# Can a compact metrizable space be determined by its Hausdorff measures?

Suppose that $(X,d)$ is a compact metric space. Now suppose that $h:[0,a]\rightarrow[0,b]$ is a continuous function with $h(0)=0$ where if $x\leq y$, then $h(x)\leq h(y)$. Then define $$L(d,h)=\lim_{\epsilon\rightarrow 0}\,\inf\{h(a_{1})+...+h(a_{n})|X=B(x_{1},a_{1})\cup...\cup B(x_{n},a_{n}),$$

$$a_{1}<\epsilon,...,a_{n}<\epsilon,x_{1},...,x_{n}\in X\}.$$

We say that a compact metric space $X$ is a Peano space if it is a compact connected locally connected metrizable space. It is well known that a Hausdorff space is the continuous surjective image of the unit interval if and only if it is a Peano space.

Suppose that $X,Y$ are Peano spaces. Furthermore, assume

1. whenever $d$ is a metric inducing the topology on $X$ there is a metric $d'$ inducing the topology on $Y$ with $L(d,h)=L(d',h)$ for each monotone mapping $h:[0,a]\rightarrow[0,b]$ with $h(0)=0$ and $x\leq y\Rightarrow h(x)\leq h(y)$, and

2. whenever $d$ is a metric inducing the topology on $Y$, there is a metric $d'$ inducing the topology on $X$ with $L(d,h)=L(d',h)$ for each monotone mapping $h:[0,a]\rightarrow[0,b]$ with $h(0)=0$ and $x\leq y\Rightarrow h(x)\leq h(y)$.

Then are there nonempty open sets $U\subseteq X,V\subseteq Y$ where $U$ and $V$ are homeomorphic?

• It kind of seems like $[0,1]\times[0,1]\cup([1,2]\times\{1\})$ may be an easy counterexample. If so, then I will have to edit this question. – Joseph Van Name Oct 7 '13 at 17:40
• The unit circle (with its geodesic distance) and the interval $[0, 2\pi]$ do have, for any $\epsilon > 0$, the same set of admissible radii $a_i < \epsilon$ for ball covering, therefore they have the same L(d,h) – Pietro Majer Oct 7 '13 at 17:48
• These metrics d , d' are not assumed to enjoy further properties as in the first version of the question, are they? – Pietro Majer Oct 7 '13 at 23:12
• For now I will not assume the metrics $d$ and $d'$ are intrinsic metrics or anything like that. – Joseph Van Name Oct 8 '13 at 1:45
• A pair of Peano spaces as a candidate counterxample: $X=[0,1]^{\mathbb{N}}$, and $Y=\mathbb{S_1}^{\mathbb{N}}$. – Pietro Majer Oct 9 '13 at 16:30