induced Alexandroff topology on a poset

Let (X,) be a poset. For any xX define following subset:


The induced Alexandroff topology τ on X is defined as a topology generated by {(-,x]}xX.

PropositionPlanetmathPlanetmathPlanetmath 1. (X,τ) is a T0, Alexandroff space.

Proof. Let x,yX be such that xy. Note that this implies that xy or yx (because is antisymmetric). Therefore x(-,y] or y(-,x]. Thus (X,τ) is T0.

Now in order to show that (X,τ) is Alexandroff it is enough to show that an arbitrary intersectionMathworldPlanetmath of base sets is open. So assume that {xi}iI is a subset of X such that


and let yA. Then (since is transitiveMathworldPlanetmathPlanetmathPlanetmathPlanetmath) it is clear that


and thus


and therefore the intersection is open, which completesPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof.

Proposition 2. Let X,Y be posets and f:XY a function. Then f preserves order if and only if f is continuous in induced Alexandroff topologies.

Proof. ,,” Assume that f preserves order and let A=(-,y]Y be an open base set. We wish to show that f-1(A) is open in X. So take any xf-1(A). Now if yx, then f(y)f(x) (since f preserves order) and thus f(y)A. Therefore yf-1(A). Since y was arbitrary we obtain that for any xf-1(A) we have (-,x]f-1(A) and thus


which implies that f-1(A) is open.

,,” Assume that f is continuous and let yx for some x,yX. Assume that f(y)f(x). Let A=(-,f(x)]. Therefore f(y)A, but A is open, so f-1(A) is open (because f is continuous). Thus (-,x]f-1(A). But yx, so yf-1(A). But this implies that f(y)A. ContradictionMathworldPlanetmathPlanetmath.

Title induced Alexandroff topology on a poset
Canonical name InducedAlexandroffTopologyOnAPoset
Date of creation 2013-03-22 18:46:01
Last modified on 2013-03-22 18:46:01
Owner joking (16130)
Last modified by joking (16130)
Numerical id 4
Author joking (16130)
Entry type Derivation
Classification msc 54A05