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A cardinal $\kappa$ is $\Sigma_n$-correct iff $V_\kappa \prec_n V$. For n>1, how to force a $\Sigma_{n+1}$-correct cardinal to be $\Sigma_{n}$-correct but not $\Sigma_{n+1}$-correct?

For $n=1$, we can force GCH below $\kappa$ and then violate GCH at $\kappa$.

If we assume some large cardinals, there are more partial answers, but the general situation is not clear to me.

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This is a very interesting question!

My student Erin Carmody (PhD 2015) had asked this question in connection with her dissertation, Force to change large cardinal strength, which contains many such killing-them-softly results. Her theorem 20 is the $\Sigma_2$-reflecting cardinal case that you mention. I recall that at the time we looked into the general $\Sigma_n$-reflecting case, but it was left open in her dissertation.

How nice finally to have an answer to this.

Theorem. If $\kappa$ is $\Sigma_n$-correct, then there is a class forcing extension preserving this in which $\kappa$ is not $\Sigma_{n+1}$-correct.

Proof. Suppose that $\kappa$ is $\Sigma_n$ correct, where $n>1$. By forcing, if necessary, let me assume that V=HOD holds and more specifically that there is a $\Delta_2$-definable well-ordering of the universe. Indeed, let us arrange this specifically by forcing to code every set into the GCH pattern on a certain definable sequence of coding points. This forcing preserves correctness.

Consider the forcing notion $\text{Add}(\kappa,1)$, which is definable in $V_\kappa$, and consider the dense sets for this forcing that are $\Sigma_n$-definable in $V_\kappa$. I claim that there is a subset $s\subset\kappa$ that is $\Sigma_n$ definable in $V_\kappa$ and generic over $V_\kappa$ with respect to $\Sigma_n$-definable dense classes for this forcing, meaning that the initial segments of $s$ meet every $\Sigma_n$-definable dense subset of this forcing.

If $\kappa$ is inaccessible, then this is easier to see, since in this case the forcing is $<\kappa$-closed and one can simply meet the dense sets one by one, using a definable enumeration of them. A careful version of this argument works even when $\kappa$ is singular, using the fact that $\kappa$ is $\Sigma_n$-correct. One simply meets the dense sets one-by-one, and at limits, the condition produced must be bounded below $\kappa$, since otherwise one would have a $\Sigma_n$-definable singularization of $\kappa$, which would violate correctness.

Now let us force with $\text{Add}(\newcommand\Ord{\text{Ord}}\Ord,1)$ to add a generic class $S\subset\Ord$ extending $s$. (We could also have used a definable $\Sigma_{n+1}$-generic class, without forcing.) In $V[S]$, let $\mathbb{P}$ be the class forcing Easton support $\Ord$-iteration that forces to code $S$ into the GCH pattern at another definable sequence of ordinals that does not interfere with coding used above.

Let $V[S][G]$ be the final extension. I claim that $\kappa$ remains $\Sigma_n$-correct, since if a $\Sigma_n$-statement $\varphi(a)$ is true in $V[S][G]$ about some parameter $a\in V_\kappa[s][G_\kappa]$, then this is forced over $V[S][G]$ by some condition, and by the correctness of $\kappa$ and the $\Sigma_n$-genericity of $s$, it follows that it holds in $V_\kappa[s][G_\kappa]$ as desired. Basically, we are lifting the relation $V_\kappa\prec_{\Sigma_n} V$ to $V[s][G_\kappa]\prec_{\Sigma_n}V[S][G]$.

But I claim that $\kappa$ is not $\Sigma_{n+1}$-correct in $V[S][G]$. Note that $V$ is $\Sigma_2$-definable in $V[S][G]$ using the first sequence of coding points, and $V[S][G]$ thinks that $S$, which is $\Sigma_2$-definable, is not $\Sigma_n$-definable in $V$, since in fact it was generic over $V$. But $V_\kappa[s][G_\kappa]$ thinks that $s$ is $\Sigma_n$-definable in $V_\kappa$. This violates $\Sigma_{n+1}$-correctness. $\Box$

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  • $\begingroup$ Since this argument is a bit finicky, I wonder whether one might try something simpler. For example, suppose we just code the $\Sigma_2$-correct cardinals into the GCH pattern? Does this make $\Sigma_{n+1}$-correctness into mere $\Sigma_n$ correctness for all larger $n$? My idea is to drop down the complexity from the bottom, rather than from the top. $\endgroup$ Commented Mar 4, 2018 at 14:36

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