# $\mathsf{ZF}$ is not finitely axiomatizable

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?

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.
• 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. – Andrés E. Caicedo Apr 4 '14 at 5:51