Well-ordering in a first-order language

  • Thread starter Thread starter jgens
  • Start date Start date
  • Tags Tags
    Language
AI Thread Summary
Well-ordering cannot be expressed in a first-order language due to limitations in quantifying over subsets of a set. The discussion highlights that while the standard model of natural numbers is well-ordered, non-standard models are not, indicating the absence of a first-order sentence for well-ordering. A more straightforward argument using compactness shows that adding constants to express well-ordering leads to contradictions. Specifically, the existence of a model satisfying the axioms contradicts the definition of well-ordering. Thus, it is concluded that well-ordering is not expressible in first-order logic.
jgens
Gold Member
Messages
1,575
Reaction score
50
Is it possible to express well-ordering in a first-order language?

For example:
If X is a set and < is a binary relation on X such that (X,<) \vDash \forall x \; \neg(x<x) and (X,<) \vDash \forall x \forall y \forall z((x<y \wedge y<z) \rightarrow x<z), then (X,<) is a partial order.

If (X,<) is a partial order such that (X,<) \vDash \forall x \forall y(x<y \vee x=y \vee y<x), then (X,<) is a total order.

I cannot think of a way to express well-ordering in this language. Since the quantifiers range over all of X and there is no obvious way to quantify over subsets of X, I am thinking it is not possible to express well-ordering in this way. But it is also possible that I am just not sufficiently clever to think of something.

Any help?
 
Physics news on Phys.org
I think I found an answer to this question. The standard model for (\mathbb{N};0,S,<,+,\times) is well-ordered but non-standard models are not. In particular, this means that there is no first-order sentence that expresses well-ordering.
 
Well that much is true, but relying on non-standard models of arithmetic in order to show that the property of being a well-ordering is not expressible in first-order classical logic seems to be a bit of an overkill. A more straightforward argument relies on compactness: call the theory of strict partial ordering with your two axioms SPO and suppose that the set of formulas T expresses the property that < is a well-ordering (M is a model of SPO+T iff < is a well-ordering on the domain of M). Add countably many constants c_0, c_1, \ldots to the language along with the axioms Ax: c_1 &lt; c_0, c_2 &lt; c_1, \ldots. Each finite subset of SPO+T+Ax is satisfiable, hence by compactness, the whole set is satisfiable in some model M. But the set of all realizations of c_i clearly has no minimum in M, which contradicts the definition of T.
 
I was reading documentation about the soundness and completeness of logic formal systems. Consider the following $$\vdash_S \phi$$ where ##S## is the proof-system making part the formal system and ##\phi## is a wff (well formed formula) of the formal language. Note the blank on left of the turnstile symbol ##\vdash_S##, as far as I can tell it actually represents the empty set. So what does it mean ? I guess it actually means ##\phi## is a theorem of the formal system, i.e. there is a...

Similar threads

Replies
2
Views
6K
Replies
6
Views
2K
Replies
22
Views
5K
Replies
2
Views
1K
Replies
1
Views
1K
Replies
3
Views
2K
Replies
8
Views
4K
Back
Top