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


*

*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.

*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. 

*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. 

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 $A$. 
