Talk:Mostowski collapse lemma

If you extend Mostowski collapse to well-founded (not just well-ordered) relations, surely the result is not an isomorphism! —Preceding unsigned comment added by 192.75.48.150 (talk) 19:36, 17 April 2008 (UTC)

Well-ordered relations are not involved in any version of Mostowski lemma. The result is an isomorphism for set-like well-founded extensional relations, which is the usual statement of the lemma, and what this article says. If you try to extend the lemma to nonextensional wellfounded relations then it obviously can't be an isomorphism. So what? — EJ (talk) 10:01, 18 April 2008 (UTC)

Actually, as a consequence of the lemma one obtains that any set-like (not necessarily extensional) well-founded relation is isomorphic to membership on a (not necessarily transitive) class. Neither the class nor the isomorphism is unique then. You can count this as an extension of Mostowski lemma to nonextensional relations which does give an isomorphism. — EJ (talk) 11:53, 18 April 2008 (UTC)

My mistake, I thought "extensional relation" meant something else entirely (basically, one which respects the intended equivalence relation). Thanks for spelling it out in the article. --192.75.48.150 (talk) 15:13, 18 April 2008 (UTC)

generalization

The recently added "homomorphism version" of the lemma, and specifically its uniqueness part, is false. Consider the relation R = {(a,b), (b,c)} on X = {a, b, c}. We can map (X, R) homomorphically onto the transitive set {∅, {∅}, {{∅}}} via f(a) = ∅, f(b) = {∅}, f(c) = {{∅}}, but we can also map it homomorphically onto the transitive set {∅, {∅}, {∅,{∅}}} via f(c) = {∅,{∅}}. The problem is even more pronounced in nonwellfounded set theory, where any relation whatsoever can be mapped homomorphically to the transitive urelement x = {x}. Let alone the fact that Mostowski collapse lemma is Mostowski collapse lemma, not any of its generalizations. — Emil J. (formerly EJ) 09:46, 14 July 2008 (UTC)

The name

Shouldn't this name Mostowski's collapse lemma? —Preceding unsigned comment added by 83.6.96.216 (talk) 17:11, 31 August 2008 (UTC)

The link on extensional

'extensional' links to the Axiom of extensionality, which is unrelated to extensional sets. —Preceding unsigned comment added by 98.207.249.254 (talk) 20:14, 14 December 2008 (UTC)

A relation is extensional iff it is a model of the axiom of extensionality, so the link is entirely appropriate. — Emil J. 15:13, 15 December 2008 (UTC)

Generalization of the lemma - embedding into an extensional relation

In the Generalizations section of the article it is said 'Every well-founded set-like relation can be embedded into a well-founded set-like extensional relation.'. Could someone add a reference or a short explanation on how this embedding is done? —Preceding unsigned comment added by 80.223.12.1 (talk) 12:50, 30 April 2011 (UTC)

Here is a proof that any (not necessarily extensional) well-founded relation that is a set is isomorphic to membership on a (not necessarily transitive) class: Shaughan Lavine (1994). Understanding the infinite. Harvard University Press. Theorem VIII.4.2. pp. 304–305. It easily extends to the full set-like case. It employs the Axiom of Choice. I'm pretty sure—though not positive enough to add it to the entry—that the Axiom of Choice is required. If one allows urelements, that seems pretty clearly true for the case of empty R on an infinite set of urelements. — Preceding unsigned comment added by 185.192.69.122 (talk) 19:10, 20 August 2020 (UTC)