5
$\begingroup$

As we know a first order theory $T$ is finitely axiomatizable if there is a finite set $F\subseteq T$ of axioms such that $F\vdash \sigma$ for every $\sigma \in T$.

How we can prove if $\mathsf{ZF}$ is consistent, then $\mathsf{ZF}$ is not finitely axiomatizable? By using the Reflection Theorem or any other if we could use?

$\endgroup$
1
$\begingroup$

You can find the proof in Kunen's famous book "Set Theory: An Introduction To Independence Proofs" (North Holland, 1980). It is Corollary IV 7.7 on page 138, and it does make use of reflection.

$\endgroup$
4
$\begingroup$

Suppose $\text{ZF}$ is consistent and $\text{ZF}$ is fnitely axiomatizable. Let $\Gamma \subset \text{ZF}$ be a finite subset such that $\Gamma \vdash \text{ZFC}$. Referring to Jech or Kunen, $\text{ZF} \vdash \text{Reflection Theorem}$. So $\text{ZF} \vdash \text{Con}(\Gamma)$. Since $\Gamma \vdash \text{ZF}$, $\text{ZF} \vdash \text{Con}(\text{ZF})$. If $\text{ZF}$ is consistent, this contradicts the Second Incompleteness Theorem.

$\endgroup$
  • 4
    $\begingroup$ You do not even need the second incompleteness: $\Gamma$ proves that there is an $\alpha$ such that $V_\alpha\models \Gamma$, so if $V_\beta\models\Gamma$, $\beta$ is not the least such ordinal. $\endgroup$ – Andrés E. Caicedo Apr 4 '14 at 5:51
  • $\begingroup$ I see that en.wikipedia.org/wiki/Reflection_principle discusses both these techniques. $\endgroup$ – MJD Apr 4 '14 at 5:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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