Is the set of $\kappa$-complete ultrafilters closed in $\beta X$?

Question

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.

Answer 1

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 $\bigcap_n\bigcup_{k\ge n}X_k=\emptyset$.

Answer 2

I claim that one can put a new topology on $\beta X$ for discrete space $X$ so that $\lambda X$ is in-fact closed in $\beta X$.

If $\kappa$ is a regular cardinal, then a $P_{\kappa}$-space is a completely regular space such that if $U_{i}$ is open for $i\in I$ and $|I|<\kappa$, then $\bigcap_{i\in I}U_{i}$ is also open.

If $\kappa$ is a regular cardinal and $(X,\mathcal{T})$ is a completely regular space, then define the $P_{\kappa}$-space coreflection $(X)_{\kappa}$ to be the topological space $(X,\mathcal{V})$ where $\mathcal{V}$ is generated by the basis consisting of all sets of the form $\bigcap_{i\in I}U_{i}$ where $|I|<\kappa$ and $U_{i}\in\mathcal{T}$ for each $i\in I$. We observe that if $\mathcal{S}$ is a subbasis for $\mathcal{T}$, then $\mathcal{V}$ is generated by the basis consisting of all intersections of the form $\bigcap_{i\in I}U_{i}$ where $|I|<\kappa$ and $U_{i}\in\mathcal{S}$ for each $i\in I$.

Proposition: If $\kappa$ is an uncountable regular cardinal and $X$ is a discrete space, then the closure of $X$ in $(\beta X)_{\kappa}$ is precisely the collection of all $\kappa$-complete ultrafilters.

Proof: Let $M$ be an ultrafilter on $X$. By Stone duality, the clopen sets $C\subseteq\beta X$ are precisely the sets $C_{R}=\{M\in\beta X\mid R\in M\}$. Therefore, $M$ is not contained in the closure of $X$ in $(\beta X)_{\kappa}$ if and only if $M\in\bigcap_{i\in I}C_{R_{i}}\subseteq X\setminus\beta X$ where $|I|<\kappa$ and $R_{i}\subseteq X$ for $i\in I$. The condition that $M\in\bigcap_{i\in I}C_{R_{i}}$ is equivalent to saying that $R_{i}\in M$ for $i\in I$, and the condition that $\bigcap_{i\in I}C_{R_{i}}\subseteq X\setminus\beta X$ is equivalent to saying that $\bigcap_{i\in I}R_{i}=\emptyset$. Therefore, $M$ is not contained in the closure of $X$ in $(\beta X)_{\kappa}$ if and only if $M$ is not $\kappa$-complete. Q.E.D.