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Timeline for Right-continuity of covering number

Current License: CC BY-SA 4.0

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Dec 2, 2022 at 16:02 comment added iom10 That makes it clear. Thanks again for your help!
Dec 2, 2022 at 14:49 comment added Fedor Petrov $N$ is the right limit of the $t$-covering number when $t$ goes to $\varepsilon$ from the right
Dec 2, 2022 at 13:39 comment added iom10 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?
Dec 2, 2022 at 9:13 comment added Fedor Petrov Bolzano-Weierstrass is for bounded subsets of $\mathbb{R}^n$, for compact metric spaces it is one of equivalent definitions (see "sequential compactness")
Dec 2, 2022 at 6:01 vote accept iom10
Dec 2, 2022 at 6:00 comment added iom10 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?
Dec 1, 2022 at 19:24 comment added Fedor Petrov 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 $X$.
Dec 1, 2022 at 19:10 comment added iom10 Thank you very much for your helpful answer (unfortunately I cannot upvote yet). Regarding your second statement, I have tried to formalize the right-continuity via convergent sequences (see initial post). Could you please have a look at the proof attempt to check if it is correct? Thanks again for the help!
Dec 1, 2022 at 9:49 history answered Fedor Petrov CC BY-SA 4.0