# Convergence of ultrafilters determines a topology - union of opens is open

I am trying to prove that an algebra of the ultrafilter monad $\mathrm{lim} : \beta X \to X$ is equivalent to a (compact hausdorff) topological space. I'm stuck at proving that it is a topological space in the first place - that a union of opens (defined by $U$ is open $:= \forall F \in \beta X, \, U \in F \implies \mathrm{lim} F \in U$) is open.

Suppose that $U_i$ is open for every $i$ in some arbitrary index set $I$. Then from $\cup_{i\in I} U_i \in F$, if I were to prove some $U_i \in F$, then by openness of $U_i$ we get $\mathrm{lim} F \in U_i \subset \cup_{i\in I} U_i$, proving the union is open.

The problem is that I can't find a way to get some $U_i \in F$. The contravariance of implication makes this definition of opens seem strange. I used the upwards closed hypothesis to prove that the intersection of two opens is open, when it seems like I should be using the finite intersection property to prove that, and upwards closure for this. Since ultrafilters are quite unconstructive, I tried assuming that each $U_i \not \in F$, but it seems like I need arbitrary intersection to prove that $\varnothing \in F$, deriving a contradiction.

Direction towards proving this, or corrections to an error I've made is preferred over a complete proof.

• That seems more likely to be the definition for "$U$ is closed". It looks straightforward to prove that the class of $U$ satisfying this condition is closed under arbitrary intersections, and under finite unions. – Daniel Schepler Dec 13 '17 at 20:20
• Hmm, yes, this seems to read "U contains all its limit points". I got this definition off nlab and it seems suspiciously mangled. Perhaps it should read "if $x \in U$ and $lim \, F = x$, then $U \in F$"? – fhyve Dec 13 '17 at 20:29
• I think that would be equivalent to the original condition on $U^c$: if $F$ is an ultrafilter, then $U^c \in F$ is equivalent to $U \notin F$, and $\lim F \in U^c$ is equivalent to $\lim F \notin U$, so this would be the contrapositive of $\lim F \in U \Rightarrow U \in F$. – Daniel Schepler Dec 13 '17 at 20:32
• I've edited the nlab page ncatlab.org/nlab/show/ultrafilter – fhyve Dec 13 '17 at 20:35

So to copy the definition in the nlab page, we have a notion of ultrafilter convergence $\xi$ that assigns to each ultrafilter $\mathcal{F}$ on $X$ a limit $\xi(\mathcal{F}) \in X$, and we define $U$ to be open iff for all $x \in U$, and all ultrafilters $\mathcal{F}$ with $\xi(\mathcal{F}) = x$ we have that $U \in \mathcal{F}$, and you want to verify the topology axioms, in particular the union axiom. So let $x \in U=\cup_i U_i$ where all $U_i$ are open, and let $\mathcal{F}$ be an ultrafilter converging to $x$, i.e. $\xi(\mathcal{F}) = x$. For some $j \in I$ we have $x \in U_j$ and by openness of $U_j$ we get that $U_j \in \mathcal{F}$; as $U_j \subseteq U$, we have $U \in \mathcal{F}$ as required.
Your definition for $U$ being open as $\forall \mathcal{F} \in \beta X$: $U \in \mathcal{F}$ implies $\xi(\mathcal{F}) \in U$ does not quite work, in this setting. The nlab one I used above is the standard way of defining a topology from a convergence.
To see intersection of two opens $U \cap V$: pick $x \in U \cap V$ and $\mathcal{F} \in \beta X$ with $\xi(\mathcal{F}) = x$. Then $x \in U$ and $U$ open gives $U \in \mathcal{F}$, and likewise $V \in \mathcal{F}$ and then $U \cap V \in\mathcal{F}$ is clear from the filter axioms.
The standard definition of $C \subseteq X$ being closed in a convergence is what you defined as open: for any $\mathcal{F} \in \beta X$, if $C \in \mathcal{F}$ then $\xi(\mathcal{F}) \in C$.
• Do note that I just edited the nlab page (see above comments). It used to read $\forall F,\, x$, if $U \in F$ and $x = \xi (F)$, then $x \in U$, which is equivalent to my definition. With the correct definition, verifying that this definition gives a topology is simple. – fhyve Dec 13 '17 at 23:03