---- > [!definition] Definition. ([[Alexandrov topology]]) > A [[topological space|topology]] $\tau$ on a set $X$ is called an **Alexandrov topology** if the intersection of any (possibly infinite) number of open sets is open. We call $(X, \tau)$ an **Alexandrov topological space**. > > Given a [[poset|preordered set]] $(X, \leq)$, there is a unique Alexandrov topology $\tau_{X}$ on $X$ whose [[the specialization preorder on a topological space|specialization preorder]] is $\leq$, namely, the collection of [[upper set|upper sets]] in $(X, \leq)$.[^1] $\tau_{X}$ is called **the Alexandrov (or specialization) topology on $X$**. Any [[monotonic map]] $f:X \to Y$ is [[continuous]] with respect to Alexandrov topologies $\tau_{X}$ and $\tau_{Y}$, giving thus a [[covariant functor]] $\tau:\mathsf{PrePos} \to \mathsf{AlexTop}$. The **'Alexandrov functor'** $\tau$ [[the category of preposets is equivalent to that of Alexandrov spaces|in fact witnesses]] that the categories $\mathsf{PrePos}$ and $\mathsf{AlexTop}$ are [[category equivalence|equivalent]]. ^definition > [!NOTE] Remark. > Crucially, any point $x$ in an Alexandrov topological space $(X, \tau)$ has a unique inclusion-minimal (open) [[neighborhood|neighborhood]] $U_{x}$ containing it. Namely, $U_{x}=\bigcap_{U \ni x \text{ open}} U$. These open sets $U_{x}$, called **stars**, form a [[basis for a topology|basis]] [[topology generated by a basis|generating]] $\tau$. > [!basicnonexample] > The [[standard topology on the real line]] $\mathbb{R}$ is not Alexandrov. Indeed, $U_{x}=\{ x \}$ for all $x \in \mathbb{R}$. ^nonexample [^1]: Clearly the collection of upper sets in $X$ forms an Alexandrov [[topological space|topology]] on $X$. Now, Suppose $\tau_{X}$ is an [[Alexandrov topology|Alexandrov]] [[topological space|topology]] on $X$ whose [[the specialization preorder on a topological space|specialization preorder]] is $\leq$. If $U \in \tau_{X}$ and $x \in U$ with $x \leq y$, then $x \in U_{y} \subset U_{x} \subset U$. In particular, $y \in U$. Hence $U$ is an [[upper set]]. Therefore, $\tau_{X}$ is necessarily the topology on $X$ consisting of upper sets. Now assume $f:X \to Y$ is a [[monotonic map]] between preordered sets $X,Y$ endowed with their canonical Alexandrov topologies. Let $V \in \tau_{Y}$, and assume $x \in f ^{-1}(V)$. Then having $x \leq y$ for some $y \in X$ implies $f(x) \leq f(y)$, and since $V$ is open in $Y$, it follows that $f(y) \in V$. $y \in f ^{-1}(V)$, witnessing $f ^{-1}(V)$ to be open. ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```