# product topology preserves the Hausdorff property

Theorem^{} Suppose ${\{{X}_{\alpha}\}}_{\alpha \in A}$ is a collection^{} of
Hausdorff spaces. Then the
generalized Cartesian product
${\prod}_{\alpha \in A}{X}_{\alpha}$
equipped with the product topology is a Hausdorff space.

*Proof.* Let $Y={\prod}_{\alpha \in A}{X}_{\alpha}$, and
let $x,y$ be distinct points in $Y$. Then there is an index $\beta \in A$
such that $x(\beta )$ and $y(\beta )$ are distinct points in
the Hausdorff space ${X}_{\beta}$. It follows that there are open sets
$U$ and $V$ in ${X}_{\beta}$ such that $x(\beta )\in U$, $y(\beta )\in V$,
and $U\cap V=\mathrm{\varnothing}$.
Let ${\pi}_{\beta}$ be the projection operator $Y\to {X}_{\beta}$ defined
here (http://planetmath.org/GeneralizedCartesianProduct). By the definition of
the product topology, ${\pi}_{\beta}$ is continuous^{}, so
${\pi}_{\beta}^{-1}(U)$ and ${\pi}_{\beta}^{-1}(V)$ are open sets in $Y$. Also,
since the
preimage^{} commutes with set operations^{} (http://planetmath.org/InverseImageCommutesWithSetOperations),
we have that

${\pi}_{\beta}^{-1}(U)\cap {\pi}_{\beta}^{-1}(V)$ | $=$ | ${\pi}_{\beta}^{-1}\left(U\cap V\right)$ | ||

$=$ | $\mathrm{\varnothing}.$ |

Finally, since $x(\beta )\in U$, i.e., ${\pi}_{\beta}(x)\in U$,
it follows
that $x\in {\pi}_{\beta}^{-1}(U)$. Similarly, $y\in {\pi}_{\beta}^{-1}(V)$.
We have shown that $U$ and $V$ are open disjoint neighborhoods^{} of
$x$ respectively $y$. In other words, $Y$ is a Hausdorff space.
$\mathrm{\square}$

Title | product topology preserves the Hausdorff property |
---|---|

Canonical name | ProductTopologyPreservesTheHausdorffProperty |

Date of creation | 2013-03-22 13:39:40 |

Last modified on | 2013-03-22 13:39:40 |

Owner | archibal (4430) |

Last modified by | archibal (4430) |

Numerical id | 7 |

Author | archibal (4430) |

Entry type | Theorem |

Classification | msc 54B10 |

Classification | msc 54D10 |