2013
DOI: 10.4064/fm222-3-1
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On ultrapowers of Banach spaces of type \mathscrL

Abstract: We prove that no ultraproduct of Banach spaces via a countably incomplete ultrafilter can contain c 0 complemented. This shows that a "result" widely used in the theory of ultraproducts is wrong. We then amend a number of results whose proofs had been infected by that statement. In particular we provide proofs for the following statements: (i) All M -spaces, in particular all C(K)-spaces, have ultrapowers isomorphic to ultrapowers of c 0 , as well as all their complemented subspaces isomorphic to their square.… Show more

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Cited by 12 publications
(4 citation statements)
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“…In fact, since c 0 is complemented in F (c 0 ), by Godefroy-Kalton's theorem [12] it would be isomorphic to a complemented subspace of X U . However that is impossible by Proposition 3.3 in [4].…”
Section: Some Remarks On the Cotype Of Lipschitz-free Spacesmentioning
confidence: 95%
“…In fact, since c 0 is complemented in F (c 0 ), by Godefroy-Kalton's theorem [12] it would be isomorphic to a complemented subspace of X U . However that is impossible by Proposition 3.3 in [4].…”
Section: Some Remarks On the Cotype Of Lipschitz-free Spacesmentioning
confidence: 95%
“…It can be proved (see [11,Theorem 6.1.8] or [12,Theorem 13.9]) that if U is countably incomplete and ℵ-good, then every family of less than ℵ internal subsets of S i U having the finite intersection property has nonempty intersection. Since |Γ × N| < ℵ and U is ℵ-good, there is x ∈ X i U in the nonempty intersection A combination of [4] and [3], see also [5], shows that there exist universally 1-separably injective spaces not isomorphic to any C(K) space. A higher cardinal generalization is as follows.…”
Section: Ultraproductsmentioning
confidence: 97%
“…It had been already observed in [AvCSCaGoMo,Proposition 4.13] that any non-trivial ultrapower of G is ultrahomogeneous. We shall now see that Fraïssé Banach spaces are locally determined.…”
Section: Consequentlymentioning
confidence: 99%