Timeline for Weak* continuity of positive parts
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| May 31, 2016 at 4:04 | comment | added | TaQ | @Nik Weaver: There are mathematicians (a) that do not care whether a particular result holds in ZF without AC, or whether it requires use of AC, and there are ones (b) that do care. I agree with you in that my answer has "little content" to mathematicians of type (a). Here Bill Johnson wrote: "What I do not know is whether the result depends on the axiom of choice." So it seems that you and he are of different type in this respect. | |
| May 30, 2016 at 20:13 | comment | added | Nik Weaver | You're right about Gram-Schmidt, good point. I will stand on the rest of my comment that there is little content here. | |
| May 30, 2016 at 19:31 | comment | added | TaQ | And, by the way, it does not matter, if it lead, since we only need to construct a vector in $\ell^{\,+\infty}\,$. | |
| May 30, 2016 at 19:01 | comment | added | TaQ | @Nik Weaver: (1) The first $e_n$ outside span$\mathcal F$ need not be orthogonal to span$\mathcal F$, and so it has to "modified" to be such in order to have the properties required. I have given, or rather sketched, the details which are needed if one does not want to use AC to get the net with the required "negative" properties. (2) Gramm−Schmidt does not lead outside $\ell^{\,1}$ since it just gives linear combinations of the original vectors, and $\ell^{\,1}$ is a vector space. | |
| May 29, 2016 at 22:22 | comment | added | Nik Weaver | It doesn't seem to work, though ... when you Gram-Schmidt you could get outside of $l^1$. | |
| May 29, 2016 at 22:22 | comment | added | Nik Weaver | This just looks like a complicated way of saying "in Bill Johnson's construction, `choose' $z_{\mathcal{F}}$ by letting it be the first $e_n$ outside of ${\rm span}(\mathcal{F})$, modified to have the desired properties." | |
| May 29, 2016 at 10:15 | history | answered | TaQ | CC BY-SA 3.0 |