|
|
|
Viewing Version
4
of
'Hausdorff's maximum principle'
|
[ view 'Hausdorff's maximum principle'
|
back to history
]
| Title of object: |
Hausdorff's maximum principle |
| Canonical Name: |
HaudorffsMaximumPrinciple |
| Type: |
Theorem |
| Created on: |
2002-09-29 19:48:23 |
| Modified on: |
2004-02-17 00:03:29 |
| Classification: |
msc:03E25 |
| Synonyms: |
Hausdorff's maximum principle=maximum principle |
Revision comment (for changes between this and next version):
| Changes for correction #7280 ('bad link'). |
Preamble:
% this is the default PlanetMath preamble. as your knowledge
% of TeX increases, you will probably want to edit this, but
% it should be fine as is for beginners.
% almost certainly you want these
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
% used for TeXing text within eps files
%\usepackage{psfrag}
% need this for including graphics (\includegraphics)
%\usepackage{graphicx}
% for neatly defining theorems and propositions
%\usepackage{amsthm}
% making logically defined graphics
%\usepackage{xypic}
% there are many more packages, add them here as you need them
% define commands here
\newcommand{\sR}[0]{\mathbb{R}}
\newcommand{\sC}[0]{\mathbb{C}}
\newcommand{\sN}[0]{\mathbb{N}}
\newcommand{\sZ}[0]{\mathbb{Z}}
% The below lines should work as the command
% \renewcommand{\bibname}{References}
% without creating havoc when rendering an entry in
% the page-image mode.
\makeatletter
\@ifundefined{bibname}{}{\renewcommand{\bibname}{References}}
\makeatother |
Content:
{\bf Theorem}
Let $X$ be a partially ordered set. Then there exists a maximal totally
ordered subset of $X$.
The Hausdorff's maximum principle is one of the many theorems equivalent
to the axiom of choice. The below proof uses Zorn's lemma, which
is also equivalent to the axiom of choice.
{\bf Proof.}
Let $S$ be the set of all totally ordered subsets of $X$.
$S$ is not empty, since the empty set is an element of $S$.
Given a subset $\tau$ of $S$, the union of
all the elements of $\tau$ is again an element of $S$,
as is easily verified. This shows that $S$, ordered by inclusion, is
inductive. The result now follows from Zorn's lemma.$\Box$ |
|
|
|
|
|