A total order $(S,<)$ is 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 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 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.