Topological characterisations of properties of posets

Question

Finite connected partially ordered sets are in bijective correspondence to connected finite topological spaces that satisfy T_0, see for example the Wikipedia article Finite topological space. Here connected for posets means that the Hasse diagram is connected.

Question 1: Is there a nice purely topological characterisation when a connected finite topological space with T_0 corresponds to a lattice and when this lattice is distributive?

Question 2: Can one describe the Dedekind–MacNeille completion of a finite connected partially ordered set purely in topological terms?

Answer 1

I will discuss

Question 1: Is there a nice purely topological characterisation when a connected finite topological space with $T_0$ corresponds to a lattice and when this lattice is distributive?

Claims. If $\mathbf{X}=\langle X; {\mathcal T}\rangle$ is a finite, $T_0$, topological space and $\langle X;\leq\rangle$ is the associated poset given by the specialization preorder, then

(Claim 1) $\langle X;\leq\rangle$ is a lattice if and only if the intersection of any family of closed irreducible subsets of $\mathbf{X}=\langle X; {\mathcal T}\rangle$ is irreducible. (Equivalently, the collection of closed irreducible subsets is closed under binary intersection and empty intersection. Equivalently, the intersection of any two closed irreducible subsets is irreducible and the whole space $\mathbf{X}$ is irreducible.)

The topological condition in Claim 1 implies that any closed subset $C\subseteq X$ has an 'irreducible closure' $C'$ (equal to the intersection of all closed irreducible subsets containing $C$). $C'$ is the least closed irreducible subset containing $C$.

(Claim 2) $\langle X;\leq\rangle$ is a distributive lattice if and only if $\langle X;\leq\rangle$ is a lattice (see Claim 1) AND the following holds: Let $C\subseteq X$ be a closed subset. Let $I\subseteq X$ be a closed, irreducible subset that is not the irreducible closure of any of its proper closed subsets. Then $I\subseteq C'$ implies $I\subseteq C$.

Reasoning. For a topological space $\mathbf{X}$, let $L_{\mathbf{X}}$ be the lattice of closed subsets of $\mathbf{X}$ ordered by inclusion. There is a map $\textrm{cl}\colon X\to L_{\mathbf{X}}\colon x\mapsto \textrm{cl}(x)$ that maps a point to its closure. A space is $T_0$ iff this map is injective. The specialization preorder $\leq$ on $X$ is obtained be pulling back the order on $L_{\mathbf{X}}$ to $X$ via this map ($x\leq y \Leftrightarrow \textrm{cl}(x)\subseteq \textrm{cl}(y)$). The image of the map $\textrm{cl}$ is the set of closures of points of $\mathbf{X}$. Since finite $T_0$ spaces are sober, we get that the image of $\textrm{cl}$ is the set of closed irreducible subsets of $\mathbf{X}$. (A closed subset of $\mathbf{X}$ is irreducible if it is not the union of two proper closed subsets.) Lattice-theoretically, the image of the map $\textrm{cl}$ is the set of join-irreducibles of $L_{\mathbf{X}}$.

For Claim 1, the poset $\langle X; \leq \rangle$ is lattice-ordered iff the isomorphic poset $\langle J(L_{\mathbf{X}}); \subseteq\rangle$ of join-irreducibles of $L_{\mathbf{X}}$ is lattice-ordered. It is not hard to see that, for any finite lattice $L$, the set $J(L)$ of join-irreducibles is lattice-ordered iff $J(L)$ is closed under the meets of $L$. (I am including the empty meet, so saying that $J(L)$ is closed under the meets of $L$ includes the statement that the top element of $L$ is join-irreducible.)

For Claim 2, assume that $\langle X; \leq \rangle$ is lattice-ordered. Equivalently, assume that the poset $\langle J(L_{\mathbf{X}});\subseteq\rangle)$ of irreducible subsets of $\mathbf{X}$ is lattice-ordered. Under what circumstances will this lattice be distributive? Here we can use the fact that a finite lattice is distributive iff every join-irreducible is join-prime. It can be checked that an irreducible set $I\in J(L_{\mathbf{X}})$ is join-irreducible in the lattice-ordered poset $J(L_{\mathbf{X}})$ of join-irreducible closed subsets of $\mathbf{X}$ iff $I$ is not the irreducible closure of any proper closed subset of $I$. (The join operation of $J(L_{\mathbf{X}})$, when it is lattice-ordered, is the irreducible closure of the union. A subset of $\mathbf{X}$ is a finite union of elements of $J(L_\mathbf{X})$ iff it is closed in the topology. So, $I$ will be the irreducible closure of a proper closed subset of $I$ iff $I$ decomposes as a proper join in $J(L_{\mathbf{X}})$.) It can be checked that such a join-irreducible $I$ is join-prime iff $I\subseteq C'$ implies $I\subseteq C$ whenever $C$ is a closed subset. (This uses the fact that any closed subset is a finite union of irreducible subsets, so the irreducible closure of a closed set, like $C'$, represents a typical join in $J(L_{\mathbf{X}})$.) This is the reason that the topological condition in Claim 2 is equivalent to distributivity. \\\

Answer 2

Recall that an Alexandrov space is a topological space where the intersection of arbitrary collection of open sets is open. Alexandrov duality states that the category of Alexandrov spaces is equivalent to the category of pre-orderings. With Alexandrov duality, we obtain a topological space from an ordered set by declaring the upwards closed sets to be open, and we obtain the specialization ordering from every topological space.

Recall that the Sierpinski space is the topological space $S$ with underlying set $\{0,1\}$ and where $\{\emptyset,\{0,1\},\{1\}\}$ are the open sets. Alternatively, the Sierpinski space is the unique topology on $\{0,1\}$ whose specialization ordering is the usual linear ordering on $\{0,1\}$.

Let $X_i$ be an Alexandrov space for each $i\in I$ with specialization ordering $\leq_i$. Then the specialization ordering on the box product $\prod_{i\in I}^{\text{box}}X_i$ is just the ordering $\leq$ where $(x_i)_{i\in I}\leq(y_i)_{i\in I}$ precisely when $x_i\leq_iy_i$ whenever $i\in I$. We shall write $(X^I)^\text{box}$ for the box topology on $X^I$.

Theorem (Knaster-Tarski theorem): Let $X$ be a complete lattice. Let $f:X\rightarrow X$ be an order preserving map. Then the set of all fixed points of $X$ is a complete lattice (though not a complete sublattice of $X$).

We therefore conclude that the complete lattices with the Alexandrov topology are up-to-homeomorphism just the subspaces of all fixed points $\{\mathbf{x}\in(S^I)^{\text{box}}:f(\mathbf{x})=\mathbf{x}\}$ where $f:(S^I)^{\text{box}}\rightarrow(S^I)^{\text{box}}$ is some continuous function.

We can include restrictions on the function $f$ and state these restrictions in purely topological terms and still obtain all complete lattices.

Suppose that $X$ is a partial ordering. A closure operator is a continuous retraction $C:X\rightarrow X$ such that $C(x)\geq x$ for each $x\in X$, and an interior operator is a continuous retraction $D:X\rightarrow X$ such that $D(x)\leq x$ for each $x\in X$. If $E$ is either a closure operator or an interior operator, then let $E^*$ denote the set of all fixed points of $E$. For each complete lattice $X$, there is some set $I$ along with a closure operator $C:\{0,1\}^I\rightarrow\{0,1\}^I$ and interior operator $D:\{0,1\}^I\rightarrow\{0,1\}^I$ such that $X$ is isomorphic the the set $C^*$ and also to the set $D^*$.

The conditions in the definition of a closure/interior operator stating that $D(x)\leq x$ and $x\leq C(x)$ and more generally $f(x)\leq g(x)$ can be translated into purely topological terms by the following proposition.

Proposition: Let $X,Y$ be topological spaces. Let $f,g:X\rightarrow Y$ be continuous functions. Then the following are equivalent:

1. $f(x)\leq g(x)$ for all $x\in X$.

2. $f[R]\subseteq\overline{g[R]}$ for each $R\subseteq X$.

3. $R\subseteq f^{-1}[\overline{g[R]}]$ for each $R\subseteq X$.

4. $f[C]\subseteq \overline{g[C]}$ for each closed $C\subseteq X$.

5. $C\subseteq f^{-1}[\overline{g[C]}]$ for each closed $C\subseteq X$.

6. $g[f^{-1}[U]]\subseteq U$ for each open set $U$.

7. $f^{-1}[U]\subseteq g^{-1}[U]$ for each open set $U$.

8. $g^{-1}[C]\subseteq f^{-1}[C]$ for each closed set $C$.

9. $f[g^{-1}[C]]\subseteq C$ for each closed set $C$.

10. $f[g^{-1}[A]]\subseteq\overline{A}$ for each $A\subseteq Y.$