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

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  • $\begingroup$ $L^p(\Bbb R)$, $p<\infty$. $\endgroup$ – David C. Ullrich Jul 18 at 12:45
  • $\begingroup$ @David C. Ullrich Can you elaborate? $\endgroup$ – less Jul 18 at 13:53
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    $\begingroup$ 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
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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 $n$.

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  • $\begingroup$ Thanks, but I don’t know anything about the inductive dimension $\endgroup$ – less Jul 18 at 13:54
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    $\begingroup$ 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
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    $\begingroup$ @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

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