Does Sigma(n)-KP, i.e. Kripke-Platek set theory with Sigma(n)-collection and Sigma(n)-separation have Sigma(n)-replacement.
$\begingroup$
$\endgroup$
1
asked Jul 19, 2013 at 16:18
-
$\begingroup$ It helps if you also give some references and context to your question, I am not an expert but would be happy to understand the problem. $\endgroup$– András BátkaiCommented Jul 19, 2013 at 16:44
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
Yes. If we have a set $A$ and for every $a\in A$ there is a unique $b$ such that $\varphi(a,b,z)$, where $\varphi$ has complexity $\Sigma_n$, then by $\Sigma_n$ collecton there is a set $B$ such that every $a\in A$ has a $b\in B$ with $\varphi(a,b,z)$. But perhaps there is extra stuff in $B$, so this set may be too large to verify the desired instance of $\Sigma_n$-replacement. But we may apply $\Sigma_n$-separation to form the set $\{b\in B\mid \exists a\in A\ \varphi(a,b,z)\}$, which does fulfill the desired instance of $\Sigma_n$-replacement.
answered Jul 19, 2013 at 16:25
-
1$\begingroup$ Thanks. That was easy and I should have seen it. Sometimes there are curtains before the windows. :) $\endgroup$ Commented Jul 19, 2013 at 16:57
-
$\begingroup$ Interestingly, to me, then Sigma(n)-KP is stronger than Sigma(n)-ZF minus power set as power set is needed to get Sigma(n)-collection from Sigma(n)-replacement. $\endgroup$ Commented Jul 19, 2013 at 17:34