# separated uniform space

Let $X$ be a uniform space with uniformity $\mathrm{\pi \x9d\x92\xb0}$. $X$ is said to be *separated* or *Hausdorff ^{}* if it satisfies the following

*separation axiom*:

^{}$$\beta \x8b\x82\mathrm{\pi \x9d\x92\xb0}=\mathrm{\Xi \x94},$$ |

where $\mathrm{\Xi \x94}$ is the diagonal relation on $X$ and $\beta \x8b\x82\mathrm{\pi \x9d\x92\xb0}$ is the intersection^{} of all elements (entourages) in $\mathrm{\pi \x9d\x92\xb0}$. Since $\mathrm{\Xi \x94}\beta \x8a\x86\beta \x8b\x82\mathrm{\pi \x9d\x92\xb0}$, the separation axiom says that the only elements that belong to every entourage of $\mathrm{\pi \x9d\x92\xb0}$ are precisely the diagonal elements $(x,x)$. Equivalently, if $x\beta \x89y$, then there is an entourage $U$ such that $(x,y)\beta \x88\x89U$.

The reason for calling $X$ separated has to do with the following assertion:

$X$ is separated iff $X$ is a Hausdorff space under the topology

^{}${T}_{\mathrm{\pi \x9d\x92\xb0}}$ induced by (http://planetmath.org/TopologyInducedByAUniformStructure) $\mathrm{\pi \x9d\x92\xb0}$.

Recall that ${T}_{\mathrm{\pi \x9d\x92\xb0}}=\{A\beta \x8a\x86X\beta \x88\pounds \text{for each\Beta}\beta \x81\u2019x\beta \x88\x88A\beta \x81\u2019\text{, there is\Beta}\beta \x81\u2019U\beta \x88\x88\mathrm{\pi \x9d\x92\xb0}\beta \x81\u2019\text{, such that\Beta}\beta \x81\u2019U\beta \x81\u2019[x]\beta \x8a\x86A\}$, where $U\beta \x81\u2019[x]$ is some uniform neighborhood of $x$ where, under ${T}_{\mathrm{\pi \x9d\x92\xb0}}$, $U\beta \x81\u2019[x]$ is also a neighborhood^{} of $x$. To say that $X$ is Hausdorff under ${T}_{\mathrm{\pi \x9d\x92\xb0}}$ is the same as saying every pair of distinct points in $X$ have disjoint uniform neighborhoods.

###### Proof.

$(\beta \x87\x92)$. Suppose $X$ is separated and $x,y\beta \x88\x88X$ are distinct. Then $(x,y)\beta \x88\x89U$ for some $U\beta \x88\x88\mathrm{\pi \x9d\x92\xb0}$. Pick $V\beta \x88\x88\mathrm{\pi \x9d\x92\xb0}$ with $V\beta \x88\x98V\beta \x8a\x86U$. Set $W=V\beta \x88\copyright {V}^{-1}$, then $W$ is symmetric^{} and $W\beta \x8a\x86V$. Furthermore, $W\beta \x88\x98W\beta \x8a\x86V\beta \x88\x98V\beta \x8a\x86U$. If $z\beta \x88\x88W\beta \x81\u2019[x]\beta \x88\copyright W\beta \x81\u2019[y]$, then $(x,z),(y,z)\beta \x88\x88W$. Since $W$ is symmetric, $(z,y)\beta \x88\x88W$, so $(x,y)=(x,z)\beta \x88\x98(z,y)\beta \x88\x88W\beta \x88\x98W\beta \x8a\x86U$, which is a contradiction^{}.

$(\beta \x87\x90)$. Suppose $X$ is Hausdorff under ${T}_{\mathrm{\pi \x9d\x92\xb0}}$ and $(x,y)\beta \x88\x88U$ for every $U\beta \x88\x88\mathrm{\pi \x9d\x92\xb0}$ for some $x,y\beta \x88\x88X$. If $x\beta \x89y$, then there are $V\beta \x81\u2019[x]\beta \x88\copyright W\beta \x81\u2019[y]=\mathrm{\beta \x88\x85}$ for some $V,W\beta \x88\x88\mathrm{\pi \x9d\x92\xb0}$. Since $(x,y)\beta \x88\x88V$ by assumption^{}, $y\beta \x88\x88V\beta \x81\u2019[x]$. But $y\beta \x88\x88W\beta \x81\u2019[y]$, contradicting the disjointness of $V\beta \x81\u2019[x]$ and $W\beta \x81\u2019[y]$. Therefore $x=y$.
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Title | separated uniform space |
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Canonical name | SeparatedUniformSpace |

Date of creation | 2013-03-22 16:42:34 |

Last modified on | 2013-03-22 16:42:34 |

Owner | CWoo (3771) |

Last modified by | CWoo (3771) |

Numerical id | 5 |

Author | CWoo (3771) |

Entry type | Definition |

Classification | msc 54E15 |

Synonym | separating |

Synonym | Hausdorff uniform space |