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Nov 17, 2021 at 12:39 comment added Abdelmalek Abdesselam Speaking about $\ell^{\infty}$, I haven't seen any evidence in this post that "it does need the Hahn-Banach" theorem. For one, it is an abstract thing, and it sound weird to say that has needs. The question rather should be: does a mathematician need to use the Hahn-Banach Theorem (or axiom here), in order to produce a continuous linear form outside $\ell^1$. I have never felt this need, but others could perhaps inform me about why they need such a thing.
Oct 15, 2021 at 21:29 history edited Martin Sleziak
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Nov 20, 2009 at 19:03 comment added Mark Meckes There are other classes of Banach spaces (e.g., uniformly convex spaces) for which Hahn-Banach can also be proved constructively. As it happens, $\ell^\infty$ is one of the first non-examples to come to mind for those other classes as well. So although it's certainly not "the" answer, it's a pretty good one from many points of view.
Nov 17, 2009 at 14:38 history edited Andrew Stacey CC BY-SA 2.5
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Nov 14, 2009 at 20:17 answer added Gerald Edgar timeline score: 33
Nov 14, 2009 at 5:20 answer added Greg Kuperberg timeline score: 50
Nov 14, 2009 at 1:49 answer added Peter Schmitt timeline score: -3
Nov 13, 2009 at 18:41 answer added Dave Penneys timeline score: 17
Nov 13, 2009 at 17:46 answer added Ari Shnidman timeline score: 3
Nov 13, 2009 at 15:25 answer added Danny Calegari timeline score: 13
Nov 13, 2009 at 15:08 answer added Mark Meckes timeline score: 24
Nov 13, 2009 at 14:36 history asked Andrew Stacey CC BY-SA 2.5