Natural numbers The standard ordering ≤ of the
natural numbers is a well ordering and has the additional property that every non-zero natural number has a unique predecessor. Another well ordering of the natural numbers is given by defining that all even numbers are less than all odd numbers, and the usual ordering applies within the evens and the odds: :\begin{matrix} 0 & 2 & 4 & 6 & 8 & \dots & 1 & 3 & 5 & 7 & 9 & \dots \end{matrix} This is a well-ordered set of
order type . Every element has a successor (there is no largest element). Two elements lack a predecessor: 0 and 1.
Integers Unlike the standard ordering ≤ of the
natural numbers, the standard ordering ≤ of the
integers is not a well ordering, since, for example, the set of
negative integers does not contain a least element. The following
binary relation is an example of well ordering of the integers:
if and only if one of the following conditions holds: • • is positive, and is negative • and are both positive, and • and are both negative, and This relation can be visualized as follows: :\begin{matrix} 0 & 1 & 2 & 3 & 4 & \dots & -1 & -2 & -3 & \dots \end{matrix} is isomorphic to the
ordinal number . Another relation for well ordering the integers is the following definition: x \leq_z y
if and only if :|x| This well order can be visualized as follows: :\begin{matrix} 0 & -1 & 1 & -2 & 2 & -3 & 3 & -4 & 4 & \dots \end{matrix} This has the
order type .
Reals The standard ordering ≤ of any
real interval is not a well ordering, since, for example, the
open interval does not contain a least element. From the
ZFC axioms of set theory (including the
axiom of choice) one can show that there is a well order of the reals. Also
Wacław Sierpiński proved that ZF + GCH (the
generalized continuum hypothesis) imply the axiom of choice and hence a well order of the reals. Nonetheless, it is possible to show that the ZFC+GCH axioms alone are not sufficient to prove the existence of a definable (by a formula) well order of the reals. However it is consistent with ZFC that a definable well ordering of the reals exists—for example, it is consistent with ZFC that
V=L, and it follows from ZFC+V=L that a particular formula well orders the reals, or indeed any set. An uncountable subset of the real numbers with the standard ordering ≤ cannot be a well order: Suppose is a subset of well ordered by . For each in , let be the successor of in ordering on (unless is the last element of ). Let A = \{(x,s(x)) \,|\, x \in X\} whose elements are nonempty and disjoint intervals. Each such interval contains at least one rational number, so there is an
injective function from to There is an injection from to (except possibly for a last element of , which could be mapped to zero later). And it is well known that there is an injection from to the natural numbers (which could be chosen to avoid hitting zero). Thus there is an injection from to the natural numbers, which means that is countable. On the other hand, a countably infinite subset of the reals may or may not be a well order with the standard . For example, • The natural numbers are a well order under the standard ordering . • The set \{1/n \,|\, n=1,2,3,\dots\} has no least element and is therefore not a well order under standard ordering . Examples of well orders: • The set of numbers \{-2^{-n} \,|\, 0 \leq n has order type . • The set of numbers \{-2^{-n} - 2^{-m-n} \,|\, 0 \leq m,n has order type . The previous set is the set of
limit points within the set. Within the set of real numbers, either with the ordinary topology or the order topology, 0 is also a limit point of the set. It is also a limit point of the set of limit points. • The set of numbers \{-2^{-n} \,|\, 0 \leq n has order type . With the
order topology of this set, 1 is a limit point of the set, despite being separated from the only limit point 0 under the ordinary topology of the real numbers. == Equivalent formulations ==