Timeline for In infinite dimensions, is it possible that convergence of distances to a sequence always implies convergence of that sequence?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jul 27, 2020 at 23:19 | comment | added | Nikhil Sahoo | Ah yes. Thank you, this is very instructive and a much cleaner statement. | |
| Jul 27, 2020 at 22:52 | history | edited | Bill Johnson | CC BY-SA 4.0 |
added 487 characters in body
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| Jul 27, 2020 at 22:45 | comment | added | Bill Johnson | Yes, but I would explain this as a consequence of facts that are standard for a course that covers metric spaces. I'll add to the answer. | |
| Jul 27, 2020 at 20:18 | comment | added | Nikhil Sahoo | Let $(x_n)$ be a $P$-sequence. To show $P$-completeness, we want $(x_n)$ to have a Cauchy subsequence. Let $d_m=\lim_{n\rightarrow \infty}d(x_m,x_n)$. Since $P$-sequences and separated sequences are preserved under taking subsequences, the assumption is that $(x_n)$ has no separated subsequences and thus $\lim_{m\rightarrow \infty}d_m=0.$ Define $n_k$ as follows. Pick $n_0$ so that $d_{n_0}<1.$ If $n_k$ is such that $d_{n_k}<2^{-k}$, we can pick $n_{k+1}>n_k$ with $d_{n_{k+1}}<2^{-(k+1)}$ and $||x_{n_k}-x_{n_{k+1}}||<2^{1-k}$. Then $(x_{n_k})$ is Cauchy by the triangle inequality? | |
| Jul 27, 2020 at 20:03 | comment | added | Nikhil Sahoo | Thanks, this is great! I feel like I somewhat understand most of the argument (sans the more advanced details). Since this is not an area in which I know a lot, I want to check that I actually understand the immediate implication that you mention in the beginning. Does the following sound right? | |
| Jul 26, 2020 at 4:34 | vote | accept | Nikhil Sahoo | ||
| Jul 25, 2020 at 23:24 | history | answered | Bill Johnson | CC BY-SA 4.0 |