# Sets of infinite Hausdorff dimension in a second countable metric space

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

• $L^p(\Bbb R)$, $p<\infty$. – David C. Ullrich Jul 18 at 12:45
• @David C. Ullrich Can you elaborate? – less Jul 18 at 13:53
• Yes. There is even a metric on the Cantor set so that the Hausdorff dimension is infinite. – GEdgar Jul 18 at 14:15
• @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 $\log2/\log3$ – less Jul 18 at 14:25
• 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 $n$. – 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 $$n$$.
• 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. – GEdgar Jul 18 at 14:14
• @GEdgar What is the metric you put on $\mathbb{R}^{\mathbb{N}}$ so that it is 2nd-countable? – less Jul 18 at 14:41
• @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 $x,y\in \prod_n X_n$. – Pedro Sánchez Terraf Jul 18 at 15:19