properties of ordinals
Recall that an ordinal is a transitive set well-ordered by the membership relation . Below we list and prove some basic properties of ordinals.
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1.
is an ordinal. This is vacuously true.
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2.
If is an ordinal, then , for otherwise , contradicting the fact that is irreflexive on . Notice that by the axiom of foundation, this statement is true even if is not an ordinal. But in this case, there is no need to resort to the axiom.
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3.
If is an ordinal, and , then is an ordinal.
Proof.
Since is transitive, implies that , and hence is well-ordered by (inherited from ). Now, suppose . Then and . Since is a linear ordering in , , as desired. ∎
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4.
If is an ordinal, so is ( is the successor of ).
Proof.
If , then either or , and in the latter case, , since is transitive. In either case, so that is transitive. Now, suppose . Then , so has a least element . Since , is also the least element of . This shows that is well-ordered by , and therefore is an ordinal. ∎
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5.
If is an ordinal and is transitive, then . As a result, if are ordinals such that , then .
Proof.
is well-ordered since is, and since is transitive by assumption, is an ordinal. Let be the least element of , and suppose . Since is a linear ordering on , either or . If the latter is true, then , and therefore since is transitive, contradicting . So , and therefore . Moreover, if , then since is the least element in . Therefore , so , and hence . ∎
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6.
(Law of Trichotomy) For any ordinals , exactly one of the following is true:
Proof.
If , then since both and are transitive. So is transitive, and hence an ordinal. If and , then by 5, which is impossible by 2. This means that either or . If , then , and hence by 5, in the former case, and in the latter. ∎
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7.
If is a non-empty class of ordinals, then is the least element of . Furthermore, is well-ordered.
Proof.
Set . There are three steps:
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is an ordinal: First, notice that is well-ordered, as the ordering is inherited from members of . Next, if , then for every . Since each is transitive, , and as a result, showing that is transitive. Finally, is a set, since it is a subclass of each element (which is a set) of . This shows that is an ordinal.
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: By definition, for each . Suppose in addition that . Since is an ordinal, then by 5. This shows that , which is impossible by 2. Therefore for some , as desired.
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is the least element in : If there is some such that . But , implying that , a contradiction.
If is any non-empty subclass of , then is the least element of by the result above. Hence is well-ordered. ∎
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8.
On, the class of all ordinals is transitive, well-ordered (by ), and proper (not a set).
Proof.
If On, then 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. ∎
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9.
If is a non-empty set of ordinals, then is an ordinal, and .
Proof.
Set . First, is a set because is. If , then for some , and therefore , showing that is transitive. If , then every element of is an element of some . Hence is a set of ordinals, and therefore well-ordered by 7. This shows that is an ordinal.
Now, pick any . Then , so either or . In the latter case, we get by 5. This shows that is an upper bound of . If is an ordinal such that for every , then , which implies , so that by 5, showing that is the least upper bound of . ∎
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10.
A transitive set of ordinals is an ordinal, true since the set is well-ordered by 7. This is a partial converse of 3. In particular, this provides an alternative proof of 4. Another corollary is that an initial segment of an ordinal is an ordinal.
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11.
Two ordinals are isomorphic iff they are the same.
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12.
Every well-ordered set is order isomorphic to exactly one ordinal.
Proof.
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 into an ordinal , then the image of is an initial segment of , which is an ordinal, and we are done. Now, assume the contrary. Since ordinals are well-ordered, the assumption implies that every On has a strict injection into (this is an important property on well-ordered sets, see a proof here (http://planetmath.org/PropertiesOfWellOrderedSets)). We shall derive a contradiction by showing that On is a set. First, for each ordinal, there is a unique order-preserving injection into . In addition, if two ordinals are such that their corresponding injections into agree, then they are isomorphic and hence identical. Thus, elements of On are completely determined by subsets (in fact, initial segments) of . Since is a set, so is On, a contradiction! ∎
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13.
An ordinal has exactly one of the following forms: , , or for some ordinals .
Proof.
Suppose an ordinal . Being a transitive set of ordinals (by 3), is also an ordinal by 9. So . Then either or . In the former case, . In the latter case, . On the one hand, since , any , either or , so that in any case. This means that . On the other hand, combined with give us the other inclusion . As a result, . ∎
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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 theory paradoxes called the Burali-Forti paradox: the collection of all ordinals is an ordinal, which is not true because the collection is not a set.
Title | properties of ordinals |
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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 |