  Let $X$ be a compact metric space and let $|A|$ denote the cardinality of a
set $A$. We prove that if $f\colon X\to X$ is a homeomorphism and $|X|=\infty$
then for all $\delta>0$ there is $A\subset X$ such that $|A|=4$ and for all
$k\in Z$ there are $x,y\in f^k(A)$, $x\neq y$, such that $dist(x,y)<\delta$. An
observer that can only distinguish two points if their distance is grater than
$\delta$, for sure will say that $A$ has at most 3 points even knowing every
iterate of $A$ and that $f$ is a homeomorphism. We show that for
hyper-expansive homeomorphisms the same $\delta$-observer will not fail about
the cardinality of $A$ if we start with $|A|=3$ instead of $4$. Generalizations
of this problem are considered via what we call $(m,n)$-expansiveness.
