I am wondering if there exists an example of a second countable metric space $X$ containing a set $A$ with infinite Hausdorff dimension.
$\begingroup$ $\endgroup$
5 
- $\begingroup$ $L^p(\Bbb R)$, $p<\infty$. $\endgroup$David C. Ullrich– David C. Ullrich2020-07-18 12:45:13 +00:00Commented Jul 18, 2020 at 12:45
- $\begingroup$ @David C. Ullrich Can you elaborate? $\endgroup$less– less2020-07-18 13:53:34 +00:00Commented Jul 18, 2020 at 13:53
- 2$\begingroup$ Yes. There is even a metric on the Cantor set so that the Hausdorff dimension is infinite. $\endgroup$GEdgar– GEdgar2020-07-18 14:15:28 +00:00Commented Jul 18, 2020 at 14:15
- $\begingroup$ @GEdgar can you tell me about this metric on the Cantor set? I know that the HausDim of the Cantor set with the euclidean metric is $\log2/\log3$ $\endgroup$less– less2020-07-18 14:25:21 +00:00Commented Jul 18, 2020 at 14:25
- $\begingroup$ Well, $L^p(\Bbb R)$ is second-countable, being a separable metric space. And it contains subspaces homeomorphic to $\Bbb R^n$, so the dimension is $\ge n$ for every $n$. $\endgroup$David C. Ullrich– David C. Ullrich2020-07-18 21:46:31 +00:00Commented Jul 18, 2020 at 21:46
Add a comment |
1 Answer
$\begingroup$ 
$\endgroup$
8 The infinite power $\mathbb{R}^{\mathbb{N}}$ is separable metrizable (hence second countable) and has infinite inductive dimension. This is a topologically defined notion, and is the smallest one among many related concepts; in particular, is less or equal to the Hausdorff dimension.
To see that $\mathrm{dim}\,\mathbb{R}^{\mathbb{N}} = \infty$: It can be proved that the inductive dimension of a space is at least equal to the sup of the dimensions of its subspaces (see Engelking, Dimension Theory, 1.1.2), and that $\mathbb{R}^{n}$ for $n\in{\mathbb{N}}$ has inductive dimension $n$.
answered Jul 18, 2020 at 13:00
Pedro Sánchez Terraf
4,8842 gold badges17 silver badges34 bronze badges
- $\begingroup$ Thanks, but I don’t know anything about the inductive dimension $\endgroup$less– less2020-07-18 13:54:38 +00:00Commented Jul 18, 2020 at 13:54
- 1$\begingroup$ The space $\mathbb R^n$ (with $n$ a positive integer) has Hausdorff dimension $n$. The above space (and Ullrich's space) contain (bi-Lipschitz equivalent) copies of $\mathbb R^n$. Therefore the Hausdorff dimension of $\mathbb R^{\mathbb N}$ is at least $n$. Since this holds for all positive integers $n$, the Hausdorff dimension is infinite. $\endgroup$GEdgar– GEdgar2020-07-18 14:14:25 +00:00Commented Jul 18, 2020 at 14:14
- $\begingroup$ @GEdgar What is the metric you put on $\mathbb{R}^{\mathbb{N}}$ so that it is 2nd-countable? $\endgroup$less– less2020-07-18 14:41:53 +00:00Commented Jul 18, 2020 at 14:41
- $\begingroup$ @Pedro Sánchez Terraf $\endgroup$less– less2020-07-18 14:42:44 +00:00Commented Jul 18, 2020 at 14:42
- 2$\begingroup$ @less There is a standard construction of a metric for a countable product of metric spaces $(X_n,d_n)$. Just take $$ d(x,y) := \sum_{n=1}^\infty \frac{d_n(x(n),y(n))}{2^n(1+d_n(x(n),y(n)))}, $$ where $x,y\in \prod_n X_n$. $\endgroup$Pedro Sánchez Terraf– Pedro Sánchez Terraf2020-07-18 15:19:22 +00:00Commented Jul 18, 2020 at 15:19