The exam is here.

Solutions follow.

Andrés Caicedo

While at Luminy, David Asperó showed me a quick proof of a nice result on Reinhardt cardinals in . It complements Grigor Sargsyan’s result discussed here.

Theorem(Asperó).Work in . Suppose is a nontrivial elementary embedding. Then there are a and an ordinal such that for all there is a and an elementary

such that and .

**Proof.** For an ordinal, set

such that and .

Note that suitable fragments of witness that is defined for all . Moreover, implies that , and therefore there is a such that for all sufficiently large. Moreover, since it is definable, we actually have .

Let be least with for . We claim that and are as wanted. For this, consider some , and pick witnessing that . All we need to do is to check that .

But note that if , then Hence, if , we have

.

But . Contradiction.

Quiz 8 is here. Please remember that the second midterm is this Wednesday.

Solutions follow.