"Normal" geometric shapes have Hausdorff dimensions equal to their topological dimensions. Mandelbrot defined fractals as shapes that have a Hausdorff dimension greater than their topological dimension. Is there a class of shapes that have a Hausdorff dimension less than their topological dimension, or is this impossible? If there is such a shape, what are common examples of them? If this is impossible, why?
Is it possible to have a Hausdorff dimension less than the topological dimension
dimension-theory-analysishausdorff-measure
Related Solutions
You are correct.
There are several ways to show that it is impossible, and it basically boils down to the fact that no open subset of ${\bf R}^n$ is homeomorphic to an open subset of ${\bf R}^m$ if $n<m$ (as $U\cap V$ would be in your case). You can assume without loss of generality that the sets in question are connected.
One way to show it is to use the fact that you can find an $(n-1)$-dimensional sphere which disconnects any open subset of ${\bf R}^n$, but it is a general fact that if a compact set disconnects ${\bf R}^m$, then it must admit a non-nullhomotopic map onto $S^{m-1}$ (in fact, it is an equivalent condition), while on the other hand, no proper compact subset of $S^{m-1}$ admits such a map.
So, summing it all up, a set homeomorphic to $S^{n-1}$ can't disconnect an open subset of ${\bf R}^m$ if $m>n$ and we're done.
(The results used can be shown by combinatorial or algebraic topology, but are not simple enough to show here.)
No, a Fibonacci spiral is not a fractal - it's a smooth curve and has dimension one
Here is a fractal spiral:
The distinction is that this set is self-similar - it consists of two copies of itself one blue copy, scaled by the factor 0.98, and one orange copy, scaled by the factor 0.08. The dimension can be computed using the theory of self-similar sets; it's approximately 1.4135.
Best Answer
The shapes you are asking about do not exist. The reason is:
Theorem. (Sznirelman) For every metric space $X$, the Hausdorff dimension of $X$ is $\ge$ the covering dimension of $X$.
See for instance section VII.2 of
W.Hurewicz, H.Wallman, Dimension Theory, Princeton University Press.