Can Lipschitz maps increase the Lebesgue dimension ? 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.
 A: 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 $[0;1]$.
A: 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 replacing 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 maps this Cantor set onto a segment, see 
(source)
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.
A: There are no Lipschitz space filling curves.  For a gentle introduction to where this question leads, see http://eprints.nuim.ie/1626/1/SBuckleycurvesfunctions41.pdf. Theorem 3 is the classical result you want.
A: A similar but even simpler example: take a Cantor set on $[0,1]$ having positive measure, and transport it optimally to a segment with the same measure.
