# induced partial order on an Alexandroff space

Let $X$ be a ${\mathrm{T}}_{0}$, Alexandroff space. For $A\subseteq X$ denote by ${A}^{o}$ the intersection^{} of all open neighbourhoods of $A$. Define a relation^{} $\le $ on $X$ as follows: for any $x,y\in X$ we have $x\le y$ if and only if $x\in {\{y\}}^{o}$. This relation will be called the induced partial order^{} on $X$.

Proposition^{} 1. $(X,\le )$ is a poset.

Proof. Of course $x\in {\{x\}}^{o}$ for any $x\in X$. Thus $\le $ is reflexive^{}.

Assume now that $x\le y$ and $y\le x$ for some $x,y\in X$. Assume that $x\ne y$. Then, since $X$ is a ${\mathrm{T}}_{0}$ space, there is an open set $U$ such that $x\in U$ and $y\notin U$ or there is an open set $V$ such that $y\in V$ and $x\notin V$. Both cases lead to contradiction^{}, because we assumed that $x\in {\{y\}}^{o}$ and $y\in {\{x\}}^{o}$. Thus every open neighbourhood of one element must also contain the other. Thus $\le $ is antisymmetric.

Finally assume that $x\le y$ and $y\le z$ for some $x,y,z\in X$. Since $y\in {\{z\}}^{o}$, then ${\{z\}}^{o}$ is an open neighbourhood of $y$ and thus ${\{y\}}^{o}\subseteq {\{z\}}^{o}$. Therefore $x\in {\{z\}}^{o}$, so $\le $ is transitive^{}, which completes^{} the proof. $\mathrm{\square}$

Proposition 2. Let $X,Y$ be two, ${\mathrm{T}}_{0}$, Alexandroff spaces and $f:X\to Y$ be a function. Then $f$ is continuous^{} if and only if $f$ preserves the induced partial order.

Proof. ,,$\Rightarrow $” Assume that $f$ is continuous and suppose that $x,y\in X$ are such that $x\le y$. We wish to show that $f(x)\le f(y)$, so assume this is not the case. Let $A={\{f(y)\}}^{o}$. Then $f(x)\notin A$. But $A$ is open, so ${f}^{-1}(A)$ is also open (because we assumed that $f$ is continuous). Furthermore $y\in {f}^{-1}(A)$ and because $x\le y$, then $x\in {f}^{-1}(A)$, but this implies that $f(x)\in A$. Contradiction.

,,$\Leftarrow $” Assume that $f$ preserves the induced partial order and let $U\subseteq Y$ be an open subset. Let $x\in U$. Then for any $y\le x$ we have $f(y)\le f(x)$ (because $f$ preserves the induced partial order) and since ${\{f(x)\}}^{o}\subseteq U$ (because $U$ is open and ${\{f(x)\}}^{o}$ is the smallest open neighbourhood of $f(x)$) we have that $f(y)\in U$. Thus

$${\{x\}}^{o}=\{y\in X|y\le x\}\subseteq {f}^{-1}(U)$$ |

which implies that ${f}^{-1}(U)$ is open because ${f}^{-1}(A)$ contains a small neighbourhood of each point. This completes the proof. $\mathrm{\square}$

Title | induced partial order on an Alexandroff space |
---|---|

Canonical name | InducedPartialOrderOnAnAlexandroffSpace |

Date of creation | 2013-03-22 18:45:55 |

Last modified on | 2013-03-22 18:45:55 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 4 |

Author | joking (16130) |

Entry type | Derivation |

Classification | msc 54A05 |