5
$\begingroup$

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, 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, 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.

$\endgroup$
  • $\begingroup$ The exercise 6Z is Proposition 17 in Rump "The absolute of a topological space and its application to abelian l-groups", Appl. Categ. Structures 17 (2009), 153–174. $\endgroup$ – Adam Przeździecki Aug 19 '14 at 18:44
  • 2
    $\begingroup$ There is also a related result: A. Błaszczyk, A construction of a Gleason spaces, Commentationes Mathematicae Universitatis Carolinae 24 (1983), 233-236. It gives a description of the projective cover as the Stone-Čech compactification of the strongest topology subject to the condition that the identity map is irreducible. $\endgroup$ – Tomek Kania Aug 25 '14 at 18:49
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.