In mathematics, especially in set theory, two ordered sets and are said to have the same order type if they are order isomorphic, that is, if there exists a bijection (each element pairs with exactly one in the other set)
f\colonX\toY
In the special case when is totally ordered, monotonicity of already implies monotonicity of its inverse.
One and the same set may be equipped with different orders. Since order-equivalence is an equivalence relation, it partitions the class of all ordered sets into equivalence classes.
If a set
X
\sigma
X
\sigma*
The order type of a well-ordered set is sometimes expressed as .[1]
The order type of the integers and rationals is usually denoted
\pi
η
n\mapsto2n
f(x)=\tfrac{2x-1}{1-\vert{2x-1}\vert}
More examples can be given now: The set of positive integers (which has a least element), and that of negative integers (which has a greatest element). The natural numbers have order type denoted by ω, as explained below.
The rationals contained in the half-closed intervals [0,1) and (0,1], and the closed interval [0,1], are three additional order type examples.
Every well-ordered set is order-equivalent to exactly one ordinal number, by definition. The ordinal numbers are taken to be the canonical representatives of their classes, and so the order type of a well-ordered set is usually identified with the corresponding ordinal. Order types thus often take the form of arithmetic expressions of ordinals.
Firstly, the order type of the set of natural numbers is . Any other model of Peano arithmetic, that is any non-standard model, starts with a segment isomorphic to ω but then adds extra numbers. For example, any countable such model has order type .
Secondly, consider the set of even ordinals less than :
V=\{0,2,4,\ldots;\omega,\omega+2,\omega+4,\ldots;\omega ⋅ 2,\omega ⋅ 2+2,\omega ⋅ 2+4,\omega ⋅ 2+6\}.
\operatorname{ord}(V)=\omega ⋅ 2+4=\{0,1,2,\ldots;\omega,\omega+1,\omega+2,\ldots;\omega ⋅ 2,\omega ⋅ 2+1,\omega ⋅ 2+2,\omega ⋅ 2+3\},
With respect to their standard ordering as numbers, the set of rationals is not well-ordered. Neither is the completed set of reals, for that matter.
Any countable totally ordered set can be mapped injectively into the rational numbers in an order-preserving way. When the order is moreover dense and has no highest nor lowest element, there even exist a bijective such mapping.