My discovery about mathematics

Naoto Meguro. Amateur.
MSC2010. Primary 03C62; Secondary 03C55.
Key Words and Phrases. The natural number theory in the set theory,
Gödel's theorem.
The abstract. The natural number theory in the set theory isn't consistent.

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I indicate that the set theory(ST) including the natural number theory(NNT) isn't consistent
by Gödel's theorem.

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Let's write ∅=0 and x∪{x}=x+1. Let X be a countable and consistent axiom system of
ST including NNT. Let d be a free individual symbol. X doesn't include d. Let N" be an individual
symbol for which (1∈N")∧∀n((n∈N")→(n+1∈N")) ∈X and
∀x((1∈x)∧∀y((y∈x)→(y+1∈x))→(N"⊂x)) ∈X.
You may set up that ∀x(x≠1+x) ∈X and ∀n((n∈N")→(n=∑_m∈n1^m)) ∈X and
∀x∀y((x∈y)→(∑_m∈y1^m=1^x+∑_m∈y-{x}1^m))∈X
for these are formed in the case of N"=N.
Theorem 1. X has a countable model M for which M |= (d∈N")∧(m∈d) for ∀m∈{0}∪N.
Proof. The axiom system X∪{d∈N",0∈d,1∈d,…,m∈d} has a model M_m for which
M_m |= N"=N and M_m |=d={0,1,…m}=m+1 and is consistent for ∀m∈N.
So X∪{d∈N",0∈d,1∈d,2∈d,…(infinitely)} is consistent and has a countable model M.
M is a model of X and M |= (d∈N")∧(m∈d) for ∀m∈{0}∪N.♦
Theorem 2. X doesn't exist.
Proof.Assume existence of X. For M in theorem 1, d becomes a countable set.
You can put M |= d={n₁,n₂,…(infinitely)}∈N". So M |= d=∑_m∈d1^m=1^n₁+1^n₂+…(infinitely).
M |= 1+d=1+(1+1+…(infinitely))=1+1+1+…(infinitely)=d. But M |= 1+d≠d.This is contradiction.♦

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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 indicate difference between consistency in the formalism and consistency in the usual mathematics.

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Let's write ∅=0, x∪{x}=x+1. Put P=(0∈N")∧∀x((x∈N")→(x+1∈N"))
and Q=∀x((0∈x)∧∀y((y∈x)→(y+1∈x))→(N"⊂x)). N" is an individual symbol.
If the set N={1,2,3,…} exists and P and Q are formed in the set theory, N"={0}∪N.
Assume that ZF is consistent.
Theorem 1. Existence of the set N leads contradiction in a consistent set theory.
Proof. Assume existence of the set N. Let d be an individual symbol.
The consistent set theory give a model of ZF.
So the axiom system (ZF)∪{d∈N",P,Q,d≠0,d≠1,d≠2,…d≠m} has a model M_m for which
M_m |= d=m+1 and M_m |= N"={0}∪N and is consistent for ∀m∈N.
So X=(ZF)∪{d∈N",P,Q,d≠0,d≠1,d≠2,…(infinitely)} is a consistent axiom system of the set theory.
By the axioms d≠0,d≠1,d≠2,…, d∉{0}∪N.
By the axioms d∈N" and P and Q and existence of the set N, d∈N"={0}∪N.
d∈{0}∪N and d∉{0}∪N are led in the consistent axiom system X if the set N exists.♦
Theorem 2. Existence of a countable set leads contadiction in the consistent set theory in theorem 1.
Proof. By the axiom of replacement and theorem 1.♦
Theorem 3. Existence of the set 2^N leads contradiction in the consistent set theory in theorem 1.
Proof. By ∪(2^N)=N and theorem 1.♦
Theorem 4. Existence of a continuum leads contradiction in the consistent set theory in theorem 1.
Proof. By the axiom of replacement and theorem 3.♦
The set theory treating a countable set or a continuum isn't consistent. You can prove any
conjectures like the millenniuum problems and abc conjecture in it.

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Theorem 2 may be the reason why Gauss didn't admit Abel's paper. The mathematics in 20th
century may be a ghastly fake for him.