Skip to main content
Stefan Mesken's user avatar
Stefan Mesken's user avatar
Stefan Mesken's user avatar
Stefan Mesken
  • Member for 10 years, 4 months
  • Last seen more than 4 years ago
comment
What is a 'power admissible model'?
In this paper, Mathias also includes the axiom of infinity and the powerset axiom in $\operatorname{KP}^{\mathcal{P}}$. Infinity is fine with me, as it will always hold in the structures I am interested in. Also, since all my models are transitive, I have full foundation for free. However, the powerset axiom typically fails. Including the powerset axiom in $\operatorname{KP}^{\mathcal{P}}$ seems curious to me, as - on first sight - it appears to render the additional predicate $\mathcal{P}$ rather pointless... I'll think about it for a little while.
asked
Loading…
awarded
awarded
comment
Defining cones and Turing cones
@Asaf Yes, I'm aware. However, since $x \in \mathcal{N}$ is not a set of ordinals in Ralf's book, this seemed more relevant to OP. The underlying reason though, that $L[x] = L(x)$, is - as you hinted - that it can be recursively coded as a subset of $\omega$.
revised
Defining cones and Turing cones
improved formatting
Loading…
revised
Defining cones and Turing cones
fixed a slight mistake
Loading…
suggested
Approve
answered
Loading…
comment
Defining cones and Turing cones
Also note that $L[x]$ is not the least inner model containing $x$ - the least inner model containing $x$ is $L(x)$. $L[x]$, in contrast, is the least inner model closed under $y \mapsto y \cap x$. However, for $x \in \mathcal{N}$, these notions agree, i.e. $L[x]= L(x)$.
awarded
revised
Loading…
awarded
awarded
awarded
Loading…
Loading…
comment
Mitchell, Steel. FSIT. Lemma 2.8: Is $k$-solidity actually needed?
Thanks @Dan. I suspected as much, spelled out the entire proof earlier today and figured out the same. Unless you want to post your proof as an answer, I'll answer my own question later today.
1
4 5
6
7 8