Why do these inequalities in metric spaces hold? The other day I stumbled across some inequalities regarding properties of metric spaces. I'm curious to see a proof of why it holds.
Suppose $(X,\rho)$ is any metric space. For a given $\epsilon\gt 0$, I let $N(X,\epsilon)$ denote the least $n$ such that $X=\bigcup\limits_{i=1}^n U_i$ where $U_i$ are sets such that $\operatorname{diam}(U_i)\leq 2\epsilon$. I also denote by $M(X,\epsilon)$ the greatest number of $m$ points $x_i$, $1\leq i\leq m$ such that $\rho(x_i,x_j)\gt\epsilon$ whenever $i\neq j$. 
With this notation, what is it that $N(X,\epsilon)\leq M(X,\epsilon)$ and $M(X,\epsilon)\leq N(X,\epsilon/2)$? Thanks.

For given $\epsilon$, pick a set of $M(X,\epsilon)$ points at distances greater than $\epsilon$, and form closed balls with radius $\epsilon$ around them. If there is a point in $X$ that belongs to none of these balls, we can add it to the set, contradicting the maximality of $M(X,\epsilon)$. Thus these $M(X,\epsilon)$ sets of diameter $2\epsilon$ cover $X$, and hence $N(X,\epsilon)\le M(X,\epsilon)$.
For the other direction, note that a set with diameter $2\epsilon/2=\epsilon$ can contain at most one point of a set of $M(X,\epsilon)$ points at distances greater than $\epsilon$; thus we need at least $M(X,\epsilon)$ such sets to cover $X$.
