Behavior of Hausdorff dimension under homeomorphisms

Question

Let $X$ and $Y$ be metric spaces, $f : X\rightarrow Y$ a homeomorphism. Denote by $\dim_{\mathcal H}$ the Hausdorff dimension. I know that it is possible that $\dim_{\mathcal H} Y < \dim_{\mathcal H} X$, e.g. if $X$ and $Y$ are two cantor sets formed by removing intervals of different lengths. I believe (but cannot prove rigorously) that if $X=\mathbb R$, then necessarily $\dim_{\mathcal H} Y\geq 1$. Is this true (perhaps under the additional assumption $Y\subset \mathbb R^n$ for some $n$)? Is there a general characterization of those metric spaces $X$ whose homeomorphic images cannot have smaller Hausdorff dimension?

Answer 1

Let's stick to separable metric spaces; without separability the Hausdorff dimension is always infinite. The following theorem can be found in Dimension Theory by Hurewicz and Wallman:

$\inf\{\dim_{\mathcal H} Y : Y \text{ homeomorphic to } X \} = \dim_{\mathcal T} X$

where $\dim_{\mathcal T}X$ is the topological dimension of $X$. Hence, the spaces whose Hausdorff dimension cannot be reduced by a homeomorphism are precisely the spaces for which the topological and Hausdorff dimension are equal (i.e., non-fractal spaces). This includes Euclidean spaces and manifolds.

Here is a direct proof that every homeomorphic image of $\mathbb R$ has Hausdorff dimension at least $1$. Let $f:\mathbb R\to Y$ be a homeomorphism. Suppose that $f([0,1])$ is covered by $N$ sets $E_1,\dots,E_N$. By the generalized triangle inequality, $$d(f(0),f(1))\le \sum_{i=1}^N \operatorname{diam} E_i\tag{1}$$ Hence, the $1$-dimensional Hausdorff measure of $f([0,1])$ is positive.

(Sketch of the proof of (1): may assume the sets are closed; they cannot be all disjoint; diameter is subadditive for non-disjoint union; induction on $N$.)