# irreflexive

A binary relation^{} $\mathcal{R}$ on a set $A$ is said to be *irreflexive ^{}* (or

*antireflexive*) if $\forall a\in A$, $\mathrm{\neg}a\mathcal{R}a$. In other words, “no element is $\mathcal{R}$-related to itself.”

For example, the relation^{} $$ (“less than”) is an irreflexive relation on the set of natural numbers.

Note that “irreflexive” is not simply the negation^{} of “reflexive^{} (http://planetmath.org/Reflexive)
.” Although it is impossible for a relation (on a nonempty set) to be both reflexive (http://planetmath.org/Reflexive)
and irreflexive, there exist relations that are neither. For example, the relation $\{(a,a)\}$ on the two element set $\{a,b\}$ is neither reflexive nor irreflexive.

Here is an example of a non-reflexive, non-irreflexive relation “in nature.” A subgroup^{} in a group is said to be *self-normalizing* if it is equal to its own normalizer. For a group $G$, define a relation $\mathcal{R}$ on the set of all subgroups of $G$ by declaring $H\mathcal{R}K$ if and only if $H$ is the normalizer of $K$. Notice that every nontrivial group has a subgroup that is not self-normalizing; namely, the trivial subgroup $\{e\}$ consisting of only the identity^{}. Thus, in any nontrivial group $G$, there is a subgroup $H$ of $G$ such that $\mathrm{\neg}H\mathcal{R}H$. So the relation $\mathcal{R}$ is non-reflexive. Moreover, since the normalizer of a group $G$ in $G$ is $G$ itself, we have $G\mathcal{R}G$. So $\mathcal{R}$ is non-irreflexive.

Title | irreflexive |
---|---|

Canonical name | Irreflexive |

Date of creation | 2013-03-22 15:41:45 |

Last modified on | 2013-03-22 15:41:45 |

Owner | CWoo (3771) |

Last modified by | CWoo (3771) |

Numerical id | 14 |

Author | CWoo (3771) |

Entry type | Definition |

Classification | msc 03E20 |

Synonym | antireflexive |

Related topic | Reflexive |