# supremum

Let $A$ be a set with a partial order $\leqslant$, and let $X\subseteq A$. Then $s=\sup X$ if and only if:

1. 1.

For all $x\in X$, we have $x\leqslant s$ (i.e. $s$ is an upper bound).

2. 2.

If $s^{\prime}$ meets condition 1, then $s\leqslant s^{\prime}$ ($s$ is the least upper bound).

There is another useful definition which works if $A=\mathbb{R}$ with $\leqslant$ the usual order on $\mathbb{R}$, supposing that s is an upper bound:

 $s=\sup X\text{ if and only if }\forall\varepsilon>0,\exists x\in X:s-% \varepsilon

Note that it is not necessarily the case that $\sup X\in X$. Suppose $X={]0,1[}$, then $\sup X=1$, but $1\not\in X$.

Note also that a set may not have an upper bound at all.

 Title supremum Canonical name Supremum Date of creation 2013-03-22 11:48:12 Last modified on 2013-03-22 11:48:12 Owner CWoo (3771) Last modified by CWoo (3771) Numerical id 11 Author CWoo (3771) Entry type Definition Classification msc 06A06 Related topic Infimum Related topic MinimalAndMaximalNumber Related topic InfimumAndSupremumForRealNumbers Related topic ExistenceOfSquareRootsOfNonNegativeRealNumbers Related topic LinearContinuum Related topic NondecreasingSequenceWithUpperBound Related topic EssentialSupremum