Bi-Lipschitz invariance of the box-counting dimension.

Question

I would like to prove that the box counting dimension is invariant under a bi-Lipschitz transformation. We have that $f$ is bi-Lipschitz if there exists $c_1, c_2$ such that $0 < c_1 \leq c_2 < \infty$ and $f$ satisfies:

$$c_1 |x-y| \leq |f(x)-f(y)| \leq c_2 |x-y|.$$

I can see how to prove that $\dim_Bf(F) \leq \dim_BF$, but I cannot see how to prove that $\dim_Bf(F) \geq \dim_BF$ in order to get that $\dim_Bf(F) = \dim_BF$.

Answer 1

Let $F'=f(F)$. Then there is an inverse $f^{-1}: F'\to F$, which is also bi-Lipschitz. Apply the inequality you already have to the inverse: $\dim_B F\le \dim_B F'$.