In Cantor's Attic, it is stated that the theories 1 and 2 are equiconsistent.
- $\mathsf{ZFC}$ + $\delta$ is inaccessible and $V_\delta\prec V$ (the latter expressed as a scheme in the language $\{\delta,\in\}$)
- $\mathsf{ZFC}$ + Ord is Mahlo
I have a question about the proof on the website and a question of why another proof doesn't just prove that 1 is strictly stronger in consistency than 2. These two questions are related so I am asking them in one post.
The proof on the website states
if $\delta$ is an inaccessible correct cardinal, then since $V_\delta\prec V$ it follows that any class club definable in $V$ with parameters below $\delta$ will be unbounded in $\delta$ and hence contain $\delta$ as an element and consequently contain an inaccessible cardinal.
My first question: I take it that this proof is meant to show that theory 1 implies theory 2, and hence Con(1) implies Con(2). I don't understand why "Ord is Mahlo" is implied. I can only see that this shows that 1 implies "every club $C\subseteq\mathsf{Ord}$ definable from parameters in $V_\delta$ contains an inaccessible". But "Ord is Mahlo" states (schematically) "every club $C\subseteq\mathsf{Ord}$ definable from parameters in $V$ contains an inaccessible". So I don't see how this latter stronger statement is proved in the proof.
My second question: why doesn't the following proof show that Theory 1 proves the consistency of Theory 2? More specifically, in Theory 1, we show that $V_\delta\vDash $ "Ord is Mahlo"
The proof: Suppose $V_\delta\vDash \varphi$ defines a club. Then by elementarity, $V\vDash \varphi$ defines a club (and this club is unbounded below $\delta$). So $V$ thinks that this club contains an inaccessible cardinal. Reflecting this down to $V_\delta$, we see that $V_\delta\vDash $ the club defined by $\varphi$ contains an inaccessible cardinal. Hence $V_\delta\vDash $ "Ord is Mahlo". And so Theory 1 proves the consistency of Theory 2.
What went wrong with this proof? Thank you!
Best Answer
Re: your second question, there's a scheme issue going on.
For each formula $\varphi$, Theory $1$ can prove that $V_\delta$ satisfies the "$\mathsf{Ord}$ is Mahlo" instance corresponding to $\varphi$. But this does not mean, however, that Theory $1$ can prove
So Theory $1$ stops short of proving that $V_\delta$ actually satisfies (all of) "$\mathsf{Ord}$ is Mahlo."
It may be easier to consider a large cardinal free example. Working in the expanded language $\{\in,\delta\}$, let $S$ be the $\{\in\}$-theory $\mathsf{ZFC}$ together with a sentence saying $V_\delta\models\varphi$ for each $\varphi\in\mathsf{ZFC}$. Then $S$ is a conservative extension of $\mathsf{ZFC}$, even though at a glance it may look like $S$ proves $V_\delta\models\mathsf{ZFC}$.
What about your first question?
Well, contrary to a previous edit of this answer - I worked too fast - we do indeed have that Theory $1$ outright implies Theory $2$ (not merely each axiom in the relativization of Theory $2$ to $V_\delta$). This is a cute application of elementarity. The parameter-free version is easy to check. Working in Theory $1$, suppose $\varphi$ defines a club in $V$. By elementarity, $\varphi$ defines a club in $V_\delta$ and so $\delta\in \varphi^V$. So in $V$ we have that $\varphi$ contains an inaccessible, namely $\delta$ itself.
Now on the face of it this breaks down when we fold in parameters: what if $\varphi$ involves a parameter in $V\setminus V_\delta$? Here we work a little more coarsely: rather than reflecting single-sentence instances, we reflect parameterized instances (we reflect the statement "There is no tuple of parameters turning $\varphi$ into a failure of "$\mathsf{Ord}$ is Mahlo"). So in fact Theory $1$ proves Theory $2$.
Reflecting to $V_\delta$ we get that Theory $1$ proves each axiom of Theory $2$ relativized to $V_\delta$ - however, Theory $1$ does not prove "$V_\delta$ satisfies Theory $2$" per the above.