If $(X,d)$ is a metric space, a homeomorphism $f\colon X\to X$ is said to be expansive if there is a constant $\varepsilon_{0}>0$, called the expansivity constant, such that for any two points of $X$, their $n$-th iterates are at least $\varepsilon_{0}$ apart for some integer $n$; i.e. if for any pair of points $x\neq y$ in $X$ there is $n\in\mathbb{Z}$ such that $d(f^{n}(x),f^{n}(y))\geq\varepsilon_{0}$.

The space $X$ is often assumed to be compact, since under that assumption expansivity is a topological property; i.e. any map which is topologically conjugate to $f$ is expansive if $f$ is expansive (possibly with a different expansivity constant).

If $f\colon X\to X$ is a continuous map, we say that $X$ is positively expansive (or forward expansive) if there is $\varepsilon_{0}$ such that, for any $x\neq y$ in $X$, there is $n\in\mathbb{N}$ such that $d(f^{n}(x),f^{n}(y))\geq\varepsilon_{0}$.

Remarks. The latter condition is much stronger than expansivity. In fact, one can prove that if $X$ is compact and $f$ is a positively expansive homeomorphism, then $X$ is finite (proof).