I was looking at some exercises in Ralf Schindlers book and I came to an equivalent of weak compactness which to me it seems like it is too weak to be true.
So as it stands, the exercise in the book is this:(Exercise $4.23$)
Let $\kappa$ be a cardinal. TFAE:
$(a)$ $\kappa$ is weakly compact.
$(b)$ If $X \subseteq \mathcal{P}(\kappa)$, $|X| \le \kappa$, then there are transitive models $H$ and $H^*$ with $X \subseteq H$, $|H| = \kappa$, $^{\lt \kappa} H \subseteq H$ and $^{\lt \kappa} H^* \subseteq H^*$ and there is some elementary embedding $\sigma:H\rightarrow H^*$ with $\operatorname{crit}(\sigma) = \kappa$.
So proving $(a) \rightarrow (b)$ wasn't that hard, I took a good elementary submodel of $H_{\kappa^+}$ and I had to use the filter property to get a good ultrapower. But on the other hand $(b) \rightarrow (a)$ is really non-trivial. The only things I managed to prove was that $\kappa$ is regular and that for all $\lambda \lt \kappa$, $2^\lambda \le \kappa$, which are immediate. I couldn't even prove that $\kappa$ is a strong limit.
And since it seemed sketchy to me that we didn't even use the $X$, we include in $H$, in the second formulation, I started to search for this. And I came about a statement in wikipedia which it is said that is equivalent to weak compactness and is really similar to this one, but seems much much stronger. It is a formulation due to K. Hauser.
Let $\kappa$ be a cardinal. TFAE:
$(a)$ $\kappa$ is weakly compact.
$(b)$ $\kappa$ is inaccessible and for every transitive set $M$ of cardinality $\kappa$ with $\kappa \in M$, $^{\lt\kappa}M \subseteq M$, and satisfying a sufficiently large fragment of ZFC, there is an elementary embedding $j$ from $M$ to a transitive set $N$ of cardinality $\kappa$ such that $^{\lt\kappa}N \subseteq N$, with critical point $\operatorname{crit}(j) = \kappa$.
So having the above characterization from Hauser in mind, is the above exercise correct? If so, how would one prove it?
EDIT I:
Surprisingly, I think I have an answer, if we additionally assume the $H$ in $(b)$ to be closed under intersections of ${\lt\kappa}$-sequences. So to see $(b) \rightarrow (a)$ in this setting, we prove that $^{\lt\kappa}\kappa = \kappa$ and that the filter property holds, which gives us the weak compactness of $\kappa$.
So first let $X = \emptyset$, and since we get an $H$ as in $(b)$ with $\kappa \subseteq H$ and $^{\lt\kappa}H \subseteq H$ and $|H| = \kappa$, we already have $^{\lt\kappa}\kappa = \kappa$.
Now to verify the filter property, assume that $X$ is a collection of subsets of $\kappa$ of size at most $\kappa$. We wish to find a $\kappa$-complete non-principal filter $F$ that measures all of $X$. As in $(b)$, we have some $H$, $H^*$ and $\sigma$ for this $X$.
So for any $A \subseteq \kappa$ let:
$$A \in F \hspace{10px}\text{ if and only if }\hspace{10px} (\exists B \in \mathcal{P}(\kappa) \cap H)\hspace{5px}B \subseteq A \hspace{5px} \land \hspace{5px} \kappa \in \sigma(B).$$
We can easily see that $F$ is a non-principal filter measuring all of $X$, since $X \subseteq \mathcal{P}(\kappa) \cap H$. To see that it is $\kappa$-complete, we can use the closure conditions on these $H$ and $H^*$'s and usual arguments for getting ultrafilters from an embedding, and here is where we use the intersection hypothesis.
So $\kappa$ is weakly compact.
Can we get rid of this additional hypothesis?