Given an arbitrary set $X$, let $\beta X$ be the set of all ultrafilters over $X$. Consider endowing $\beta X$ with a topology consisting of the following open sets:
$$
\{\mathcal{U} \in \beta X : A \in \mathcal{U}\}
$$
where $A$ ranges over subsets of $X$. Now let $\kappa < |X|$ be an infinite cardinal, and assume there exists a $\kappa$-complete ultrafilter over $X$. Let
$$
\lambda X := \{\mathcal{U} \in \beta X : \mathcal{U} \text{ is $\kappa$-complete}\}
$$
My question is (working in $\mathsf{ZFC}$, if that matters):



> 
> Is $\lambda X$ a closed subset of $\beta X$ under the topology above?
> 
> 
> 


If this question is false for general $\kappa$, I would also like to know what are the assumptions I should impose for this assertion to be true. I welcome any large cardinal axioms, if necessary.



Is the set of $\kappa$-complete ultrafilters closed in $\beta X$?
If $\kappa=\aleph\_0$ then yes: every ultrafilter is $\aleph\_0$-complete.


If $\kappa>\aleph\_0$ then no, if $\lambda X$ is nonempty. Split $X$ into countably many sets $\{X\_n:n\in\mathbb{N}\}$, of the same cardinality as $X$ itself.
Take a $\kappa$-complete $u\_n$ on $X\_n$ for each $n$.
For every free ultrafilter $p$ on $\mathbb{N}$ the $p$-limit $u\_p$ of the sequence $\langle u\_n\rangle\_n$ is in the closure of $\lambda X$, but not in $\lambda X$, because $\bigcup\_{k\ge n}X\_k$ is in $u\_p$ for all $n$, yet $\bigcap\_n\bigcup\_{k\ge n}X\_k=\emptyset$.


