounds like the only important thing, in order to have convergence, is that the sum of the $r_i$'s is less or equal than $D$, NOT that $r_i\le R_i$ for every $i$. And this sounds wrong to me: in this way I'd get out of $\Omega$.
The sentence "at the expences of the other radii" sounds like if we would handle one variable with as many comfort as I can, we must pay this with an uncomfortable use of the other variables. But $z_0\in\Omega$ iff $z_0^i\in\Omega_i\;\;\forall i=1,\dots,n$, so the choice of a variable should be indipendent from the other! Or did I lost something in the way?
On the other hand, if I interpret the sentence "the distance of $z_0$ to $\partial\Omega$ along the $z_j$-plane" as I wrote before, i.e. $r_j=R_j-(n-1)\varepsilon$ and $r_k=\varepsilon$, than, the remark seems trivial.
Where's the truth?
If the domain is of the form $\Omega=\Omega_1\times\ldots\times\Omega_n$ then the best you can do is just to take $r_j=R_j-\varepsilon$ for $\varepsilon>0$ small. The book comment applies to cases where $\Omega$ cannot be written as a tensor product. This can easily be visualized if you replace $\mathbb{C}^2$ by $\mathbb{R}^2$ (this is fine for visualization purposes and power series reasonings) and consider for example, $\Omega=\{|x|+|y|<1\}$ and $z_0=(0,0)$. In this case, any "polydisk" $P(0,r)$ with $r=(a,1-a)$ fits into $\Omega$, for $0<a<1$. If we want to stress attention to $z_1$, then we would choose $a\approx1$, which would then force $r_2=1-a$ to be very small.