Well-order and well-ordered set

Let $W$ be a set. A well-order on $W$ is a total order $\le$ on $W$ (Partial order and partially ordered set) with the property that

$\text{every nonempty }S\subseteq W\text{ has a least element, i.e. an }\mathrm{\ell }\in S\text{ with }\mathrm{\ell }\le s\text{ for all }s\in S\mathrm{.}$

The pair $(W,\le )$ is then a well-ordered set, and $W$ is well-ordered by $\le$.

A least element of $S$ is unique when it exists: two of them are below each other, hence equal by antisymmetry (Partial order and partially ordered set). We may therefore write $\min S$ for it.

Strict form. Everything on this page is more convenient in terms of the associated strict order $x<y:⟺(x\le y\text{ and }x\mathrm{\ne }y)$ (Partial order and partially ordered set). Spelled out strictly, a well-order on $W$ is a relation $<$ that is

• irreflexive: $x<x$ holds for no $x\in W$;
• transitive: $x<y$ and $y<z$ imply $x<z$;
• trichotomous: for all $x,y\in W$ exactly one of $x<y$, $x=y$, $y<x$ holds;
• and such that every nonempty $S\subseteq W$ has an element $\mathrm{\ell }$ with no $s\in S$ satisfying $s<\mathrm{\ell }$.

The two presentations determine each other by $x\le y⟺(x<y$ or $x=y)$, and we write $(W,<)$ or $(W,\le )$ as convenient.

• Totality is not an extra hypothesis. If $\le$ is a partial order on $W$ in which every nonempty subset has a least element, then $\le$ is already total: apply the hypothesis to the two element subset $\{x,y\}$, whose least element is below the other. Totality is nevertheless stated, because in the strict presentation trichotomy has to be written down explicitly.
• A well-order is total, so every subset of a well-ordered set is a chain (Chain in a poset), and $W$ itself is one. Chains are therefore not the interesting invariant here; the least element property is.
• The model case is $(\mathbb{N},\le )$, which is a linear order ($\le$ is a linear order on $\mathbb{N}$) in which every nonempty subset has a least element (The well-ordering principle). Ordinals, defined later on this page, are the exact generalisation of that picture.
• $(\mathbb{Z},\le )$ and $(\mathbb{R},\le )$ are total orders but not well-orders: $\mathbb{Z}$ has no least element at all, and the bounded set $\{x\in \mathbb{R}:0<x<1\}$ has none either. Being bounded below does not help, which is exactly why well-ordering is a strong condition.
• The empty set carries exactly one well-order, the empty relation, vacuously. Every one element set carries exactly one.
• A well-order admits no infinite strictly decreasing sequence $x_0>x_1>x_2>\cdots$, since the set of its terms would have no least element. This direction follows directly from the definition; no converse is used here.