This is a continuation of the question [covering properties of strongly compact embedding](https://mathoverflow.net/questions/173263/covering-properties-of-strongly-compact-embedding).


Recall that a cardinal $\kappa$ is $\nu$-strongly compact cardinal if there is an elementary embedding $j:V\rightarrow M$, with critical point $\kappa$, $j(\kappa) > \nu$, such that $j^{\prime \prime} \nu \subset s$ where $|s|^M < j(\kappa)$.


**Question 1:** Is it consistent (relative to the existence of large cardinals, of course), that there is a $\nu$-strongly compact cardinal $\kappa$, $\nu > 2^{\kappa}$ regular, such that for some $j:V\rightarrow M$, (a $\nu$-strongly compact embedding elementary embedding with $crit(j)=\kappa$), $j^{\prime\prime} \nu$ cannot be covered by any set from $M$ with order-type $\nu$?


Assuming that the answer to the first question is positive, how far can we get? namely:


**Question 2:** With the same notations, how large can be the gap between the minimal $M$-cardinality of a set in $M$ that covers $j^{\prime\prime} \nu$ and $\nu$? Does there is any non trivial restriction on this $M$-cardinal?



Strongly compact cardinal with bad covering properties
The answer is yes, and indeed the situation is not merely
consistent with strong compactness, but rather every strongly
compact cardinal has such embeddings with no small covers.


**Theorem.** Suppose that $\kappa$ is $\delta^+$-strongly compact,
where $\delta^{\lt\kappa}=\delta$. Then there is a
$\delta^+$-strong compactness embedding $j:V\to M$ such that there
is no $s\in M$ with $j''\delta^+\subset s$ and $|s|^M=\delta^+$.


Proof. Let $j\_0:V\to M\_0$ be any $\delta^+$-strong compactness
embedding. Let $s\_0\in M\_0$ be a cover, so that $j\_0''\delta^+\subset
s\_0$, and we may assume without loss of generality that
$|s\_0|^{M\_0}=\delta^+$, for if there is no such cover then we are
already done. Let $j\_1:V\to M\_1$ be any $\delta$-strong
compactness ultrapower. We may consider $h=j\_1\upharpoonright
M\_0$, so that $h:M\_0\to M$ is an elementary embedding of $M\_0$
into $M=j\_1(M\_0)=\bigcup\_\alpha j\_1(V\_\alpha^{M\_0})$. Note that we
do not necessarily have the measure for $j\_1$ inside $M\_0$, and so
this embedding $h$ is defined in $V$ rather than in $M\_0$. Let
$j=h\circ j\_0:V\to M$ be the composition elementary embedding.
(Although the argument here will not use it, in fact one can prove
that $j:V\to M$ is the ultrapower by the product measure
$\mu\times\nu$, where $j\_0$ is the ultrapower by $\nu$ and $j\_1$
is the ultrapower by $\mu$.)


I claim that $j$ is a $\delta^+$-strong compactness embedding. It
clearly has critical point $\kappa$. Define $s=j\_0(s\_0)$. It
follows easily that $j''\delta^+=h''j\_0''\delta^+\subset
h''s\_0\subset h(s\_0)=s$, and so $s$ covers $j''\delta^+$. And
$|s|^M=h(|s|^{M\_0})=h(\delta^+)$, which is less than
$h(j\_0(\kappa))=j(\kappa)$.


Meanwhile, since $s\_0$ has size $\delta^+$ in $M\_0$, it follows
that $\sup j\_0''\delta^+$ has cofinality $\delta^+$ in $M\_0$. But
now the key point is that $j\_1$ and hence $h$ is continuous at
ordinals of cofinality $\delta^+$. Thus, $h(\sup
j\_0''\delta^+)=\sup h''j\_0''\delta^+=\sup j''\delta^+$. Therefore,
any covering set of $j''\delta^+$ in $M$ must have size at least
the cofinality of this supremum, which is $h(\delta^+)$, which is
the same as $j\_1(\delta^+)$, which is larger than $j\_1(\kappa)$
which is significantly larger than $\delta^+$ in $M$.


So every cover of $j''\delta^+$ in $M$ has size much larger than
$\delta^+$. QED


