The topological entropy of a continuous map $f : X \to X$ is a non-negative number that measures its dynamical complexity. In order to define the topological entropy it is not enough to have a topological space — a uniform structure compatible with the topology is needed. The value of the topological entropy usually depends on the choice of uniform structure. In the case of a compact space there is a unique uniform structure, so everything is very simple. Otherwise a uniform structure needs to be specified, which is typically done by specifying a metric.

The choice of uniform structure can have a radical effect on the value of the topological entropy. For example consider the map of ${\mathbb R}^n$ to itself given by multiplying by $\lambda > 1$. In the uniform structure induced by the standard Euclidean metric the topological entropy is $n \log(\lambda)$. On the other hand, if $R^n$ is compactified by adding an $(n-1)$-sphere at infinity, resulting in the compact space $\widehat{R^n}$, then the uniform structure of $\widehat{R^n}$ induces a uniform structure on $R^n$. In this uniform structure the topological entropy of the multiplication-by-$\lambda$ map, in fact of any affine map, is zero.

Now consider a continuous piecewise affine map $f : {\mathbb R}^n \to {\mathbb R}^n$. "Piecewise affine" means that there should exist a finite number of disjoint open sets $U_1, U_2, \ldots U_m$ such that the union $U_1 \cup U_2 \cdots \cup U_m$ is dense in ${\mathbb R}^n$, and such that the restriction of $f$ to each $U_i$ should coincide with an invertible affine map $A_i$. For example, $f$ could be a member of the Lozi family with $b \neq 0$.

It is not hard to show that $f$ extends (uniquely) to a continuous map $\widehat{f} : \widehat{R^n} \to \widehat{R^n}$. What is the topological entropy of $\widehat{f}$?