# Tag Info

1

Something is not clear to me about your question. Global choice implies the axiom of choice for sets. Of course it does. Every set is a subclass of $V$, and if $V$ can be well-ordered, then every set can be well-ordered. Requiring that there is an injection from the class of ordinals into $V$ is provable in $\sf ZF$. This function is the identity function. ...

6

First of all, by elementarity $\operatorname{rank}(j(x))=j(\operatorname{rank}(x))$. So it suffices to show that there is some $x$ whose rank is moved. Now we can prove the following by induction: Suppose that $\operatorname{rank}(j(x))=\operatorname{rank}(x)$ and for all $y$ such that $\operatorname{rank}(y)<\operatorname{rank}(x)$, $j(y)=y$, then ...

4

Not every second-order theory has to be categorical. It doesn't work like that, and it has nothing to do with the Lowenheim-Skolem theorems. The failure of categoricity of $\sf ZFC_2$ is simply because there are no additional axioms of infinity. What do I mean by that? The axiom of infinity is a special axiom in $\sf ZFC$ because when replacing this axiom ...

Top 50 recent answers are included