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 $X$.
