# the category of T0 Alexandroff spaces is equivalent to the category of posets

Let $\mathcal{A}\mathcal{T}$ be the category of all ${\mathrm{T}}_{0}$, Alexandroff spaces and continuous maps between them. Furthermore let $\mathcal{P}\mathcal{O}\mathcal{S}\mathcal{E}\mathcal{T}$ be the category of all posets and order preserving maps.

Theorem. The categories $\mathcal{A}\mathcal{T}$ and $\mathcal{P}\mathcal{O}\mathcal{S}\mathcal{E}\mathcal{T}$ are equivalent^{}.

Proof. Consider two functors:

$$T:\mathcal{A}\mathcal{T}\to \mathcal{P}\mathcal{O}\mathcal{S}\mathcal{E}\mathcal{T};$$ |

$$S:\mathcal{P}\mathcal{O}\mathcal{S}\mathcal{E}\mathcal{T}\to \mathcal{A}\mathcal{T},$$ |

such that $T(X,\tau )=(X,\le )$, where $\le $ is an induced partial order^{} on an Alexandroff space and $T(f)=f$ for continuous map. Analogously, let $S(X,\le )=(X,\tau )$, where $\tau $ is an induced Alexandroff topology^{} on a poset and $S(f)=f$ for order preserving maps. One can easily show that $T$ and $S$ are well defined. Furthermore, it is easy to verify that equalities $T\circ S={1}_{\mathcal{P}\mathcal{O}\mathcal{S}\mathcal{E}\mathcal{T}}$ and $S\circ T={1}_{\mathcal{A}\mathcal{T}}$ hold, which completes^{} the proof. $\mathrm{\square}$

Remark. Of course every finite topological space is Alexandroff, thus we have very nice ,,interpretation^{}” of finite ${\mathrm{T}}_{0}$ spaces - finite posets (since functors $T$ and $S$ do not change set-theoretic properties of underlying sets such as finitness).

Title | the category of T0 Alexandroff spaces is equivalent to the category of posets |
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Canonical name | TheCategoryOfT0AlexandroffSpacesIsEquivalentToTheCategoryOfPosets |

Date of creation | 2013-03-22 18:46:04 |

Last modified on | 2013-03-22 18:46:04 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 8 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 54A05 |