Let $(X,d)$ be a metric space and $T\colon X\to X$ uniformly continuous. A set $E\subset X$ is said to be $(n,\varepsilon)$-separated if for any distinct $x,y\in E$ there is a $0\leq j< n$ such that $d(T^j(x),T^j(y))>\varepsilon$.
For a compact set $K\subset X$ let $s_n(\varepsilon,K)$ denote the largest cardinality of any $(n,\varepsilon)$-separated set $E$ contained in $K$.
Is there a lower bound for $s_n(\varepsilon,K)$?
Doesn't from the definition follow, that $s_n(\varepsilon,K)\geq 2$?
This is basically related to the hausdorff measure of $K$.
For any $\varepsilon >0$ there is $\varepsilon>\delta(\varepsilon)>0$ such that $d(x,y)\leq \delta$ implies $d(Tx,Ty)<\varepsilon$
Let $\delta^j(\varepsilon)=\delta(\delta(\delta(....\delta(\varepsilon))$ $j$-times.
Thus, if $d(x,y)\leq \delta^j(\varepsilon)$ then $x,y$ are not $(j,\varepsilon)$-separated.
It folows that if a subset of $K$ is $(n,\varepsilon)$-separated, then it is a $\delta(\varepsilon)^n$-packing (according to the wikipedia definition)
In particular if a subset of $K$ is a maximal $\delta^n(\varepsilon)$-packing then it is a $\delta^n(\varepsilon)$-net.
Thus you get estimate on the cardinality depending on the Hausdorff measure of $K$.