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An Alexandrov topology on a set $X$ is a topology in which arbitrary intersections of open sets are open. Equivalently, every point has a smallest open neighborhood.

Given a partition on a set $X$, one can form the corresponding partition topology by taking the partition as a base for the topology.

It is clear that a partition topology is an Alexandrov topology (the smallest nbhd of a point is the partition block containing the point), and is regular.

The converse should be true:

Every regular Alexandrov topology is a partition topology.

In particular, every finite regular space has a partition topology.

Can anybody provides a proof?

PatrickR
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1 Answers1

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Suppose $X$ is regular and Alexandrov. If $x \in X$, let $U_x$ be the minimal open neighbourhood of $x$. As Randall says, it would be nice if we could show that the family of all $U_x$ is a partition. It is pretty clear that this family is a basis for the topology on $X$, as for any open $U$, we have that $U = \bigcup_{x \in U} U_x$. It's clear that the $U_x$ cover all of $X$, so all we need to show is that the $U_x$ are pairwise either equal or disjoint.

Equivalently, we need to show that if a pair of points in $X$ is topologically distinguishable, then they can be separated by open sets.

Indeed, suppose that $x, y \in X$ are topologically distinguishable. Without loss of generality, assume there is an open set $U$ with $x \in U$ but $y \notin U$. Then $X \setminus U$ is a closed set which contains the point $y$, but does not contain the point $x$. By regularity, we can find open sets that separate $X \setminus U$ and $x$. In particular, they separate $x$ and $y$. So we are done! $X$ must have the partition topology derived from the family $U_x$, or equivalently, from the equivalence relation of topological indistinguishability.

Izaak van Dongen
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    Nice. Your proof also shows the more general result: [Alexandrov + $R_1$ $\implies$ partition topology], where $R_1$ = preregular = topologically distinguishable points can be separated by nbhds. That's a property implied by both Hausdorff and regular. So also [Alexandrov + Hausdorff $\implies$ partition topology]. – PatrickR Feb 28 '24 at 03:36
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    Actually my previous comment does not say much, as Alexandrov preregular spaces are automatically regular, as seen here. – PatrickR Feb 28 '24 at 03:41
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    @PatrickR Alexandroff $T_1$ spaces are discrete, since for any set $A\subseteq X$ we have $A = \bigcap_{x\notin A} X\setminus{x}$. In fact this result is on pi-base. Here the power of this argument lies with not assuming separation properties that relate to points ($T_1$, Hausdorff etc.), otherwise this is trivial. – Jakobian Feb 28 '24 at 16:39