# Partial orders arising from $l$-spaces

Let $X$ be a $l$-space, i.e. a locally compact totally disconnected hausdorff space, which is not compact. Then $P = \{K : K \subseteq X \text{ compact-open}\}$ is a basis for the topology. Regard $P$ as a partial order with respect to "$\subseteq$".

Question: Which partial orders are isomorphic to partial orders which arise from $l$-spaces as above? Note that they have finite infima and suprema and a smallest element, but not a maximal element. But I doubt that this is already the whole characterization.

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I don't understand the definition of $P$: what is $U$? –  Mariano Suárez-Alvarez Oct 28 '10 at 17:18
One idea might be: Call a decreasing sequence $K_0\supset K_1\supset\ldots$ less or equal to another decreasing sequence $L_0\supset L_1\supset \ldots$ , iff $\forall n\exists i(n) : L_{i(n)}\subset K_n$. Call two sequences equivalent, iff the are less or equal to each other. Then the equivalence classes are ordered and I hope, that the minimal classes correspond to the points of $X$. –  HenrikRüping Oct 28 '10 at 19:49
to make this work one hass of course to remove the smallest element, i.e. the emptyset. –  HenrikRüping Oct 28 '10 at 20:09
The lattice of compact-open sets is also distributive. –  François G. Dorais Oct 28 '10 at 21:55
You should change "which is not compact" to "which is not necessarily compact", just to emphasize you don't assume compactness. The condition "which is not compact" just makes your question ugly. –  Andrej Bauer Nov 3 '10 at 16:09

Note that, a $l$-space has a basis consisting of compact, open sets. Especially any two points can be separated by compact open sets.

Consider the functor $P$ from the category of $l$-spaces (with open maps as morphisms) to the category of partially ordered sets with order preserving maps as morphisms, defined by

$X\mapsto P(X):=\{K\subseteq X| K$ compact, open, non-empty$\}$.

I want to construct a assignment in the other direction. So let $P$ be any partially ordered set. Consider the set $A(P)$ of all subsets, with the property, that greatest lower bounds (of two elements) always exist. Thus the greatest lower bound for any finite subset $F$ exists and is denoted by $glb(F)$.

Define a order on $A(P)$ via $S_1\le S_2\Leftrightarrow \forall K\in S_2 \exists K'\in S_1:K'\le K$. This ordering is not antisymmetric. So one passes to equivalence classes under $S_1\sim S_2\Leftrightarrow S_1\le S_2\wedge S_2\le S_1$. Let $B(P):=A(P)/\sim$, and let $M(P)$ denote the set of all minimal elements in $B(P)$. There is a inclusion $i:P\rightarrow B(P)\qquad p\mapsto \{p\}$. Choose $\{\{x\in M(X)|x\le i(p)\}|p\in P\}$ as a basis for a topology on $M(P)$. I don't know how one could turn $M(-)$ into a functor (morphisms cause problems).

Lemma: For a $l$-space $X$, we have $M(P(X))\cong X$.

Proof: Given any $M$ in $M(P(X))$, consider its intersection $\bigcap M$. I want to show, that this intersection is a one point set. Assume it is empty. Choose any $m\in M$. Then $\{m-m'|m'\in M\}$ is a open covering of $m$. By compactness it has a finite subcovering $\{m-m_i| m_i\in M,i=1,..,n \}$. But we get, that $glb_M(\{m,m_1,\ldots,m_n\})\subset m_i$ for all $i=1,..,n$ and $glb_M(\{m,m_1,\ldots,m_n\})\subset m$. Thus $glb_M(\{m,m_1,\ldots,m_n\})\subset m\cap \bigcap_{1\le i\le n}m_i$. But the fact, that the complements of $m_i$ in $m$ cover $m$ implies, that this intersection is empty. Contradiction! So $\bigcap M$ can't be empty.

Assume $\{x_1,x_2\}\subset \bigcap M$ (with $x_1\neq x_2$). Choose a compact, open neighbourhood $U$ of $x_1$, that doesn't contain $x_2$. Then the partially ordered set $\{U\cap m|m\in M\}$ is really smaller than $M$, which contradicts the minimality.

So we get a map $f:M(P(X))\rightarrow X \qquad M\mapsto x$, with $\bigcap M=\{x\}$. It is surjective. A preimage of $x$ is given by $\{U\subset X|U$compact, open, $x\in U\}$. As any two points can be separated by compact, open sets, the intersection of this System is $\{x\}$. Greatest lower bounds exist in this system; they are given by intersection (which can't be empty, as it contains $x$).

Injectivity is a bit more tricky. We have to show: Given any two subsets $A_1,A_2\in A(P)$ with the property $\bigcap A_1=\{x\}=\bigcap A_2$, then there is for each $K_2\in S_2$ a $K_1\in S_1$ with $K_1\le K_2$. Assume, there is no such $K_1$. Then $\{K-K_2|K\in S_1\}$ is a partially ordered system of nonempty compact, open sets allowing finite greatest lower bounds whose intersection is $\bigcap S_1 \setminus K_2=\{x\}\setminus K_2=\emptyset$, which cannot exist (see above). Thus there is a $K_1\in S_1$ with $K_1\subseteq K_2$ and hence the map is injective.

We have to show that the map and its inverse are continuous. Thatfor we have to verify that a basis for the topology gets mapped to a basis. The family of all (nonempty) compact, open sets forms a basis for the topology of a $l$-space. For any compact, open set C, we get $f(\{x\in M(X)|x\le i(C)\})=C$, which gives continuity in both directions. $\square$.

To decide, whether any given partial ordered set $Q$ arises as $P(X)$, we just have to decide, whether $P(M(Q))\cong Q$. ($M(Q)$ is the only candidate). I want to give a list of properties, that every partial order of the form $P(X)$ (for a $l$-space $X$) has and that imply, that the given partial order can be obtained that way:

1) For any finite set $F\subset Q$ the set of all common upper bounds has a smallest element, which is called $lub(F)$ (least upper bound).

2) Any finite set $F\subset Q$, that has a lower bound, also has a greatest lower bound.

3) Assume, that a given set $S\subset Q$ there has a least upper bound. Then this bound is already the least upper bound of a finite subset of $S$.

4) If $q_1\not\le q_2$, then there is a $m\in M(Q)$ with $m\le i(q_1)\wedge m\not\le i(q_2)$.

5) For $m\in M(Q), C,C'\in Q$, we have $x\le i(lub(C,C'))\Leftrightarrow x\le i(C)\vee x\le i(C')$ (and $lub(C,C')$ exists).

Consider the map $g:Q\rightarrow P(M(Q))\qquad q\mapsto \{x\in M(Q)|x\le i(q)\}$. First, we have to show, that for any $x\in X$ the set $\{x\in M(Q)|x\le i(q)\}$ is open, nonempty and compact. It is open by definition of the topology on $M(X)$. Either $q$ is a minimal element in $Q$ (then $\{q\}\in M(Q)$) or there is a smaller element $q'\lneq q$, in which case by (4) $\exists m\in M(Q):m\le i(q)$. In both cases the set in question is not empty. It is much harder to show the compactness: Given any covering $\{\{x\in M(Q)|x\le i(C)\}|C\in S\}$ (for some $S\subset Q$) with basic open sets, we have to show, that there is a finite subcovering. Without restriction, we can assume that $\forall C\in S: S\subseteq q$, otherwise replace $C$ by $C\cap q$. This implies, that $q$ is a upper bound for $S$.

First we want to show, that it is a minimal upper bound. Assume $l\lneq q$ is also an upper bound. Then by (4), there is a $x\in M(X),$ with $x\le i(q)\wedge x\not\le i(l)$. The covering condition

$\{x\in M(Q)|x\le i(q)\}=\bigcup_{C\in S}\{x\in M(Q)|x\le i(C)\}$

tells us, that there is a $C\in S$ with $x\le C$. But as $l$ is a upper bound for $S$, we know that $x\le C\le l$, which contradicts $x\not\le C$. So $q$ is really a minimal upper bound. Furthermore it is a least upper bound: For any other upper bound $u$ consider $glb(u,q)$. It is another upper bound and $glb(u,q)\le q$. By minimality $glb(u,q)=q$ and so $q\le u$.

By (3), we know, that there is a finite set $F\subset S$ with $q=lub(F)$. Using (5), we get

$\{x\in M(Q)|x\le i(q)\}=\bigcup_{C\in F}\{x\in M(Q)|x\le i(C)\}$

So the map $g$ is really well defined. Furthermore it is injective (by (4)). It obviously preserves the ordering. We still have to show the surjectivity. So let any nonempty, compact, open subset $q$ of $P(M(Q))$ be given. It is a finite union of basic open sets:

$q=\bigcup_{C\in F}\{x\in M(Q)|x\le i(C)\}=\{x\in M(Q)|x\le i(lub(F))\}=g(lub(F))$. Thus the map is surjective.

The second last equality follows from (5).

Summarizing: A partially ordered set $Q$ satisfies (1)-(5), if and only if it is isomorphic to $P(X)$ for a $l$-space $X$.

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There is a problem. Property (3) needn't be satisfied. (Consider the Cantor set and let $C_0$ be the right half and $C_1$ the right half of the left half etc. then the $lub$ of this family is the whole space. It is not the lowest upper bound of any finite subfamily. –  HenrikRüping Nov 4 '10 at 9:50

In

M. H. Stone. Applications of the Theory of Boolean Rings to General Topology. Transactions of the American Mathematical Society Vol. 41, No. 3 (May, 1937), pp. 375-481

Marshall Stone proved (see Theorem 4) that there is a duality between locally compact totally disconnected Hausdorff spaces and Boolean rings (possibly without unit). Boolean rings are in turn equivalent to generalized Boolean algebras. The duality assigns to each space the generalized Boolean algebra of the compact open subsets. So the answer to your question seems to be that such posets are precisely the generalized Boolean algebras.

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Hm, I know this duality (maddin.110mb.com/pdf/boolean.pdf; of course, Stone's terminology is older). But can we describe boolean rings as special partial ordered sets? Of course the boolean structure induces the partial order, but how can we determine if a partial order comes from a boolean ring? –  Martin Brandenburg Nov 3 '10 at 19:48
I thought I answered that: a poset comes from a Boolean ring precisely when it is a generalized Boolean algebra. –  Andrej Bauer Nov 3 '10 at 23:05
Ah, of course! I thought that generalized boolean algebras have more structural data than partial orders, but they don't. –  Martin Brandenburg Nov 3 '10 at 23:25
I still can't believe that the powerset of the naturals is obtained in this way. $\mathbb{N}=\bigcup_{n\in \NN}\\{n\\}$ and there is no finite subcovering. –  HenrikRüping Nov 4 '10 at 16:24
@Henrik: I don't quite understand what it is you don't believe. Are you doubting the Stone representation theorem for Boolean algebras? The powerset $P(\mathbb{N})$ of the natural numbers is a Boolean algebra. The corresponding Stone space is the space of ultrafilters on $P(\mathbb{N})$, whereas you seem to think that the corresponding Stone space is $\mathbb{N}$, which is false. –  Andrej Bauer Nov 4 '10 at 22:39