Timeline for Is there an infinite-dimensional Banach space with a compact unit ball?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Jun 14, 2013 at 20:30 | answer | added | Delio Mugnolo | timeline score: 4 | |
| Oct 11, 2011 at 2:07 | vote | accept | Mark Meckes | ||
| Oct 10, 2011 at 23:13 | answer | added | godelian | timeline score: 12 | |
| Oct 10, 2011 at 22:33 | comment | added | user5810 | For all members $p$ of $[1,\scriptsize{+}\normalsize{\infty}]$, $\; \ell^p($amorphous set$) \;$ is an infinite dimensional Banach space without a sequence of finite dimensional subspaces whose dimensions are unbounded. $\hspace{1.25 in}$ | |
| Oct 10, 2011 at 22:30 | comment | added | godelian | Note that it is important to take a specific notion of compactness, since in the absence of choice the usual notions may not longer be equivalent. For instance, in N. Brunner's article "Sequential compactness and the axiom of choice" (available online), it is mentioned that the statement "A Hilbert space is finite dimensional iff its closed unit ball is compact" is provable in ZF without choice (even without foundation), but that such a result may no longer be valid in case of sequential compactness. | |
| Oct 10, 2011 at 21:55 | comment | added | Bill Johnson | It would be sufficient to know that $X$ contains a sequence $(E_n)$ of finite dimensional subspaces whose dimensions tend to infinity. Must such a sequence exist in the absence of AC? | |
| Oct 10, 2011 at 21:38 | history | asked | Mark Meckes | CC BY-SA 3.0 |