Timeline for Is there an infinite-dimensional Banach space with a compact unit ball?
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6 events
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| Jun 20, 2013 at 15:51 | comment | added | Mark Meckes | Delio: You're right, I was overlooking that you don't need HB to prove Riesz. But as Martin says, Riesz needs to be applied an infinite number of times, requiring dependent choice. | |
| Jun 15, 2013 at 14:05 | comment | added | Nate Eldredge | @Ian: How do you construct an orthonormal basis without AC? The usual construction uses Zorn's lemma to produce a maximal orthonormal set. | |
| Jun 14, 2013 at 22:03 | comment | added | godelian | Martin's comment is indeed correct. Dependent choice fails in the model described in my answer. I also agree with Ian's comment, see also my comment to the post above. | |
| Jun 14, 2013 at 21:09 | comment | added | Ian Morris | It seems to me that a key point here might be the difference between compactness and sequential compactness, at least in Hilbert spaces such as in godelian's example. Let $H$ be a Hilbert space and $\{e_\alpha \colon \alpha \in A\}$ an orthonormal basis. The set $\{e_\alpha\}$ is clearly closed, so its complement in the closed unit sphere $S$ - call this complement $U_0$ - is open. Each of the sets $U_\alpha:=\{x \in S \colon \|x-e_\alpha\|<1\}$ is also open, and the collection of all the $U_\alpha$'s together with $U_0$ is an open cover without a finite subcover. | |
| Jun 14, 2013 at 21:05 | comment | added | Martin | The Riesz lemma itself holds true without the axiom of choice. But you will apply it countably many times to choose an "almost orthonormal" sequence of vectors. On the face of it this seems to require an application of the axiom of countable dependent choice: en.wikipedia.org/wiki/Axiom_of_dependent_choice | |
| Jun 14, 2013 at 20:30 | history | answered | Delio Mugnolo | CC BY-SA 3.0 |