Does finite Hausdorff dimension imply finite packing dimension? In other words, does there exist a metric space $(E,\rho)$ with finite Hausdorff dimension but infinite packing dimension?
Here are my thoughts:

*

*I know that it is generally hard to relate Hausdorff and packing measures and dimensions (other than the fact that Hausdorff dimension is always less than or equal to packing dimension), and I was not able to find a direct proof using the definitions.

*Assume that $(E,\rho)$ is separable and complete. It is known that $n:=\dim_C E \le \dim_H E\le \dim_P E$ (respectively denoting topological covering dimension, Hausdorff dimension, and packing dimension). In particular, if $(E,\rho)$ has finite Hausdorff dimension, then it has - as a topological space - finite covering dimension, and is thus homeomorphic to a subset of $\mathbb{R}^{2n+1}$ (see book by Hurewicz below; a similar embedding result is in the proof of Theorem 6.3.10 in Edgar's book). In particular, there exists a metric $\rho'$ on $E$ that induces the same topology as $\rho$ such that $(E,\rho')$ has packing dimension at most $2n+1$. Then the identity map $\textrm{id}\colon (E,\rho') \to (E,\rho)$ is bi-continuous (because the metrics induce the same topology), but this is not enough to relate their packing dimensions. If it was $\eta$-Hölder continuous for some $\eta \in (0,1)$, then we could deduce $\dim_P (E,\rho) \le (2n+1) / \eta < \infty$, but this must a priori not be the case.

I am also interested in any results that require more specific assumptions on $E$, say separable, complete, compact, etc.

Edit: I refer to packing dimension as defined here.

Hurewicz, W.; Wallman, H., Dimension theory., 165 p. Princeton University Press (1941). ZBL67.1092.03.
Edgar, Gerald A., Measure, topology, and fractal geometry, Undergraduate Texts in Mathematics. New York etc.: Springer-Verlag. ix, 230 p. DM 58.00/hbk (1990). ZBL0727.28003.
 A: A construction used (repeatedy) in the paper
Edgar, G. A., Centered densities and fractal measures, New York J. Math. 13, 33-87 (2007). ZBL1112.28004.
For more information, see that paper.

We construct a compact metric space $E$ for this purpose.
For $n=1,2,3,\dots$ let $k_n \in \{2,3,4,5,\dots\}$ and $r_n \in (0,1/2)$.  More precise properties are to be specified later.
Let $T_n$ be a set with $k_n$ elements (with the discrete topology).  Let
$$
K_n = k_1 k_2\dots k_n,\qquad R_n = r_1 r_2\dots r_n.
$$
so $K_n \nearrow \infty$ and $R_n \searrow 0.$  Also let $R_0 = 1$.
Our space is
$E = \prod_{n=1}^\infty T_n$ with the product topology.  So $E$ is separable, compact.  Define metric $\rho$ on $E$ as follows:
Let $x = (x_1,x_2,\dots), y=(y_1,y_2,\dots) \in E$.  For $x=y$, define $\rho(x,y) = 0$.  For $x \ne y$, let $m \in \mathbb N$ be such that
$$
x_j=y_j\quad\text{for }1 \le j \le m,\quad\text{and }
x_{m+1} \ne y_{m+1}
$$
and define $\rho(x,y) = R_m$.
Then
$$
E \text{ has diameter } R_0 .
$$
$E$ is the disjoint union of $k_1$ closed subsets $E[a_1], a_1 \in T_1$,
$$
E[a_1] = \{(x_1,x_2,\dots) \in E : x_1 = a_1\}
\text{ has diameter } R_1 .
$$
Each set $E[a_1]$ is the disjoint union of $k_2$ closed subsets
$E[a_1,a_2], a_2 \in T_2$,
$$
E[a_1,a_2] = \{(x_1,x_2,\dots) \in E : x_1 = a_1, x_2=a_2\}
\text{ has diameter } R_2
$$
Continue in this way, so that each $E[a_1,\dots,a_m]$ is the disjoint union
of $k_{m+1}$ closed subsets $E[a_1,\dots,a_m,a_{m+1}], a_{m+1} \in T_{m+1}$, and $E[a_1,\dots,a_m]$ has diameter $R_{m}$.
For each $m$, the space $E$ is the disjoint union of $K_m$ closed sets of diameter $R_m$.
Define a "uniform" measure $\mu$ on $T$ so that
$$
\mu\big(E[a_1,\dots,a_m]\big) = \frac{1}{K_m} .
$$
Now, let $s\in(0,\infty)$.  The upper $s$-density of $\mu$ at a point $x \in E$ is
$$
\overline{D}_\mu^s(x) =
\limsup_{\eta\searrow 0} \frac{\mu(B_\eta(x))}{(\operatorname{diam} B_\eta(x))^s} =
\limsup_{n \to \infty} \frac{1/K_n}{R_n^s}=
\limsup_{n \to \infty} \frac{1}{K_nR_n^s} .
$$
The lower density is $$
\underline{D}_\mu^s(x)
= \liminf_{n \to \infty} \frac{1}{K_nR_n^s} .
$$
Now what we need to do is select $k_n, r_n$ at the beginning so that, for all $s \in (0,\infty)$ we have
$$
\limsup_{n \to \infty} \frac{1}{K_nR_n^s} = +\infty,\qquad
\liminf_{n \to \infty} \frac{1}{K_nR_n^s} = 0
$$
Then our metric space $(E,\rho)$ will have Hausdorff dimension $0$ and packing dimension $\infty$.
