2002
DOI: 10.1090/dimacs/058/10
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Some applications of superrigidity to Borel equivalence relations

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Cited by 9 publications
(5 citation statements)
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“…However, no countable Borel equivalence relations beyond E 0 were known even to be indivisible into two pieces until we showed this for E ∞ in [30]. In contrast, we note that Thomas [39] has shown the existence of a countable Borel equivalence relation E on an uncountable standard Borel space that is not 2-indivisible. Indeed, there is a countable Borel equivalence relation E where E < B E ⊕E, where E ⊕ E is the direct sum of two disjoint copies of E.…”
Section: Theorem 325 Given Any Uniformly Universal Countable Borel Eq...mentioning
confidence: 74%
See 1 more Smart Citation
“…However, no countable Borel equivalence relations beyond E 0 were known even to be indivisible into two pieces until we showed this for E ∞ in [30]. In contrast, we note that Thomas [39] has shown the existence of a countable Borel equivalence relation E on an uncountable standard Borel space that is not 2-indivisible. Indeed, there is a countable Borel equivalence relation E where E < B E ⊕E, where E ⊕ E is the direct sum of two disjoint copies of E.…”
Section: Theorem 325 Given Any Uniformly Universal Countable Borel Eq...mentioning
confidence: 74%
“…For instance, this is a typical proof strategy in applications of cocycle superrigidity to the field of countable Borel equivalence relations. See for example [1,16,39,41,46].…”
Section: Introductionmentioning
confidence: 99%
“…Condition 2.1(ii) is reminiscent of the conclusion of the "unique ergodicity argument" first introduced by Adams [1] in the measure-theoretical setting and later exploited by Thomas [16,17] and Hjorth-Kechris [7]. Of course, the following result is an immediate consequence of Theorem 2.1 and implies that ≡ T is not countable universal.…”
Section: Introductionmentioning
confidence: 89%
“…In fact [Th,Lemma 3.4] gives a countable Borel equivalence relation E ∈ N SC, which is not divisible by any n > 1, i.e., there is no F ∈ C with nF ∼ B F . It follows, using Corollary 2.5, that there is a fam on [N SC], + for which ϕ([E]) = 1, so ϕ takes a finite value.…”
Section: Cardinal Algebrasmentioning
confidence: 99%