-
Notifications
You must be signed in to change notification settings - Fork 4
/
Copy path06A06-Supremum.tex
72 lines (62 loc) · 2.27 KB
/
06A06-Supremum.tex
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
\documentclass[12pt]{article}
\usepackage{pmmeta}
\pmcanonicalname{Supremum}
\pmcreated{2013-03-22 11:48:12}
\pmmodified{2013-03-22 11:48:12}
\pmowner{CWoo}{3771}
\pmmodifier{CWoo}{3771}
\pmtitle{supremum}
\pmrecord{11}{30340}
\pmprivacy{1}
\pmauthor{CWoo}{3771}
\pmtype{Definition}
\pmcomment{trigger rebuild}
\pmclassification{msc}{06A06}
%\pmkeywords{real analysis}
\pmrelated{Infimum}
\pmrelated{MinimalAndMaximalNumber}
\pmrelated{InfimumAndSupremumForRealNumbers}
\pmrelated{ExistenceOfSquareRootsOfNonNegativeRealNumbers}
\pmrelated{LinearContinuum}
\pmrelated{NondecreasingSequenceWithUpperBound}
\pmrelated{EssentialSupremum}
\endmetadata
% This is Cosmin's preamble.
% Packages
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsfonts}
\usepackage{amsthm}
\usepackage{mathrsfs}
%\usepackage{graphicx}
%%%%%\usepackage{xypic}
%\usepackage{babel}
% Theorem Environments
\newtheorem*{thm}{Theorem}
\newtheorem{thmn}{Theorem}
\newtheorem*{lem}{Lemma}
\newtheorem{lemn}{Lemma}
\newtheorem*{cor}{Corollary}
\newtheorem{corn}{Corollary}
\newtheorem*{prop}{Proposition}
\newtheorem{propn}{Proposition}
% Other Commands
\renewcommand{\geq}{\geqslant}
\renewcommand{\leq}{\leqslant}
\newcommand{\vect}[1]{\boldsymbol{#1}}
\newcommand{\mat}[1]{\mathsf{#1}}
\renewcommand{\div}{\!\mid\!}
\begin{document}
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.
%%%%%
%%%%%
%%%%%
%%%%%
\end{document}