Prove that the Von Koch snowflake curve $E$ and the set $[0,1] \times \{0\}$ are not lipeomorphic (bi-Lipschitz equivalent). Are they homeomorphic?
I've done a fair bit of looking around I've struggled to find anything that demonstrates a method of proving two sets are not bi-Lipschitz equivalent outright. I believe that the two sets are indeed homeomorphic though, as there are some good examples on this site that show this to be true.
I don't really know where to start, if I could find a bijection between the two sets then this would help, but nothing jumps out at me as an obvious choice for such a map.
You don't need the notion of Hausdorff dimension for this purpose, the more intuitive concept of the length of a curve suffices. The set $[0,1]\times \{0\}$ is the image of a curve $\gamma:[0,1]\to\mathbb R^2$, which has finite length, meaning that the supremum of sums $\sum_{i}|\gamma(t_i)-\gamma(t_{i-1})|$ over all partitions $\{t_i\}$ of $[0,1]$ is finite.
On the other hand, the snowflake has infinite length. Indeed, the vertices of its $n$th generation: are at distance about $3^{-n}$ from each other, and there are $4^n$ of them. This gives a partition with the sum $(4/3)^n$, and since $n$ can be arbitrarily large, the length is infinite.
It remains to observe that a bi-Lipschitz map distorts the lengths by a bounded amount: that is, if $L^{-1}\le |F(a)-F(b)|/|a-b|\le L$ for all $a,b$, then the image of a curve of length $\ell$ under $F$ is a curve of length between $\ell/L$ and $\ell L$.