# uniformities on a set form a complete lattice

###### Theorem.

The collection^{} of uniformities on a given set ordered by set inclusion forms a complete lattice^{}.

###### Proof.

Let $X$ be a set. Let $\U0001d518(X)$ denote the collection of uniformities on $X$. The coarsest uniformity on $X$ is $\{X\times X\}$, and the finest is the *discrete uniformity*:

$$\{S\subset X\times X:\mathrm{\Delta}(X)\subseteq S\}.$$ |

Hence $\U0001d518(X)$ is bounded^{}. To show that $\U0001d518(X)$ is complete^{}, we must prove that it has the least upper bound property.

Suppose ${\{{\mathcal{U}}_{\alpha}\}}_{\alpha \in I}$ is a nonempty family of uniformities on $X$. Let $\mathcal{B}$ consist of all finite intersections^{} of elements of the ${\mathcal{U}}_{\alpha}$. Let us check that $\mathcal{B}$ is a fundamental system of entourages for a uniformity on $X$.

(B1) Let $S$, $T\in \mathcal{B}$. Each of $S$ and $T$ is a finite intersection of elements of the ${\mathcal{U}}_{\alpha}$, so their intersection is as well. Hence $S\cap T\in \mathcal{B}$.

(B2) Every element of $\mathcal{B}$ is a finite intersection of subsets of $X\times X$ containing $\mathrm{\Delta}(X)$. So every element of $\mathcal{B}$ contains the diagonal.

(B3) Let $S\in \mathcal{B}$. Without loss of generality, $S={S}_{\alpha}\cap {S}_{\beta}$, where ${S}_{\alpha}\in {\mathcal{U}}_{\alpha}$ and ${S}_{\beta}\in {\mathcal{U}}_{\beta}$. Since ${S}_{\alpha}\in {\mathcal{U}}_{\alpha}$, ${S}_{\alpha}^{-1}\in {\mathcal{U}}_{\alpha}$. Similarly, ${S}_{\beta}^{-1}\in {\mathcal{U}}_{\beta}$. Since the process of taking the inverse^{} of a relation^{} commutes with taking finite intersections, ${({S}_{\alpha}\cap {S}_{\beta})}^{-1}\in \mathcal{B}$.

(B4) Let $S\in \mathcal{B}$. Again suppose $S={S}_{\alpha}\cap {S}_{\beta}$ with ${S}_{\alpha}\in {\mathcal{U}}_{\alpha}$ and ${S}_{\beta}\in {\mathcal{U}}_{\beta}$. Then there exist ${T}_{\alpha}\in {\mathcal{U}}_{\alpha}$ and ${T}_{\beta}\in {\mathcal{U}}_{\beta}$ such that ${T}_{\alpha}\circ {T}_{\alpha}\subseteq {S}_{\alpha}$ and ${T}_{\beta}\circ {T}_{\beta}\subseteq {S}_{\beta}$. The set $T={T}_{\alpha}\cap {T}_{\beta}$ is in $\mathcal{U}$, and since $T\circ T$ is a subset of both ${S}_{\alpha}$ and ${S}_{\beta}$, it is a subset of $S$.

The fundamental system $\mathcal{B}$ generates a uniformity $\mathcal{U}$. By construction, $\mathcal{U}$ is an upper bound of the ${\mathcal{U}}_{\alpha}$. But any upper bound of the ${\mathcal{U}}_{\alpha}$ would have to contain all finite intersections of elements of the ${\mathcal{U}}_{\alpha}$. So $\mathcal{U}={sup}_{\alpha \in I}{\mathcal{U}}_{\alpha}$. ∎

This theorem^{} is useful because it allows us to assert the existence of the coarsest uniform space satisfying a particular property.

###### Corollary.

Let $X$ be a set and let ${\mathrm{\{}{Y}_{\alpha}\mathrm{\}}}_{\alpha \mathrm{\in}I}$ be a family of uniform spaces. Then for any family of functions $\mathrm{\{}{f}_{\alpha}\mathrm{:}X\mathrm{\to}{Y}_{\alpha}\mathrm{\}}$, there is a coarsest uniformity on $X$ making all the ${f}_{\alpha}$ uniformly continous.

The coarsest uniformity making a family of functions uniformly continuous^{} is called the *initial uniformity* or *weak uniformity*.

## References

- 1 Nicolas Bourbaki, Elements of Mathematics: General Topology: Part 1, Hermann, 1966.

Title | uniformities on a set form a complete lattice |
---|---|

Canonical name | UniformitiesOnASetFormACompleteLattice |

Date of creation | 2013-03-22 16:30:46 |

Last modified on | 2013-03-22 16:30:46 |

Owner | mps (409) |

Last modified by | mps (409) |

Numerical id | 4 |

Author | mps (409) |

Entry type | Derivation |

Classification | msc 54E15 |

Classification | msc 06B23 |

Defines | discrete uniformity |

Defines | initial uniformity |

Defines | weak uniformity |