Is the following statement true, and if it is, does someone have a reference?



> 
> Let $X$ be a compact (i.e., compact and Hausdorff) topological space. Then the Gleason space (=Iliadis absolute, =Stone dual of the Boolean algebra of regular open sets) of $X$ is (naturally homeomorphic to) the projective limit of $\beta V$ for $V$ ranging over dense open subsets of $X$, where $\beta$ is the Stone-Čech compactification functor (and, of course, the maps $\beta V'\to \beta V$ are obtained from the inclusion $V'\to V$ by functoriality).
> 
> 
> 


Since every $V$ and also every $\beta V$ has the same regular open algebra, hence the same Gleason space $E(X)$, there is at least a continuous map from $E(X)$ to the projective limit in question, and I am asking whether it is a homeomorphism commuting with the obvious maps from both of these spaces to $X$. I don't suppose it should be very difficult, but I'd prefer a reference.


I looked in Porter & Woods's book *Extensions and Absolutes of Hausdorff Spaces* with no success. There is, however, a related exercise 6Z (p.521) which describes $E(X)$ as as different projective limit (whose morphisms are finite-to-one; essentially by "splitting" finite systems of regular closed sets).


The projective limit of the $\beta V$ for $V$ open dense occurs in [Fine, Gillman & Lambek's *Rings of Quotients of Rings of Functions*](https://archive.org/details/flooved3383), but the words "absolute" or "Gleason" don't seem to appear anywhere.


**Edit:** It turns out that I didn't look carefully enough in Fine, Gillman & Lambek's book: the statement does appear there (even if the term "absolute" is not mentioned), by combining theorem 6.9 with theorem 11.15. I much prefer the proof that Adam Przeździecki gives [below](https://mathoverflow.net/questions/178806/is-the-absolute-of-a-compact-space-the-projective-limit-of-the-stone-%C4%8Cech-compac/179375#179375), however, because it is "direct" (describing the arrow on the spaces themselves) rather than "dual" (through maximal ideals and rings of functions), so it is more enlightening.



Is the absolute of a compact space the projective limit of the Stone-Čech compactifications of its open dense subsets?
Yes, your space, let's call it $GX$, is the projective cover $EX$ of $X$ in the category of compact spaces. Recall that the projective cover $p:EX\to X$ can be characterized by the properties:


* $EX$ is projective, that is every epimorphism $f:Y\to EX$ has a right inverse.
* $p$ is irreducible, that is for every closed subset $S\subsetneq EX$ we have $p(S)\neq X$.


In the category of compact spaces this is the same as the *absolute* you mentioned. I will compare your space to the construction by Rump ("The absolute of a topological space and its application to abelian l-groups", Appl. Categ. Structures 17 (2009), 153–174), Proposition 17:


Let $\mathcal{B}X$ be the set of regular subsets of $X$ ($U$ is regular if $U={\rm int}\bar U$. Let $\Phi X$ be the family of finite subsets $A\subseteq\mathcal{B}X$ such that: 


* $A$ is pairwise disjoint.
* $\bigcup A$ is dense in $X$.


Let $\bar A=\coprod\{\bar U\mid U\in A\}$. The set $\Phi X$ is ordered by inclusions. Rump proves that the map


$$p:\lim\_{A\in\Phi X}\bar A\longrightarrow X $$


is the projective cover of $X$. The map is the projection from the limit onto the axis $A=\{X\}$ - the terminal element of $\Phi X$.


We have obvious epimorphisms $\beta\bigcup A\to\bar A$ however, the indexing sets of this and your limits are not yet compatible. Let $\Phi\_0X=\{\bigcup A\mid A\in\Phi X\}$ - this is a set of certain dense subsets of $X$. The terminal member $\{X\}$ of $\Phi X$ belongs to $\Phi\_0X$. We have an epimorphism $\pi:\Phi X\longrightarrow\Phi\_0 X$ given by $\pi(A)=\bigcup A$. If $A,B\in\Phi X$ are such that $\bigcup A=\bigcup B$ then we define $C=\{U\cap V\mid U\in A, V\in B, U\cap V\neq\emptyset\}$. Obviously $\bigcup A=\bigcup C=\bigcup B$ and we have natural homeomorphisms $\bar A\cong \bar C\cong\bar B$ so that $\pi$ induces homeomorphism of the limits:


$$
 \pi^\*:\lim\_{A\in \Phi\_0X}\bar A\stackrel{\cong}{\longrightarrow}\lim\_{A\in \Phi X}\bar A
$$


Let $\mathcal{V}X$ be the set of dense open subsets of $X$, as in you question. Since $\Phi\_0X\subseteq\mathcal{V}X$, we can now assemble our epimorphisms into a single epimorphism of the limits


$$
 f:GX=\lim\_{V\in \mathcal{V}X}\beta V{\longrightarrow}\lim\_{A\in \Phi\_0 X}\beta A{\longrightarrow}\lim\_{A\in \Phi\_0 X}\bar A=EX
$$


The left arrow is the projection induced by the inclusion. The right arrow is induced by the epimorphisms $\beta A\to\bar A$ where $\bar A$ can be defined as $\bar A\_0$ where $A\_0$ is any element of $\Phi X$ such that $\pi(A\_0)=A$.


It remains to show that $f$ is one-to-one. Suppose that $x\neq y$ in $GX$. Then there exists $V\in\mathcal{V}X$ such that their projections on $\beta V$ are different: $x\_V\neq y\_V$. Then there exist disjoint regular open subsets of $\beta V$, namely $W\_x\ni x\_V$ and $W\_y\ni y\_V$, we may assume that $W\_x\cup W\_y$ is dense in $\beta V$. Then $V\_x=W\_x\cap V$ and $V\_y=W\_y\cap V$ are regular open disjoint subsets of $V$. Then $U\_x={\rm int}\bar V\_x$ and $U\_y={\rm int}\bar V\_y$ are disjoint regular open subsets of $X$ and $U\_x\cup U\_y$ is dense in $X$ hence it belongs to $\Phi\_0X$.


The inclusions
$$
 V\supseteq V\_x\cup V\_y \subseteq U\_x\cup U\_y
$$
induce epimorphisms
$$
 \beta V \stackrel{a}{\longleftarrow}\beta (V\_x\cup V\_y)\stackrel{b}{\longrightarrow}\bar U\_x\sqcup \bar U\_y
$$
We have $a^{-1}(x)\subseteq\beta V\_x$ and then $b(a^{-1}(x))\subseteq\bar U\_x$. Similarly for $y$ hence $f(x)$ and $f(y)$ are separated on the $U\_x\cup U\_y$ axis of $\lim\_{A\in \Phi\_0 X}\bar A$ hence the injectivity of $f$.


Well, it took much longer than I expected. Let me mention that we may have a much smaller limit: $EX\cong\lim\_{A\in\Phi X}A$ where $A$'s are viewed as finite sets.


