What does positive co-dimension mean in this proof? I'm just trying to understand a line in a specific proof but I'm a little unfamiliar with some of the terminology. The proof in question is Lemma 5 on Page 3 of Piecewise Contractions are Asymptotically Periodic, but I'll paraphrase for convenience.
The result is in regards to the dynamics of a piecewise contraction on a disc in the plane $\mathbb{C}$ (that is, the disc is partitioned into finitely many $\{X_k\}_{k=1}^K$ such that for each $k$, $G|_{X_k}$ is an affine map contracting distances). Thus the piecewise continuous map is:
$$
G(z):=G_k(z)=\lambda_k z + (1-\lambda_k)w_k:\text{       $z \in X_k$}
$$
For $\lambda = (\lambda_1, \ldots,\lambda_K)\in \mathbb{D}^K$ ($\mathbb{D}$ is the unit disc) and $w = (w_1, \ldots,w_K)\in \mathbb{C}^K$. Define $S$ as the union of the boundaries $\partial X_k$ of each of the segments.
At one point in a proof by contradiction, the author shows that there is some $x \in S$ such that $G^{(r_i)}(x) \in S$ for $i \in \{1,2\}$ and $r_i \in \mathbb{N}$. That is, there is a point in the boundaries that maps to another point in the boundaries at least twice. (Page 3, last line).
It is then argued that this is a contradiction by stating:

This is a condition that happens with positive co-dimension, so for Lebesgue
  a.e. $w$, it will not occur.

I'm assuming this is stating something to the effect of: since the boundaries are one-dimensional slice in a two-dimensional plane, there is no finite sequence of affine functions that can map the boundary set to itself more than once.
I was hoping someone could elaborate on what the exact argument is, perhaps with a reference to a generalisation to higher dimensions? For context: I am trying to see how to apply the Lemma to a higher dimensional but identical piecewise contraction map in $\mathbb{R}^m$ with a known set of intercepts.
 A: First of all, this is a statement about a tuple of points $w=(w_1,...,w_K) \in \mathbb C^K = \mathbb R^{2K}$. 
After that, I believe what they are saying is that the set of points $w$ for which this condition occurs is a set of dimension $\le 2K-1$ (or of codimension $\ge 1$, i.e. of positive codimension), and then they are using the fact that the Lebesgue measure of every set of dimension $\le 2K-1$ is equal to zero. The point seems to be that the boundary of region $X_k \subset \mathbb C$ is a rectifiable curve, which has "dimension $1$", and therefore has Lebesgue measure zero; then one is working with the images of the boundaries of the $X_k$ under iterates of the contraction mappings, all of which still have dimension $1$ and therefore Lebesgue measure zero; and one is perhaps taking products of at least one of these boundaries with other stuff to get a region in $\mathbb R^{2K}$, and so the dimension is $\le 2K-1$ and so has Lebesgue measure zero.
It is also possible that they are saying something ever so slightly more complicated, namely that the set for which this condition occurs is a union of countably many sets of dimension $\le 2K-1$, but a union of countably many sets of Lebesgue measure zero has Lebesgue measure zero.
