Timeline for Right-continuity of covering number
Current License: CC BY-SA 4.0
9 events
when toggle format | what | by | license | comment | |
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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 |