# Right-continuity of covering number

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 $$$$\Vert x - x_i\Vert_\infty\leq\varepsilon$$$$ From the observation that the ball is closed and the covering number is defined with $$\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?

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 $$1+1/m$$.
• I do not understand your notations and choosing quantifiers seems suspicial, but the idea is very simple: if for all $n$ you may cover $X$ by $N$ balls with radius $\varepsilon+1/n$ centered in points $x_1(n),\dots,x_N(n)$, then choose a subsequence $n_1<n_2<\dots$ such that $x_i(n_k)$ converges to some $x_i$ for all $i=1,\dots,N$. The balls cenrered in $x_i$'s of radius $\varepsilon$ cover $X$. Commented Dec 1, 2022 at 19:24
• Thanks for the clarifications. I was wondering if we can always find a converging subsequence $(x_i(n_k))_k$. Then I found the Bolzano–Weierstrass theorem. Do we deduce the existence of the converging subsequence from this theorem? Commented Dec 2, 2022 at 6:00
• Bolzano-Weierstrass is for bounded subsets of $\mathbb{R}^n$, for compact metric spaces it is one of equivalent definitions (see "sequential compactness") Commented Dec 2, 2022 at 9:13
• Great, I appreciate your help. Maybe one more question: how is $N$ defined in your proof? Do we take $N$ to be the $\varepsilon$-covering number? Commented Dec 2, 2022 at 13:39