properties of ordinals

Recall that an ordinalMathworldPlanetmathPlanetmath is a transitive set well-ordered by the membership relationMathworldPlanetmathPlanetmath . Below we list and prove some basic properties of ordinals.

  1. 1.

    is an ordinal. This is vacuously truePlanetmathPlanetmath.

  2. 2.

    If α is an ordinal, then αα, for otherwise ααα, contradicting the fact that is irreflexiveMathworldPlanetmath on α. Notice that by the axiom of foundationMathworldPlanetmath, this statement is true even if α is not an ordinal. But in this case, there is no need to resort to the axiom.

  3. 3.

    If α is an ordinal, and βα, then β is an ordinal.


    Since α is transitiveMathworldPlanetmathPlanetmath, βα implies that βα, and hence β is well-ordered by (inherited from α). Now, suppose δγβ. Then γα and δα. Since is a linear orderingPlanetmathPlanetmath in α, δβ, as desired. ∎

  4. 4.

    If α is an ordinal, so is α+:=α{α} (α+ is the successorMathworldPlanetmathPlanetmathPlanetmath of α).


    If γα+, then either γ=α or γα, and in the latter case, γα, since α is transitive. In either case, γα+ so that α+ is transitive. Now, suppose δα+. Then δ-{α}α, so has a least element η. Since ηα, η is also the least element of δ. This shows that α+ is well-ordered by , and therefore α+ is an ordinal. ∎

  5. 5.

    If α is an ordinal and βα is transitive, then βα. As a result, if αβ are ordinals such that βα, then βα.


    β is well-ordered since α is, and since β is transitive by assumptionPlanetmathPlanetmath, β is an ordinal. Let γ be the least element of α-β, and suppose δβ. Since is a linear ordering on α, either δγ or γδ. If the latter is true, then γδβ, and therefore γβ since β is transitive, contradicting γα-β. So δγ, and therefore βγ. Moreover, if δγ, then δα-β since γ is the least element in α-β. Therefore δβ, so γβ, and hence β=γα. ∎

  6. 6.

    (Law of Trichotomy) For any ordinals α,β, exactly one of the following is true:

    α=β,αβ,and  βα.

    If θδαβ, then θαβ since both α and β are transitive. So αβ is transitive, and hence an ordinal. If αβα and αββ, then αβαβ by 5, which is impossible by 2. This means that either αβ=α or αβ=β. If αβ, then αβ, and hence by 5, αβ in the former case, and βα in the latter. ∎

  7. 7.

    If A is a non-empty class of ordinals, then A is the least element of A. Furthermore, A is well-ordered.


    Set α:=A. There are three steps:

    • α is an ordinal: First, notice that α is well-ordered, as the ordering is inherited from members of A. Next, if βα, then βγ for every γA. Since each γ is transitive, βγ, and βα as a result, showing that α is transitive. Finally, α is a set, since it is a subclass of each element (which is a set) of A. This shows that α is an ordinal.

    • αA: By definition, αγ for each γA. Suppose in additionPlanetmathPlanetmath that αγ. Since α is an ordinal, then αγ by 5. This shows that α{γγA}=α, which is impossible by 2. Therefore α=γ for some γA, as desired.

    • α is the least element in A: If there is some δA such that δα. But α={γγA}δ, implying that δδ, a contradictionMathworldPlanetmathPlanetmath.

    If B is any non-empty subclass of A, then B is the least element of B by the result above. Hence A is well-ordered. ∎

  8. 8.

    On, the class of all ordinals is transitive, well-ordered (by ), and proper (not a set).


    If αβ On, then α On by 3, so On is transitive. It is well-ordered by 7. Finally, if On were a set, then On is an ordinal since it is transitive and well-ordered. This means that On On, contradicts 2 above. ∎

  9. 9.

    If A is a non-empty set of ordinals, then A is an ordinal, and A=supA.


    Set α:=A. First, α is a set because A is. If γβα, then γβδ for some δA, and therefore γδα, showing that α is transitive. If xα, then every element of x is an element of some δA. Hence x is a set of ordinals, and therefore well-ordered by 7. This shows that α is an ordinal.

    Now, pick any δA. Then δA=α, so either δ=α or δα. In the latter case, we get δα by 5. This shows that α is an upper bound of A. If θ is an ordinal such that δθ for every δA, then δθ, which implies αθ, so that αθ by 5, showing that α is the least upper bound of A. ∎

  10. 10.

    A transitive set of ordinals is an ordinal, true since the set is well-ordered by 7. This is a partial converseMathworldPlanetmath of 3. In particular, this provides an alternative proof of 4. Another corollary is that an initial segment of an ordinal is an ordinal.

  11. 11.

    Two ordinals are isomorphic iff they are the same.


    One direction is obvious. Suppose now that ϕ:αβ is an order isomorphism between ordinals α and β. Suppose αβ. Then either αβ or βα, which means that αβ or βα. In either case, we get a contradiction since ϕ is a bijection. ∎

  12. 12.

    Every well-ordered set is order isomorphic to exactly one ordinal.


    Uniqueness is provided by 12. Now, let us prove the existence of such an ordinal. First, notice that if there is an order preserving injection W into an ordinal α, then the image of W is an initial segment of α, which is an ordinal, and we are done. Now, assume the contrary. Since ordinals are well-ordered, the assumption implies that every α On has a strict injection into W (this is an important property on well-ordered sets, see a proof here ( We shall derive a contradiction by showing that On is a set. First, for each ordinal, there is a unique order-preserving injection into W. In addition, if two ordinals are such that their corresponding injections into W agree, then they are isomorphic and hence identical. Thus, elements of On are completely determined by subsets (in fact, initial segments) of W. Since W is a set, so is On, a contradiction! ∎

  13. 13.

    An ordinal has exactly one of the following forms: , α, or β+ for some ordinals α,β.


    Suppose an ordinal α. Being a transitive set of ordinals (by 3), β:=α is also an ordinal by 9. So βα. Then either β=α or βα. In the former case, α=α. In the latter case, βα. On the one hand, since β=supα, any γα, either γ=β or γβ, so that γβ+ in any case. This means that αβ+. On the other hand, βα combined with βα give us the other inclusion β+α. As a result, α=β+. ∎

  14. 14.

    A set is an ordinal iff it is a von Neumann ordinal. Every von Neumann ordinal is clearly an ordinal by 1, 4, and 9. The converse is true by 13.

Remark. Property 8 above also resolves one of the set theoryMathworldPlanetmath paradoxesMathworldPlanetmath called the Burali-Forti paradoxMathworldPlanetmath: the collectionMathworldPlanetmath of all ordinals is an ordinal, which is not true because the collection is not a set.

Title properties of ordinals
Canonical name PropertiesOfOrdinals
Date of creation 2013-03-22 18:48:58
Last modified on 2013-03-22 18:48:58
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 15
Author CWoo (3771)
Entry type Result
Classification msc 03E10