# induced Alexandroff topology on a poset

Let $(X,\le )$ be a poset. For any $x\in X$ define following subset:

$$(-\mathrm{\infty},x]=\{y\in X|y\le x\}.$$ |

The induced Alexandroff topology $\tau $ on $X$ is defined as a topology generated by ${\{(-\mathrm{\infty},x]\}}_{x\in X}$.

Proposition^{} 1. $(X,\tau )$ is a ${\mathrm{T}}_{0}$, Alexandroff space.

Proof. Let $x,y\in X$ be such that $x\ne y$. Note that this implies that $x\nleqq y$ or $y\nleqq x$ (because $\le $ is antisymmetric). Therefore $x\notin (-\mathrm{\infty},y]$ or $y\notin (-\mathrm{\infty},x]$. Thus $(X,\tau )$ is ${\mathrm{T}}_{0}$.

Now in order to show that $(X,\tau )$ is Alexandroff it is enough to show that an arbitrary intersection^{} of base sets is open. So assume that ${\{{x}_{i}\}}_{i\in I}$ is a subset of $X$ such that

$$A=\bigcap _{i\in I}(-\mathrm{\infty},{x}_{i}]\ne \mathrm{\varnothing}$$ |

and let $y\in A$. Then (since $\le $ is transitive^{}) it is clear that

$$(-\mathrm{\infty},y]\subseteq A$$ |

and thus

$$\bigcap _{i\in I}(-\mathrm{\infty},{x}_{i}]=A=\bigcup _{y\in A}(-\mathrm{\infty},y]$$ |

and therefore the intersection is open, which completes^{} the proof. $\mathrm{\square}$

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

Proof. ,,$\Rightarrow $” Assume that $f$ preserves order and let $A=(-\mathrm{\infty},y]\subseteq Y$ be an open base set. We wish to show that ${f}^{-1}(A)$ is open in $X$. So take any $x\in {f}^{-1}(A)$. Now if $y\le x$, then $f(y)\le f(x)$ (since $f$ preserves order) and thus $f(y)\in A$. Therefore $y\in {f}^{-1}(A)$. Since $y$ was arbitrary we obtain that for any $x\in {f}^{-1}(A)$ we have $(-\mathrm{\infty},x]\subseteq {f}^{-1}(A)$ and thus

$${f}^{-1}(A)=\bigcup _{x\in {f}^{-1}(A)}(-\mathrm{\infty},x],$$ |

which implies that ${f}^{-1}(A)$ is open.

,,$\Leftarrow $” Assume that $f$ is continuous and let $y\le x$ for some $x,y\in X$. Assume that $f(y)\nleqq f(x)$. Let $A=(-\mathrm{\infty},f(x)]$. Therefore $f(y)\notin A$, but $A$ is open, so ${f}^{-1}(A)$ is open (because $f$ is continuous). Thus $(-\mathrm{\infty},x]\subseteq {f}^{-1}(A)$. But $y\le x$, so $y\in {f}^{-1}(A)$. But this implies that $f(y)\in A$. Contradiction^{}. $\mathrm{\square}$

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 |