# Is there a concept of uniform Hausdorff dimension?

Let $M$ be a metric space and let $U \subset M$ be open. Then the Hausdorff dimension of $U$ is defined in the usual way. If there is a single dimension number $d$ that is the Hausdorff dimension of every open set $U$ in $M$ I say the metric space $M$ has uniform Hausdorff dimension $d$.

What I want to know is whether some concept like this already exists, possible under a different name.

I want to use this as a property that 'nice' metric spaces have. A simple example: Let $I$ denote the unit interval and $D$ the unit ball in $\mathbb{R}^2$. Then both $I$ and $D$ have a uniform Hausdorff dimension but the disjoint union of $I$ and $D$ does not.

• Well, the unit interval $I$ is not open as a subset of $\mathbb{R}^2$, so it doesn't invalidate the "uniform Hausdorff dimension" of the disjoint union of $I$ and $D$ being 2 according to your definition. – N Unnikrishnan Sep 6 '16 at 7:01

There is a significantly stronger property that is commonly used and does have a name: A metric space is called Ahlfors $n$-regular if there is a constant $K$ such that for each closed ball $B(x,r)$ in the space ($0<r\leq\text{diam}(X)$), $$K^{-1} r^n \leq H^n(B(x,r)) \leq K r^n$$ where $H^n$ is $n$-dimensional Hausdorff measure.
In your example, $I$ and $D$ are of course Ahlfors $1$-regular and Ahlfors $2$-regular, respectively, while the union is not Ahlfors regular for any $n$.