# example of an Alexandroff space which cannot be turned into a topological group

Let $\mathbb{R}$ denote the set of real numbers and $\tau =\{[a,\mathrm{\infty})|a\in \mathbb{R}\}\cup \{(b,\mathrm{\infty})|b\in \mathbb{R}\}$. One can easily verify that $(\mathbb{R},\tau )$ is an Alexandroff space.

Proposition^{}. The Alexandroff space $(\mathbb{R},\tau )$ cannot be turned into a topological group^{}.

Proof. Assume that $\mathbb{R}=(\mathbb{R},\tau ,\circ )$ is a topological group. It is well known that this implies that there is $H\subseteq \mathbb{R}$ which is open, normal subgroup^{} of $\mathbb{R}$. This subgroup^{} ,,generates” the topology^{} (see the parent object for more details). Thus $H\ne \mathbb{R}$ because $\tau $ is not antidiscrete. Let $g\in \mathbb{R}$ such that $g\notin H$ (and thus $gH\cap H=\mathrm{\varnothing}$). Then $gH$ is again open (because the mapping $f(x)=g\circ x$ is a homeomorphism). But since both $H$ and $gH$ are open, then $gH\cap H\ne \mathrm{\varnothing}$. Indeed, every two open subsets in $\tau $ have nonempty intersection^{}. Contradiction^{}, because diffrent cosets are disjoint. $\mathrm{\square}$

Title | example of an Alexandroff space which cannot be turned into a topological group |
---|---|

Canonical name | ExampleOfAnAlexandroffSpaceWhichCannotBeTurnedIntoATopologicalGroup |

Date of creation | 2013-03-22 18:45:46 |

Last modified on | 2013-03-22 18:45:46 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 4 |

Author | joking (16130) |

Entry type | Example |

Classification | msc 22A05 |