Timeline for Combinatorial Hilbert spaces
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 8, 2011 at 10:22 | comment | added | David Feldman | @Stefan Please explain your remark about ℓ∞(ω). | |
| Jan 6, 2011 at 22:05 | comment | added | Stefan Geschke | You are absolutely right. I missed the point that since you are looking at closed subspaces you practically get closure under countably infinite sums. | |
| Jan 6, 2011 at 8:15 | comment | added | David Feldman | > Is the collection of finite and cofinite subsets of ω a combinatorial Hilbert space? If you have all the finite subsets, you have all the singleton subsets, so you have a basis, and you get all of $\ell^2(\omega)$, including many subsets neither finite nor cofinite. You could ask about all the cofinite subset of $\omega$. Off the top of my head (sketch) generate a subspace by $v_1, v_2, \ldots$ with each $v_i$ having full support but decreasing much more rapidly then all the previous. | |
| Jan 6, 2011 at 7:57 | comment | added | Stefan Geschke | Is the collection of finite and cofinite subsets of $\omega$ a combinatorial Hilbert space? In the analogous question for $\ell^\infty(\omega)$ you can obviously get all subalgebras of $\mathcal P(\omega)$, but here this is not clear at all. | |
| Jan 6, 2011 at 4:08 | comment | added | David Feldman | @Ricky No, see arxiv.org/PS_cache/arxiv/pdf/0806/0806.1957v1.pdf for example. An infinite Dedekind finite set of reals would contain no countable sequence at all. | |
| Jan 6, 2011 at 2:35 | comment | added | user5810 | Now, is that provable in ZF? | |
| Jan 6, 2011 at 2:05 | comment | added | Chris Eagle | @Mariano: yes, but if a subset of $\omega$ is the union of a collection of subsets of $\omega$, then it is in fact the union of countably many of them, since $\omega$ itself is countable. | |
| Jan 6, 2011 at 1:52 | comment | added | Mariano Suárez-Álvarez | There are uncountable families of subsets of $\omega$, so your «and thus here arbitrary» needs some justification, no? | |
| Jan 6, 2011 at 1:27 | history | asked | David Feldman | CC BY-SA 2.5 |