# bounded complete

Let $P$ be a poset. Recall that a subset $S$ of $P$ is called *bounded from above* if there is an element $a\in P$ such that, for every $s\in S$, $s\le a$.

A poset $P$ is said to be *bounded complete* if every subset which is bounded from above has a supremum^{}.

Remark. Since it is not required that the subset be non-empty, we see that $P$ has a bottom. This is because the empty set^{} is vacuously bounded from above, and therefore has a supremum. However, this supremum is less than or equal to every member of $P$, and hence it is the least element of $P$.

Clearly, any complete lattice^{} is bounded complete. An example of a non-complete bounded complete poset is any closed subset of $\mathbb{R}$ of the form $[a,\mathrm{\infty})$, where $a\in \mathbb{R}$. In addition^{}, arbitrary products of bounded complete posets is also bounded complete.

It can be shown that a poset is a bounded complete dcpo iff it is a complete semilattice.

Remark. A weaker concept is that of *Dedekind completeness*: A poset $P$ is *Dedekind complete* if every *non-empty* subset bounded from above has a supremum. An obvious example is $\mathbb{R}$, which is Dedekind complete but not bounded complete (as it has no bottom). Dedekind completeness is more commonly known as the least upper bound property.

Title | bounded complete |
---|---|

Canonical name | BoundedComplete |

Date of creation | 2013-03-22 17:01:08 |

Last modified on | 2013-03-22 17:01:08 |

Owner | CWoo (3771) |

Last modified by | CWoo (3771) |

Numerical id | 8 |

Author | CWoo (3771) |

Entry type | Definition |

Classification | msc 06A12 |

Classification | msc 06B23 |

Classification | msc 03G10 |

Related topic | CompletenessPrinciple |

Defines | Dedekind complete |