Consider an ambient metric space $(\mathcal{X},\Vert\cdot\Vert\_\infty)$. Let $\mathcal{B}\_1 = \mathcal{B}\_{\Vert\cdot\Vert\_K}(0,1)\subseteq\mathcal{X}$ be the closed unit ball with respect to some norm $\Vert\cdot\Vert\_K$. Denote the $\varepsilon$-covering number of $\mathcal{B}\_1$ with respect to $\Vert\cdot\Vert\_\infty$ by $\mathcal{N}(\varepsilon, \mathcal{B}\_1,\Vert\cdot\Vert\_\infty)$. That is, we can find a set of points $\{x\_1,\dots,x\_n\}\subseteq\mathcal{X}$ with $n = \mathcal{N}\left(\varepsilon, \mathcal{B}\_1,\Vert\cdot\Vert\_\infty\right)$ such that for all $x\in\mathcal{B}\_1$, there exists $i\in[n]$ with
\begin{equation}
\Vert x - x\_i\Vert\_\infty\leq\varepsilon
\end{equation}
From the observation that the ball is closed and the covering number is defined with $\leq$ instead of $<$ (see equation above), I am tempted to assume that the covering number is a **right-continuous** function of $\varepsilon$. Is this assumption correct?



Right-continuity of covering number
Without additional assumptions on the metric space, it may appear that for every $\varepsilon>1$ the covering number equals 1, but for $\varepsilon=1$ it is infinite. For example, let positive integers be the points and the distance between $n$ and $m>n$ be equal $1+1/m$.


For compact metric space, it is right-continuous as you may take a convergent subsequence of cover sets.


