supremumSupremum2001-10-18 22:53:562007-08-11 04:54:37Definitionreal analysis% This is Cosmin's preamble.
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\renewcommand{\div}{\!\mid\!}The \emph{supremum} of a set $X$ having a partial order is the least upper bound of $X$ (if it exists) and is denoted $\sup{X}$.
Let $A$ be a set with a partial order $\leq$, and let $X \subseteq A$. Then $s = \sup X$ if and only if: \begin{enumerate}
\item[\textbf{1.}]{For all $x\in X$, we have $x \leq s$ (i.e. $s$ is an upper bound).}
\item[\textbf{2.}]{If $s^{\prime}$ meets condition \textbf{1}, then $s \leq s^{\prime}$ ($s$ is the \emph{least} upper bound).}
\end{enumerate}
There is another useful definition which works if $A = \mathbb{R}$ with $\leq$ the usual order on $\mathbb{R}$, supposing that s is an upper bound: \[s = \sup X \text{ if and only if } \forall \varepsilon > 0, \exists x\in X : s-\varepsilon < x.\]
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.