Holomorphic function on the unit disk $f$, show the set $z,w\in \mathbb{C}$ such that $f(z)=f(w)$ is not countable

Question

Here is the problem statement:

Suppose $f$ is a holomorphic function on the unit disk. Show that the set $A=\lbrace (z,w) \in \mathbb{C}^2\;|\; |z|,|w| \leq \frac{1}{2}, z\neq w, f(z)=f(w)\rbrace$ is either finite or uncountably infinite.

I'm pretty stuck, but here are some thoughts:

1. The bound on the $z,w$ is a ltitle odd. The only thing I take away from it is that $|z+w| \leq |z| + |w| = 1$ so the sum stays in the closure of the disk. But this doesn't seem relevant.

2. I tried to find a clever way to use the identity principle but I couldn't. For example, suppose to the contrary the set $A$ is countably infinite, then it is pointless to consider the function $g(z) = f(z) - f(w_0)$ for some $w_0 \in A$ since there doesn't need to be countably many $z_0$ matching up with $w_0$ in the sense $f(z_n) = f(w_0)$, only countably many pairs $z,w$ with the same images.

3. I tried representing $f(z)$ as a power series centered at $z_0=0$: $$ f(z) = \sum_{n=0}^\infty a_nz^n $$ and then considering expressions like $$ f(z_0) - f(w_0) = \sum_{n=0}^\infty a_n(z_0^n - w_0^n) $$ to learn something about the coefficients but this didn't lead anywhere.

I think there must be some slick solution but I can't see it, so I'd prefer hints rather than full solutions right now.

Answer 1

I believe I may have a proof that uses well-known results about holomorphic functions of several complex variables and the Baire Category Theorem.

First, recall that zeros of holomorphic functions of more than one variable are not isolated. This is a consequence of a well known theorem due to Hartogs. Now, suppose that $A$ is countable, say $A=\{(z_1,w_1),(z_2,w_2),\dots\}$. The set $$S:=\{(z,w) : |z|,|w| \leq 1/2 : f(z)=f(w)\}$$ is closed, and we have $$S= \{(z,z):|z| \leq 1/2\} \cup \{(z_1,w_1)\} \cup \{(z_2,w_2)\} \cup \dots.$$ By the Baire Category Theorem, one of the sets in the union has non-empty interior. Clearly it cannot be the diagonal, and so we have that one of the points $(z_j,w_j)$ is isolated in $S$. But this point would be an isolated zero of the holomorphic function $g(z,w):=f(z)-f(w)$, which is a contradiction.

Answer 2

Here's a solution that was pointed out to me by a fellow student:

The claim is that the set is either empty or uncountable, in fact. Assume there is one such pair $(z,w) \in A$ such that $f(z) = f(w)$. But since $\mathbb{C}$ is Hausdorff, we can separate each point by an open set such that the open sets are disjoint: $z\in U, w\in V, U\cap V = \varnothing$.

But by the open mapping theorem, the image contains a ball around $f(z)=f(w)$ which is contained in $f(U) \cap f(V)$. That is, $f(U)$ is open, $f(V)$ is open and they have a nontrivial intersection which must too be open and thus contains a ball $B$ which contains $f(z)=f(w)$.

Now we pick some other point $f(z) \neq \alpha \in B$. Then $f^{-1}(\alpha) \supset \lbrace z_0 \neq z\in U, w_0 \neq w \in V \rbrace$ where $f(z_0) = f(w_0)$. Thus, $f^{-1}(B) \subset A$ where $f^{-1}(B)$ is uncountable.

Thus, if there is one point in $A$, there are uncountably many points in $A$.