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About the set theory

2018-07-15 22:19:32 | Mathematics
Naoto Meguro. Amateur.
MSC2010. Primary 03E30;Secondary 03E20.
Key Words and Phrases. A countable set, a continuum.
The abstract. Existence of a countable set or a continuum leads contradiction in the set theory.

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I show a new paradox of the set theory. The set theory which you know isn't consistent.
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Let's write ∅=0, x∪{x}=x+1. Let X be a consistent axiom system of
the set theory. You may set up that (0∈N")∧∀x((x∈N")→(x+1∈N")) ∈X
and ∀x((0∈x)∧∀y((y∈x)→(y+1∈x))→(N"⊂x)) ∈X.
If N={1,2,…} exists,N"={0}∪N. Let d be a free individual symbol. X doesn't
include d.
Theorem 1. Existence of the set N leads contradiction in the set theory.
Proof. Assume existence of the set N. The axiom system X∪{d⊂N",d∈N",0∈d,…m∈d}
has a model Mm for which Mm |= d={0,…,m}=m+1 and Mm |= N"={0}∪N
and is consistent for ∀m∈N. So Y=X∪{d⊂N",d∈N",0∈d,1∈d,…(infinitely)} is consistent.
Y is a consistent axiom system of the set theory. The elements of Y are theorems of
the set theory. By the theorems and existence of N,
N"={0}∪N⊂d⊂N". N"=d∈N". {0}∪N∈{0}∪N.
{0}∪N is an infinite set but the elements of {0}∪N are finite sets.
This is contradiction.♦
Theorem 2. Existence of a countable set leads contradiction in the set theory.
Proof. By the axiom of replacement and theorem 1.♦
Theorem 3. Existence of the set 2N leads contradiction in the set theory.
Proof. By ∪(2N)=N and theorem 1.♦
Theorem 4. Existence of a continuum leads contradiction in the set theory.
Proof. By the axiom of replacement and theorem 3.♦
The set theory treating a countable set or a continuum isn't consistent. Conjectures
treating a countable set or a continuum like the millenniuum problems and abc conjecture
are provable nonsense in the set theory.
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The mathematics in 20th century may be a fake.
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