# If $\kappa$ is weakly inaccessible and $A\subset\kappa$, can $L[A]$ violate $\kappa^{\lt\kappa}=\kappa$?

In some current work, my co-authors and I had wanted in a certain argument to appeal to $\kappa^{\lt\kappa}=\kappa$ in $L[A]$, in a situation where $A\subset\kappa$ and $\kappa$ was weakly inaccessible in $V$, but $\kappa$ was below the continuum in $V$ (and so $\kappa^{\lt\kappa}\neq\kappa$ in $V$). But we have lost confidence in this statement. Is it consistent that there is a counterexample?

Question. Is it (relatively) consistent that $\kappa$ is weakly inaccessible, but there is $A\subset\kappa$ with $L[A]\models\kappa^{\lt\kappa}>\kappa$?

The model $L[A]$ here is the relative constructible universe, defined as in the constructible universe, but with a predicate for the set $A$.

We had at first thought a condensation argument might show $\kappa^{\lt\kappa}=\kappa$ in every $L[A]$, but upon looking at it closely, I can't quite make it work. Condensation works fine above $\kappa$, but I don't see why, for example, every real number in $L[A]$ must be added by a stage before $\kappa$. It would be interesting to me even to resolve the question of how big $2^\omega$ must be or can be in comparison with $\kappa$ in $L[A]$, where $A\subset\kappa$ and $\kappa$ is a regular limit cardinal in $V$.

(Meanwhile, we've saved our application by finding another route for our argument that does not rely on this issue.)

Claim. If $A\subseteq \kappa,$ and if $Y\in L[A]$ is a bounded subset of $\kappa,$ then there exists a proper initial segment $A'$ of $A$ such that $Y\in L[A'].$
From this, $\kappa^{<\kappa}=\kappa$ in $L[A].$
• Great! Let me take a look. It seems you use a kind of catch-up club argument that I had tried, but didn't quite push through, getting the collapses of elementary substructures to increasingly align with and become an initial segment of $A$. Mar 16 '15 at 12:19