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The claim on this page that the second part of the definition of elementary substructure can be expressed by stating simply that is incorrect if refers to the set of first-order -sentences satisfied by . In this case, the latter is a statement only of elementary equivalence, and it is possible for a structure to have elementarily equivalent substructures that are not elementary substructures. As an example, consider the structure consisting of the naturals with the usual ordering. It is easily seen that the substructure consisting of the evens with the induced ordering is isomorphic, and hence elementarily equivalent, to ; but it is not an elementary substructure by Tarski's criterion, since every natural is definable in (the natural by the formula stating 'there exist exactly elements less than me').
On the other hand, the author may be using a slightly nonstandard notation, where refers to something else.
"More generally, any first-order theory has non-isomorphic, elementary equivalent models, which can be obtained via the Löwenheim–Skolem theorem. Thus, for example, there are non-standard models of Peano arithmetic, which contain other objects than just the numbers 0, 1, 2, etc., and yet are elementarily equivalent to the standard model"
Isn't this claim false? Don't we need the condition that the first-order theory have infinite models? (I.e. doesn't the theory that (forall x)(forall y)(x = y) have a single model?) Or are we not assuming that each model interprets '=' as the diagonal relation? — Preceding unsigned comment added by 128.135.100.105 ( talk) 19:24, 26 November 2012 (UTC)
The current version states:
An elementary embedding of a structure N into a structure M of the same signature σ is a map h: N → M such that for every first-order σ-formula φ(x1, …, xn) and all elements a1, …, an of N,
Every elementary embedding is a strong homomorphism, and its image is an elementary substructure.
The previous version had "implies" in place of "if and only if". It does not really make any difference but just for the purposes of understanding the issue, it seems to me that "implies" is also accurate. This is because we quantify over all φ, so in particular the negation of it will give the opposite implication. Just wondering if this is right. Tkuvho ( talk) 14:36, 12 June 2013 (UTC)
It would we useful to add (to this article) that
(For now, just copy-pasted from
modeltheory.wikia.com.)
A source:
"A note on definable Skolem functions" by D. A. Anapolitanos.
Boris Tsirelson (
talk) 13:52, 5 December 2018 (UTC)
This
level-5 vital article is rated Start-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | |||||||||||
|
A summary of this article appears in Embedding#Universal algebra and model theory. |
The claim on this page that the second part of the definition of elementary substructure can be expressed by stating simply that is incorrect if refers to the set of first-order -sentences satisfied by . In this case, the latter is a statement only of elementary equivalence, and it is possible for a structure to have elementarily equivalent substructures that are not elementary substructures. As an example, consider the structure consisting of the naturals with the usual ordering. It is easily seen that the substructure consisting of the evens with the induced ordering is isomorphic, and hence elementarily equivalent, to ; but it is not an elementary substructure by Tarski's criterion, since every natural is definable in (the natural by the formula stating 'there exist exactly elements less than me').
On the other hand, the author may be using a slightly nonstandard notation, where refers to something else.
"More generally, any first-order theory has non-isomorphic, elementary equivalent models, which can be obtained via the Löwenheim–Skolem theorem. Thus, for example, there are non-standard models of Peano arithmetic, which contain other objects than just the numbers 0, 1, 2, etc., and yet are elementarily equivalent to the standard model"
Isn't this claim false? Don't we need the condition that the first-order theory have infinite models? (I.e. doesn't the theory that (forall x)(forall y)(x = y) have a single model?) Or are we not assuming that each model interprets '=' as the diagonal relation? — Preceding unsigned comment added by 128.135.100.105 ( talk) 19:24, 26 November 2012 (UTC)
The current version states:
An elementary embedding of a structure N into a structure M of the same signature σ is a map h: N → M such that for every first-order σ-formula φ(x1, …, xn) and all elements a1, …, an of N,
Every elementary embedding is a strong homomorphism, and its image is an elementary substructure.
The previous version had "implies" in place of "if and only if". It does not really make any difference but just for the purposes of understanding the issue, it seems to me that "implies" is also accurate. This is because we quantify over all φ, so in particular the negation of it will give the opposite implication. Just wondering if this is right. Tkuvho ( talk) 14:36, 12 June 2013 (UTC)
It would we useful to add (to this article) that
(For now, just copy-pasted from
modeltheory.wikia.com.)
A source:
"A note on definable Skolem functions" by D. A. Anapolitanos.
Boris Tsirelson (
talk) 13:52, 5 December 2018 (UTC)