Set theory - chapter 1: Axioms of Set Theory
1.1
⇐ Trivial
⇒ Let . We shall use the axiom of extensionality to prove
and . Suppose one of the pair contains the same elements, for instance
. Then is equal to and so has one element. Hence
and the same argument gives
. Finally and . Otherwise, both sets have two elements. The elements of same
cardinality are equal : and . The first equality gives
. Using this in the second equal, we get
.
1.2
Suppose that . Since we get which contradicts the axiom of regularity :
has no ∈-minimal element. Alternative proof (using result of chapter
3) : contradicts Cantor's theorem.
1.3
Let be inductive and . We shall show that is inductive :
- and so
- Let then and . Because is inductive, . Moreover, and , hence . Consequently, .
If then for all inductive set
, . By the previous result,
is inductive thus and . Hence, and is transitive.
Now let and . Clearly, by definition of <. Conversely,
is transitive, so . Finally .
1.4
Let be inductive and . and is transitive so . Let . Since X is inductive,. If the either or . In the former case, by transitivity of so . In the latter case, . Hence we showed that is transitive and . Finally, is inductive.
Let and be inductive. Then . Hence is transitive.
1.5
Let be inductive and . We shall show that is inductive :
- If then as in 1.4 is transitive. Suppose
. Then or . Because is transitive, the first case implies
. In any case, we get
which is a contradiction. Hence
.
Now let and be inductive. Then so . But so .
1.6
Let be inductive and . We shall show that is inductive :
- As in 1.4, if is transitive, then so is
. Let be a nonempty subset of
. Suppose is nonempty and let
one of its ∈-minimal elements. Then
is also ∈-minimal in
for otherwise which contradicts the axiom of regularity. If
then obviously has an ∈-minimal element. Finally
.
1.7
Let be a nonempty set and . If then is clearly a ∈-minimal element of
and we are done.
Suppose that is nonempty. is inductive and its elements are transitive by 1.4. The set of all
such that every nonempty subset of
has an ∈-minimal element is inductive, so is
itself. Hence the subset
of has an ∈-minimal element
. There is no such that , otherwise it would belong to
by transitivity of . Finally is ∈-minimal in .
1.8
Let be inductive and . We shall show that is inductive :
- If then and consequently
Hence and for all there is such that . We have so . Finally .
1.9 (Induction)
Suppose and let be ∈-minimal in (such an element exists by 1.7).
so for some . But for otherwise would not be∈-minimal in
. Thus . A contradiction.
1.10
Let . We shall show that by induction (1.9) i.e. every set is T-finite.
- . If is an non-empty subset of
, then and is ⊂-maximal in .
- Let and a non-empty set. Let
be a maximal element of the non-empty set
. If is maximal in then we are done. Otherwise, let show that
is maximal in . If then hence by maximality of in . Thus .
1.11
For all , shows that not maximal. Hence has no ⊂-maximal element i.e.
is T-infinite.
1.12
Let be 1-1 onto. Let a non-empty set. Then is non-empty and there exists a maximal
. But . Hence is ⊂-maximal in and is T-finite.
1.13
Let be infinite and . is non-empty since . Suppose has an ⊂-maximal element
. Then and there exists . Then . A contradiction. Then is T-infinite.
1.14
Let be two sets and be a formula. Let . By replacement, is set. and the separation axiom holds for
.
1.15
- For (1.8), let such that let . Then .
- For (1.9) consider .
- For (1.10) consider .