Hausdorff dimension of radial set

Question

Let $A\subset [0,1]$ be a compact set of Hausdorff dimension $\alpha$. Let also $A_n:=\{x\in \mathbb{R}^n~:~ \|x\|_2 \in A\}$.

Is it true that $\text{dim}_H(A_n)=n-1+\alpha$? I believe that this should be the case because the dimension of $S^{n-1}$ is $n-1$, so the dimension of $A_n$ should depend in an essential way on the dimension of $A$.

I thought that this could be proved by reasoning in polar coordinates, where a radial set becomes a product set, but I'm not sure whether there is a simpler way to prove it. Does it follow trivially from some well-known theorem?

Answer 1

"Fractal Geometry: Mathematical Foundations and Applications", 2nd ed.
by Kenneth Falconer
Chapter 7: Products of fractals

The key "product formula" is Corollary 7.4, which is:

If $\dim_H F = \overline{\dim_B} F$ then $$ \dim_H (E \times F) = \dim_H E + \dim_H F $$

An application of this is:

Example 7.7 The "Cantor target"

The calculation proceeds via polar coordinate transform, which is a Lipschitz mapping (used to prove the upper bound), and bi-Lipschitz on a restricted domain (used to prove the lower bound). The bounds are equal, thus the dimension is calculated.