Are there elementary substructures of $V_{\lambda+1}$ with the following property?

72 Views Asked by At

Straight to the point with this one. Is there a transitive class $M$ such that:

$$\exists A\in \mathcal{P}(M)\setminus M\exists S\in M(\langle S;\in,A\rangle\prec\langle\mathcal{P}(M);\in, A\rangle)$$

The existence of these guarantees the consistency of the following seemingly strong reflection principle:

There is a proper class $W$ and a set $S$ such that $\phi^S(W)$ iff $\phi(W)$ for any first-order formula $\phi$. This guarantees that $S$ strongly approximates $W$. This is why I doubt such $M$ could exist. But are they consistent nonetheless?*

*The original question specified $M$ to a $V_\lambda$. Noah Schweber showed this was false.

1

There are 1 best solutions below

3
On

No, this is not consistent. Even ignoring $A$, we can never have an $S\in V_\lambda$ such that $(S,\in)\prec (V_{\lambda+1},\in)$. This is because $\lambda$ is a definable element of $V_{\lambda+1}$ (it's the largest ordinal) so any elementary substructure of $V_{\lambda+1}$ has $\lambda$ as an element, but if $S\in V_\lambda$ then $\lambda\not\in S$.


EDIT: addressing your new question (this is really just expanding on Andres' comment below), the answer is still no:

  • $M$ is a definable element of $(P(M); \in)$. This is because $M$ is the unique element of $P(M)$ such that every other element of $P(M)$ is a subset of $M$.

  • This means that if $(S;\in)\preccurlyeq(P(M);\in)$, then $M\in S$.

  • But then we can't have $S\in M$.

Basically, the problem is that structures like $V_{\lambda+1}$ or $P(M)$ have a "top level," and any elementary substructure must contain elements (often a canonical element, like $\lambda$ or $M$ respectively) of that level.