A question on isolated singularity involving analytic varieties.

Question

Suppose $V\subset \mathbb{D}^2$ ($V$ is a subset of the unit bidisc) is a set. Also, suppose that the following conditions hold:

1. If $\hat{V}$ is the polynomial convex hull of $V$, then $\hat{V}\cap \mathbb{D}^2 =V$. So $V$ is a closed subset of $\mathbb{D}^2$.
2. For any $\lambda\in \mathbb{D}^2-V$, there is a bounded holomorphic function $h\in H^\infty (\mathbb{D}^2)$ such that $h(\lambda)\neq 0$ and $h|_V \equiv 0$.
3. There is a connected open set $G\subset\mathbb D$ and a holomorphic function $f:G\rightarrow \mathbb D$ such that $(z,f(z))\in V$ for all $z\in G$. Also, there is a point $w\in \partial G\cap \mathbb D$ which is a singular point of $f$. Furthermore, modulus values of $f$ are bounded away from $1$ in a neighborhood of $w$.
4. (I believe this condition is not required but I can be wrong) For a $g\in H^\infty (\mathbb{D}^2)$ and some $(z_1,z_2)\in \mathbb{D}^2$ if $g(z_1,z_2)=0$, then there is a nonnegative integer $n$ and two positive numbers $\varepsilon_1,\varepsilon_2$ such that for any $z\in B(z_1,\varepsilon_1)-\{z_1\}$ there is a neighborhood $U_z$ of $z$ with $U_z\subset B(z_1,\varepsilon_1)-\{z_1\}$ and there are $n$ holomorphic functions $f_1,\ldots,f_n:U_z\rightarrow \mathbb D$ such that $$g^{-1}\{0\}\cap(U_z\times B(z_2,\varepsilon_2))=\cup_{1\leq l\leq n}\{(z,f_l (z)):z\in U_z\}.$$

I am reading the article Norm preserving extensions of holomorphic functions from subvarieties of the bidisk by Agler and McCarthy where it is stated that condition 2. and 4. imply the point $w$ in 3. is an isolated singularity of the function $f$.

My questions are the following:

1. Why does the singularity have to be isolated?
2. Even if the above claim is hard to prove, is it possible to show that if $f$ in 3. has a non-removable singularity at $w$, then there is a neighborhood $W$ of $w$ such that the set $\{(z,f(z)):z\in W\cap G\}$ has Hausdorff dimension greater than $2$?

Thank you.

Edited later: There is another property of the function $f$ in 3. I should have mentioned:

5. If $m$ stands for the Möbius distance on $\mathbb D$, then for all distinct $z$ and $w$ in $G$, $m(z,w)> m(f(z),f(w))$.

Also, the actual goal is to show that $f$ can not have a non-removable singularity at $w$.

Answer 1

Assuming 5. above, we shall show that $f$ can not have a non-removable singularity at $w$. If $f$ has a non-removable singularity at $w$, then we can find two disjoint sequences $\{w^1 _n\}$ and $\{w^2 _n\}$ in $G$ that converge to $w$, but $lim_{n\rightarrow \infty} m(f(w^1 _n),f(w^2 _n))>0$. This contradicts condition 5. for all large $n$. So $f$ does not have a non-removable singularity at $w$.