Set theory - Chapter 2: Ordinal Numbers
2.1
- is an isomorphism
- If is an isomorphism then so is
- If and are isomorphisms then so is
2.2
⇒Suppose there is such that . For all , either or . Hence and . Hence would be a successor ordinal.
⇐If is a successor ordinal, let
such that . Clearly, is not satisfied.
2.3
Let be inductive. Then is a set (since it is included in
) and is inductive:
- and implies
- If then also belongs to .
By (1.3) is transitive and by (1.7),
is well-founded. Moreover by lemma 2.11 (and because
) is a linear ordering. Hence
is a transitive set well-ordered by ∈ i.e. an ordinal number.
so . If is a non-zero ordinal then by (1.8)
i.e. is a successor ordinal. By (2.2)
is limit. Finally is the least non-zero ordinal.
2.4
(Without the Axiom of Infinity)
(i) ⇒ (ii) If there is an inductive set, then we can construct the
intersection of all inductive sets. It is T-infinite (1.11) so it is
infinite.
(ii) ⇒ (iii) Suppose there exists an infinite set
and that i.e. all ordinals are successor. Let
be be set of all finite subset of
. For all there is a 1-1 mapping of y onto some (unique)
. By replacement, is a set. and . Hence is nonempty and . If then there is such that . Let (such an element exists since
is infinite an is finite). Clearly and . Hence by induction, . Then is an ordinal and . A contradiction.
(iii) ⇒ (i) If is a set, then since is non-zero and for all
, since is a limit ordinal. Hence
is inductive.
2.5
The set is a non-empty subset of
. Let and such that . Then , a contradiction.
2.6
Let and define , and . If , there is such that . Thus . Hence β is limit.
2.7
Let be normal. Let , and . Then (because is continuous). Since is increasing, by Lemma 2.4. Hence .
2.8
By induction on .
- For all limit,
- For all limit,
- For all limit,
2.9
2.10
Let . We shall show by induction on
than , and .
- so
- Let be limit. For all
. Thus
- and so (use 2.10-i for the last inequality). Hence
.
- Same as 2.10-i
- (use 2.10-ii for the last inequality).
- Same as 2.10-i
2.11
- and
- and
- and
2.12
Define and for all . If , then by continuity of . We shall show that is the least ordinal that satisfies this property. Suppose
satisfies . Clearly, . Hence the least such that satisfies . If we let then and . A contradiction.
2.13
Let be a limit ordinal and consider the following properties :
- is indecomposable i.e there is no
such that
- For all
- for some
(ii) ⇒ (i) For all there exists (an unique)
such that . Since is indecomposable, . . so by (2.10) . Hence .
(i) ⇒ (iii) Let be indecomposable and be its Cantor's normal form
(and ). If or then is a decomposition of . Clearly . Moreover so . The inequality is strict for otherwise the Cantor's normal form of
would not be unique. Hence
and .
(iii) ⇒ (ii) First we shall prove by induction on
that for all , . It is clear for . If it is the case for then . Finally for a limit ordinal
such that the property is verified for all the
,. Now let and consider its Cantor's normal form
(and ). We have and for all , . Hence . where satisfies . But we showed that so . If we repeat this several times, we get
.
2.14
The set is a non-empty subset of
. Let be a minimal element of
(for the well-founded relation
) and such that . Then , a contradiction.
2.15
See Theorem 2.15