TITLE: Holomorphic maps preserve Hausdorff dimension.
QUESTION [1 upvotes]: In a paper I read there is the following claim:
Let $f:\mathbb{C}\to \mathbb{C}$ be a non-constant entire transcendental function(essential singularity at infinity) and $A\subset \mathbb{C}$ a set in the complex plane. Then $f^{-1}(A)$, $A$ and $f(A)$ have the same Hausdorff dimension.
I know that bi-Lipschitz maps preserve Hausdorff dimension but I dont see why entire maps in the complex plane should too. Perhaps because entire maps are locally bi-Lipschitz away from critical points. But do locally bi-Lipschitz maps preserve the dimension?
Can someone prove this or provide a reference for a proof?

REPLY [1 votes]: It's exactly as you said. Since $f$ is locally bi-Lipschitz, then if $A_n=A\cap \tilde{D}(0,n)$, where $\tilde{D}(0,n)$ is the closed disk of radius $n$, then $\dim_H(f(A_n))=\dim_H(f^{-1}(A_n))=\dim_H(A_n)$, $\forall \ n\in\mathbb{N}$.
Now use that in general, $\dim_H(\cup_{n\geq 1} B_n)= \sup_{n\geq 1} \dim_H(B_n)$.
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