# topology induced by uniform structure

Let $\mathcal{U}$ be a uniform structure on a set $X$. We define a subset $A$ to be open if and only if for each $x\in A$ there exists an entourage $U\in \mathcal{U}$ such that whenever $(x,y)\in U$, then $y\in A$.

Let us verify that this defines a topology^{} on $X$.

Clearly, the subsets $\mathrm{\varnothing}$ and $X$ are open. If $A$ and $B$ are two open sets, then for each $x\in A\cap B$, there exist an entourage $U$ such that, whenever $(x,y)\in U$, then $y\in A$, and an entourage $V$ such that, whenever $(x,y)\in V$, then $y\in B$. Consider the entourage $U\cap V$: whenever $(x,y)\in U\cap V$, then $y\in A\cap B$, hence $A\cap B$ is open.

Suppose $\mathcal{F}$ is an arbitrary family of open subsets. For each $x\in \bigcup \mathcal{F}$, there exists $A\in \mathcal{F}$ such that $x\in A$. Let $U$ be the entourage whose existence is granted by the definition of open set. We have that whenever $(x,y)\in U$, then $y\in A$; hence $y\in \bigcup \mathcal{F}$, which concludes the proof.

Title | topology induced by uniform structure |
---|---|

Canonical name | TopologyInducedByUniformStructure |

Date of creation | 2013-03-22 12:46:44 |

Last modified on | 2013-03-22 12:46:44 |

Owner | Mathprof (13753) |

Last modified by | Mathprof (13753) |

Numerical id | 7 |

Author | Mathprof (13753) |

Entry type | Derivation |

Classification | msc 54E15 |

Related topic | UniformNeighborhood |

Defines | uniform topology |