Timeline for Can we take a supremum over all Hilbert spaces?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 14, 2021 at 22:40 | vote | accept | Ivan Feshchenko | ||
| Apr 14, 2021 at 22:41 | |||||
| Apr 12, 2021 at 18:22 | comment | added | Dmitri Pavlov | @ReidBarton: This is already indicated in the other answer, so I see no point in duplicating it here. The point of this answer is that the classical proof works just fine for proper classes, without any modifications. | |
| Apr 12, 2021 at 17:57 | comment | added | Reid Barton | sup $I$ exists because $I$ is a nonempty, bounded-above subset of the reals. The point being that you don't have to reexamine the proof to check whether it works for class-sized families; instead you can immediately reduce to the standard fact (and not care about how it was proved). | |
| Apr 12, 2021 at 15:55 | history | edited | LSpice | CC BY-SA 4.0 |
f's to g's
|
| Apr 12, 2021 at 15:51 | history | edited | Dmitri Pavlov | CC BY-SA 4.0 |
added 246 characters in body
|
| Apr 12, 2021 at 15:09 | comment | added | Dmitri Pavlov | @EmilJeřábek: More directly than what? How do you show that sup I exists in your proof? Note that inf U exists because U is an upper Dedekind cut. | |
| Apr 12, 2021 at 6:00 | comment | added | Emil Jeřábek | More directly, use separation to show that the image of $g$, $I=\{u\in\mathbb R:\exists c\in C\,g(c)=u\}$, is a set, and then $\sup_{c\in C}g(c)=\sup I$. | |
| Apr 12, 2021 at 0:27 | history | answered | Dmitri Pavlov | CC BY-SA 4.0 |