I am wondering if there exists an example of a second countable metric space $X$ containing a set $A$ with infinite Hausdorff dimension.

  • $\begingroup$ $L^p(\Bbb R)$, $p<\infty$. $\endgroup$ – David C. Ullrich Jul 18 at 12:45
  • $\begingroup$ @David C. Ullrich Can you elaborate? $\endgroup$ – less Jul 18 at 13:53
  • 2
    $\begingroup$ Yes. There is even a metric on the Cantor set so that the Hausdorff dimension is infinite. $\endgroup$ – GEdgar Jul 18 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 Jul 18 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 Jul 18 at 21:46

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 $n$.

| cite | improve this answer | |
  • $\begingroup$ Thanks, but I don’t know anything about the inductive dimension $\endgroup$ – less Jul 18 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 Jul 18 at 14:14
  • $\begingroup$ @GEdgar What is the metric you put on $\mathbb{R}^{\mathbb{N}}$ so that it is 2nd-countable? $\endgroup$ – less Jul 18 at 14:41
  • $\begingroup$ @Pedro Sánchez Terraf $\endgroup$ – less Jul 18 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 Jul 18 at 15:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.