It is known that there exist some isomorphism between $L_\infty$ and $\ell_\infty$, which is not explicit at all.
Could someone tell me whether there exist an isometric isomorphism between $L_\infty$ and $\ell_\infty$?
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Side note (sorry): existence of this isomorphism depends essentially on the Axiom of Choice. Which is why you found it to be "not explicit at all". – GEdgar Oct 05 '12 at 14:11
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They aren't isometric.
We have $\ell_\infty = C(K)$ and $L_\infty = C(L)$ for $K = \beta \mathbb N$ and $L$ the Stone space of the Lebesgue algebra on $[0,1]$. By the Banach-Stone theorem an isometry would involve a homeomorphism between $K$ and $L$. Such a homeomorphism can't exist: For example, $K = \beta\mathbb{N}$ is separable while $L$ isn't separable, being the Stone space of an atomless probability algebra. In fact, every countable subset of $L$ is nowhere dense.
A more explicit and probably more elementary approach is outlined in these two exercises from page 268 in M. Fabian, P. Habala et al. Functional Analysis and Infinite-Dimensional Geometry, Springer 2001, depending on recognizing points of differentiability of the norms:
Xander Henderson
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Could you give a reference for separability and non-separability of $K$ and $L$ compactifications. I'm not familiar with Stone-Cech compactification. – Norbert Oct 05 '12 at 22:54
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@Nate: See e.g. Fremlin, measure theory, volume 3. Combine 322F with 316I to get that every meager set in the Stone space of a probability algebra is nowhere dense. 316L says that atoms correspond to isolated points in the Stone space so that countable unions of points are meager and hence nowhere dense. See also the last paragraph on page 68 of loc. cit. – commenter Oct 06 '12 at 02:08
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@Norbert: For non-separability of $L$ see my comment to Nate. The separability of the Stone-Cech compactification follows from the definition of a compactification: $\mathbb{N}$ is dense in $\beta\mathbb{N}$. – commenter Oct 06 '12 at 02:10
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Sorry, I meant to say: "countable unions of points in the Stone space of an atomless probability algebra are meager". – commenter Oct 06 '12 at 02:36
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Thanks for the reference. I guess it will take me some work to translate Fremlin's discussion into language that I am familiar with. – Nate Eldredge Oct 06 '12 at 02:59
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@Nate: A more easily accessible reference might be §24 of Halmos's Lectures on Boolean Algebra (=chapter 41 of Givant-Halmos Introduction to BA) where the statement is Exercise 2 (=Ex. 5). The hardest part is in establishing the regularity property of the measure on the Stone space: $\mu(B) = \sup \mu(F) = \inf \mu(G)$ where $B$ is Borel and $F \subset B \subset G$ with $F,G$ clopen (Lemma 3). Once you have that (strange) property then it follows that taking the closure doesn't increase the measure and thus null sets and meager sets are nowhere dense (Lemmas 4,5,6). – commenter Oct 06 '12 at 07:41
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@NateEldredge: I found a nice pair of exercises in Fabian et al. containing a probably more accessible argument. See the second part of the answer. – commenter Jun 29 '13 at 21:37
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@XanderHenderson: when you addressed my flag, it looks like you didn't pay much attention to the edit history. The part of the answer with an image was edited in by the original answerer 9 years ago. My edit was totally unrelated to that part of the answer, and only related to the previous part of the answer; that's why I kept the division line between my edit and the part with the image. – Martin Argerami Aug 16 '22 at 08:33
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@MartinArgeramia In fact, I did understand the edit you made, and felt that it added theory and ideas that are not present anywhere in the original answer. Again, It would be better to post a new answer. – Xander Henderson Aug 16 '22 at 12:58