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Question: Can we have a set theory in which there exists a $\kappa$-Suslin tree with $\kappa$ larger than the least measurable cardinal?

A $\kappa$-Suslin tree is a tree with levels indexed by $\kappa$, the cardinality of each level is less than $\kappa$, and all chains and antichains have cardinality less than $\kappa$. Essentially, the only thing I could find is the Jensen theorem that if $V=L$ then there exists a $\kappa$-Suslin tree for every infinite successor cardinal $\kappa$. But $V=L$ excludes measurable cardinals.

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2 Answers 2

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Yes, one can add Suslin trees above a measurable cardinal, by forcing that adds no subsets to the measurable cardinal and therefore preserves its measurability.

For example, if $\delta$ is measurable and $\kappa=\delta^+$, then one can add a $\kappa$-Suslin tree by forcing with the conditions consisting of ${\lt}\delta$-closed normal $\beta$-trees for $\beta<\kappa$, ordered by end-extension. This adds a $\kappa$-Suslin tree for similar reasons as in the usual Suslin tree forcing on $\omega_1$, and since the forcing is $\leq\delta$-closed, it preserves the measurability of $\delta$.

Similar arguments work more generally to add Suslin trees at other cardinals above $\delta$.

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To complement Joel's answer and tie into the issue of $L$, consider a model $L[U]$, where $U$ is a normal measure on a cardinal $\kappa$. It is well-known that $L[U]$ satisfies GCH and $\diamondsuit_\mu(S)$ for every regular $\mu$ and every stationary $S \subseteq \mu$. These principles suffice to construct $\mu$-Suslin trees whenever $\mu$ is a successor of a regular cardinal.

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