Suppose $V$ is the universe and $\kappa$ is supercompact, let $\beta>\kappa$. Then there are $M$ and $j$ such that $[M]^\beta\subseteq M$ and $j:V\prec M$ with $crit(j)=\kappa$ and $j(\kappa)>\beta$. ($j$ elementary. $[M]^\beta$ is ‘the collection of all $\beta$-sequences of elements of $M$’.)
In this situation it is apparently true that $M_\beta =V_\beta$, but I am not able to see why. I imagine it is pretty simple, but please take me out of my misery.
This is the lemma:
Screenshot from: New proofs of the consistency of the normal Moore space conjecture I.
$M^\beta\subseteq M$ does not imply $M_\beta=V_\beta$.
However, for a given $\beta$ you can find $\lambda$ large enough so that $M^\lambda\subseteq M$ implies $M_\beta = V_\beta.$
(And this is enough for the present purpose, which is to find for a fixed $\beta$ a supercompact embedding and a class $M$ with $j:V\rightarrow M$ and $V_\beta=M_\beta$ and $M^\beta\subseteq M.$)
For fixed $\beta$, let $$\lambda=\bigcup\{|x|\mid x\in V_\beta\}.$$ Suppose $M^\lambda\subseteq M$ and suppose by contradiction that $V_\beta\setminus M_\beta$ is non-empty and let $x\in V_\beta\setminus M_\beta$ be $\in$-minimal.
By choice of $x$ and definition of $\lambda$ we have $x\subseteq M_\beta$ and $|x|\leq\lambda.$ As $M^\lambda\subseteq M$, this implies $x\in M$. But rank$(x)<\beta$ gives $x\in M_\beta$ which contradicts choice of $x$.
This is based on Andrés Caicedo's comment but there were a couple of details to fill in.