Timeline for Group homomorphisms and maps between function spaces
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 22, 2010 at 18:31 | comment | added | Matthew Daws | Well, I've decided to accept this: I'm a soft touch for these sorts of things! And Victor has been very helpful... | |
| Jun 22, 2010 at 18:31 | vote | accept | Matthew Daws | ||
| Jun 20, 2010 at 21:12 | comment | added | Yemon Choi | @Matt: I think there's nothing to stop you accepting your own answer. (If you don't want votes for the answer to count as credit then you could make your answer community-wiki.) Consensus seems to be that it's better to have the most complete/relevant answer accepted, than to abide by "real-world" etiquette - but I may have misunderstood, others are welcome to correct me if this is the case. | |
| Jun 20, 2010 at 20:51 | comment | added | Matthew Daws | Unless I get a massively inspired answer, I'll accept Victor's, as it's nice, and gave me a lot of inspiration... | |
| Jun 20, 2010 at 16:47 | comment | added | Matthew Daws | I've deleted some comments which no longer apply. This looks promising. I was worried for a bit as to why G' being locally compact implies that it was closed in H. However, I think this is because of the following. WLOG, we may suppose that G' is dense (and it has the subspace topology). By secure.wikimedia.org/wikipedia/en/wiki/… we have that G' is thus open in H, but as G' is a subgroup, it must also be closed, and hence equal to H as required. | |
| Jun 20, 2010 at 15:00 | comment | added | Victor Protsak | Transferred comments to the proof. | |
| Jun 20, 2010 at 14:59 | history | edited | Victor Protsak | CC BY-SA 2.5 |
typo
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| Jun 20, 2010 at 10:19 | history | edited | Victor Protsak | CC BY-SA 2.5 |
note: doesn't work
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| Jun 20, 2010 at 10:16 | comment | added | Victor Protsak | Matthew: You are right, this simple trick doesn't work under the local compactness assumption. To get a "bad" open cover, just consider the complements of shrinking closed neighborhoods of an irrational point (this is equivalent to your Baire space argument). | |
| Jun 20, 2010 at 9:24 | comment | added | Konrad Voelkel | If you take a point x in Q, and some compact interval [a,b] in R which contains Q, then every open cover of [a,b] cap Q is dense in an open cover of [a,b], which has a finite subcover, so the associated subcover over Q is finite, too (and still a cover). So Q with the induced topology is still locally compact. Right? | |
| Jun 20, 2010 at 8:54 | history | answered | Victor Protsak | CC BY-SA 2.5 |