Indestructibility and destructible measurable cardinals

Abstract

Say that $${\kappa}$$?'s measurability is destructible if there exists a $${\kappa}$$?. It then follows that $${A_{1} = \{\delta < \kappa \mid \delta}$$A1={?<??? is measurable, ? is not a limit of measurable cardinals, ? is not ?+ strongly compact, and ?'s measurability is destructible when forcing with partial orderings having rank below ??} is unbounded in $${\kappa}$$?. On the other hand, under the same hypotheses, $${A_{2} = \{\delta < \kappa \mid \delta}$$A2={?<??? is measurable, ? is not a limit of measurable cardinals, ? is not ?+ strongly compact, and ??s measurability is indestructible when forcing with either Add(?, 1) or Add(?, ?+)} is unbounded in $${\kappa}$$? as well. The large cardinal hypothesis on ? is necessary, as we further demonstrate by constructing via forcing two distinct models in which either $${A_{1} = \emptyset}$$A1=? or $${A_{2} = \emptyset}$$A2=?. In each of these models, both of which have restricted large cardinal structures above $${\kappa}$$?, every measurable cardinal ? which is not a limit of measurable cardinals is ?+ strongly compact, and there is an indestructibly supercompact cardinal $${\kappa}$$?. In the model in which $${A_{1} = \emptyset}$$A1=?, every measurable cardinal ? which is not a limit of measurable cardinals is <?? strongly compact and has its <?? strong compactness (and hence also its measurability) indestructible when forcing with ?-directed closed partial orderings having rank below ??. The choice of the least beth fixed point above ? is arbitrary, and other values of ?? are also possible.