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Given a map $f:X\rightarrow Y$ of compact metric spaces, such that there is a $C\in \mathbb{R}$ with $d(f(x),f(x'))\le C\cdot d(x,x')$.

Does this already imply, that the Lebesgue dimension of $f(X)$ is at most the Lebesgue dimension of $Y$?

My motivation were space filling curves. All constructed examples were not rectifiable, so I was wondering, whether the additional assumption of being Lipschitz rules out the existence of space filling curves.

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up vote 17 down vote accepted

The answer to your first question is No. It is natural, as Lebesgue dimension is a topological invariant.

There is a very simple example, which took me however a little time to figure out. Consider the Cantor set obtained by recursively replace a square by four squares of $1/4$-th side length, one at each corner, endowed with the metric inherited from the Euclidean one. There is an orthogonal projection that map this Cantor set onto a segment, see

Such a map is $1$-Lipschitz, but a segment has Lebesgue dimension $1$ while a Cantor set has Lebesgue dimension $0$.

What you need to rule out Lipschitz space-filling curves is for example Hausdorff dimension.

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Thanks a lot. And its even possible to achieve an arbitrary dimension of the image by crossing this map with itself. – HenrikRüping Aug 24 '10 at 12:59
This can be done more easily with the usual middle third Cantor set $C:$ since $C+C=[0,2]$ (nice exercise!), the image of $C\times C$ under the map $$\mathbb{R}^2\to\mathbb{R},\quad (x,y)\mapsto x+y$$ is a closed interval. – Victor Protsak Aug 24 '10 at 17:59
I think that your example is too thin. If $D=\{\sum_{k\geq 1} a_k 4^{-k}: a_k=0,3\}$ then $D+D$ doesn't cover the interval $[0,2]:$ for example, the point $\sum_{k\geq 1} 4^{-k}$ is not in the image (more generally, the subword 11 cannot occur in the 4-ary expansion of an element of $D+D$). – Victor Protsak Aug 24 '10 at 18:42
@Victor Protsak: I think our examples are mostly the same; beware that i am not taking the same projection than you (look at the image), and my example is classical. – Benoît Kloeckner Aug 24 '10 at 19:27
Benoît, I am taking the same projection and the picture corresponds to what I wrote in the comment (which is indeed classical and involves the usual Cantor set), but not to what you wrote in the answer. You said "Consider the Cantor set obtained by recursively replace a square by four squares of 1/4-th side length, one at each corner". Thus the set you get is $D\times D$ in the notation of my previous comment, and its projection $D+D.$ Well, that set not only doesn't cover the interval, it has measure 0, because it can be characterized as $$D+D=\{ 3\sum_{k\geq 1} a_k 4^{-k}: a_k=0,1,2\}$$ – Victor Protsak Aug 24 '10 at 19:41

There are no Lipschitz space filling curves. For a gentle introduction to where this question leads, see Theorem 3 is the classical result you want.

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Let me add to the examples by Benoit and Victor another Cantor example, this time a straightforward naive one rather than ingenious.

Consider at and just after stage $0$ a closed interval $I$ with the standard (Euclidean) metrics but of length $\frac{11}{10}$. At stage $k>0$ remove the center open interval of length $\frac 1{2^{k-1}\times 11^k}$ of each interval left after the previous stage $k-1$. After all stages $0\ 1\ \ldots$, in the remaining set $C$ in addition to the Euclidean metrics consider also the following pseudo-metrics:

$$d(x\ y) = |x-y| - s_{x\ y}$$

where $s_{x\ y}$ is the sum of the lengths of all removed intervals which are between points $x\ y$. The identity map from Euclidean $C$ to $C$ with the pseudo-metric $d$ is Lipschitz with constant 1. Let $C'$ be the metric space induced by $C$. Then $C'$ is homeomorphic to a nondegenerated closed interval, and the map induced by the identity on $C$ is Lipschitz with constant 1.

Actually, C' is isometric with the unit Euclidean interval $[0;1]$.

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