overkill proofs
No.14404887 ViewReplyOriginalReport
Quoted By: >>14404889 >>14404915 >>14405367 >>14405428
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There is no largest natural number. The reason is that by Cantor's theorem, the power set of a finite set is a strictly larger set, and one can prove inductively that the power set of a finite set is still finite.
All numbers of the form 2n for natural numbers n?1 are even. The reason is that the power set of an n-element set has size 2n, proved by induction, and this is a Boolean algebra, which can be decomposed into complementary pairs {a,¬a}. So it is a multiple of 2.
Every finite set can be well-ordered. This follows by the Axiom of Choice via the Well-ordering Principle, which asserts that every set can be well-ordered.
Every non-empty set A has at least one element. The reason is that if A is nonempty, then {A} is a family of nonempty sets, and so by the Axiom of Choice it admits a choice function f, which selects an element f(A)?A.
There is no largest natural number. The reason is that by Cantor's theorem, the power set of a finite set is a strictly larger set, and one can prove inductively that the power set of a finite set is still finite.
All numbers of the form 2n for natural numbers n?1 are even. The reason is that the power set of an n-element set has size 2n, proved by induction, and this is a Boolean algebra, which can be decomposed into complementary pairs {a,¬a}. So it is a multiple of 2.
Every finite set can be well-ordered. This follows by the Axiom of Choice via the Well-ordering Principle, which asserts that every set can be well-ordered.
Every non-empty set A has at least one element. The reason is that if A is nonempty, then {A} is a family of nonempty sets, and so by the Axiom of Choice it admits a choice function f, which selects an element f(A)?A.
