# a space is ${T}_{1}$ if and only if distinct points are separated

###### Theorem 1.

Let $X$ be a topological space^{}. Then $X$ is a ${T}_{\mathrm{1}}$-space
if and only if sets $\mathrm{\{}x\mathrm{\}}$, $\mathrm{\{}y\mathrm{\}}$
are separated for all distinct $x\mathrm{,}y\mathrm{\in}X$.

###### Proof.

Suppose $X$ is a ${T}_{1}$-space. Then every singleton is closed and if $x,y\in X$ are distinct, then

$\mathrm{\{}x\}\cap \overline{\{y\}}$ | $=$ | $\mathrm{\{}x\}\cap \{y\}=\mathrm{\varnothing},$ | ||

$\overline{\{x\}}\cap \{y\}$ | $=$ | $\mathrm{\{}x\}\cap \{y\}=\mathrm{\varnothing},$ |

and $\{x\}$, $\{y\}$ are separated. On the other hand, suppose that $\{x\}\cap \overline{\{y\}}=\mathrm{\varnothing}$ for all $x\ne y$. It follows that $\overline{\{y\}}=\{y\}$, so $\{y\}$ is closed and $X$ is a ${T}_{1}$-space. ∎

Title | a space is ${T}_{1}$ if and only if distinct points are separated |
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Canonical name | ASpaceIsT1IfAndOnlyIfDistinctPointsAreSeparated |

Date of creation | 2013-03-22 15:16:49 |

Last modified on | 2013-03-22 15:16:49 |

Owner | matte (1858) |

Last modified by | matte (1858) |

Numerical id | 6 |

Author | matte (1858) |

Entry type | Theorem |

Classification | msc 54D10 |