Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For each first-order $\sigma \,$-formula $\varphi(y,x_1, \ldots, x_n) \,,$ the axiom of choice implies the existence of a function $f_\varphi: M^n\to M$ such that, for all $a_1, \ldots, a_n \in M$, either $M\models\varphi(f_\varphi (a_1, \dots, a_n), a_1, \dots, a_n)$ or $M\models\neg\exists y \, \varphi(y, a_1, \dots, a_n) \,.$

This is a quote from the Wikipedia page on the Löwenheim–Skolem theorem.

What I am confused about is the last part of the quote — starting from $$M\models\varphi(f_\varphi (a_1, \dots, a_n), a_1, \dots, a_n)\,.$$ Specifically, $$M\models\neg\exists y \, \varphi(y, a_1, \dots, a_n) \,.$$ What does it mean? There does not exist $y$ that satisfies the predicate and what? How does it relate to a model - what is a model satisfying?

Thanks.

share|improve this question
    
"Either $M$ believes that $\phi(f(\ldots), \ldots)$ holds or $M$ believes that $\lnot \exists y \phi(y, \ldots)$ holds." This is basically just the law of excluded middle + the axiom of choice. –  Zhen Lin May 6 '12 at 13:41
add comment

1 Answer

up vote 3 down vote accepted

The claim says that given a formula $\varphi(y,x_1,\ldots,x_n)$ then for every $n$ parameters $a_1,\ldots,a_n$ exactly one of the following happens:

  1. There exists some $y\in M$ such that $M\models\varphi(y,a_1,\ldots,a_n)$.
  2. For all $y\in M$ the above is false, therefore $M\models\lnot\exists y\varphi(y,a_1,\ldots,a_n)$. This is because in classical logic the model thinks that there is no $y$ such that $\varphi(y,a_1,\ldots,a_n)$ is true if and only if it things that $\lnot\exists y\varphi(y,a_1,\ldots,a_n)$ is true.

Using the axiom of choice we can define a function $f_\varphi$ that for every $n$-tuple, $(a_1,\ldots,a_n)$ for which the first thing holds, $f(a_1,\ldots,a_n)$ returns such $y$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.