dense total order DenseTotalOrder 2007-02-08 11:18:42 2007-02-08 14:32:15 Definition total order dense dense in dense in itself % 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 A total order $(S,<)$ is \emph{dense} if whenever $x < z$ in $S$, there exists at least one element $y$ of $S$ such that $x < y < z$. That is, each nontrivial closed interval has nonempty interior. A subset $T$ of a total order $S$ is \emph{dense in} $S$ if for every $x,z\in S$ such that $x<z$, there exists some $y\in T$ such that $x<y<z$. Because of this, a dense total order $S$ is sometimes said to be \emph{dense in itself}. For example, the integers with the usual order are not dense, since there is no integer strictly between $0$ and $1$. On the other hand, the rationals $\mathbb{Q}$ are dense, since whenever $r$ and $s$ are rational numbers, it follows that $(r+s)/2$ is a rational number strictly between $r$ and $s$. Also, both $\mathbb{Q}$ and the irrationals $\mathbb{R}\setminus\mathbb{Q}$ are dense in $\mathbb{R}$. It is usually convenient to assume that a dense order has at least two elements. This allows one to avoid the trivial cases of the one-point order and the empty order.