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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.
I indicate difference between consistency in the formalism and consistency in the usual mathematics.
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 Mm for which
Mm |= d=m+1 and Mm |= 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 2N leads contradiction in the consistent set theory in theorem 1.
Proof. By ∪(2N)=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.
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.
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.
1
I indicate difference between consistency in the formalism and consistency in the usual mathematics.
2
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 Mm for which
Mm |= d=m+1 and Mm |= 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 2N leads contradiction in the consistent set theory in theorem 1.
Proof. By ∪(2N)=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.
3
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.