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when toggle format what by license comment
Sep 6, 2016 at 7:37 vote accept quarague
Sep 6, 2016 at 7:01 comment added N Unnikrishnan Well, the unit interval $I$ is not open as a subset of $\mathbb{R}^2$, so it doesn't invalidate the "uniform Hausdorff dimension" of the disjoint union of $I$ and $D$ being 2 according to your definition.
Sep 5, 2016 at 13:55 answer added user98074 timeline score: 5
Sep 5, 2016 at 12:18 history asked quarague CC BY-SA 3.0