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Naoto Meguro. Amateur.
MSC 2010. Primary 03E30;Secondary 03E20.
Key Words and Phrases. A countable set,a continuum,a direct sum.
The abstract. Existence of a countable set or a continuum leads contradiction in the set theory.
I indicte that the set theory treating a countable set isn't consistent.
It may be the reason why Gauss didn't admit Abel's paper.
Let's write ∅=0,x∪{x}=x+1. Let X be a set of consistent axioms of the set theory.
Let d be a free individual symbol. Let n(x) be a free function symbol.
The mathematical axioms in X don't include d and n(x).
Put X"=X∪{n(∅)=0, ∀x∀y((x∈d)∧(y⊂d)∧(x∉y)→(n({x}∪y)=n(y)+1)).
Theorem 1. Existence of a countable set C leads contradiction in the set theory.
Proof. Put C={c0}∪{c1}∪…(infinitely) (ci≠ck for i≠k) and C"=C-{c0}.
X"∪{c0∈d,…,cm∈d, c0≠c1,…,cm-1≠cm} has a model Mm for which Mm |= d={c0,…,cm}
and Mm |= n(c)=0 when ¬(c⊂d) and Mm |= n(c)=the number of the elements of c when c⊂d
and is consistent for ∀m∈N. So X"∪{c0∈d,c1∈d,…(infinitely),c0≠c1,c0≠c2,…(infinitely)} is consistent.
By these consistent axioms,
n({c1})=1, n({c1}∪{c2})=1+1,…n({c1}∪{c2}∪…(infinitely))=1+1+…(infinitely)
={0}∪{1}∪{2}∪…(infinitely)={0}∪N. n(C")={0}∪N.
Similarly,n(C)=n({c0}∪{c1}∪…(infinitely))=1+1+…(infinitely)={0}∪N.
n(C")={0}∪N=n(C)=n({c0}∪C")=n(C")+1=n(C")∪{n(C")}∋n(C"). {0}∪N∈{0}∪N
{0}∪N is an infinite set but the elements of {0}∪N are finite sets. This is contradiction.♦
The mathemaatics treating a countable set isn't consistent. You need the countable set Q to express BSD
conjecture and Hodge conjecture. So they are provable nonsense.
Theorem 2. Existence of a continuum leads contradiction in the set theory.
Proof. If a continuum exists,R exists as a set by the axiom of replacement. Let [ ] be the symbol of Gauss.
∀x((x∈[0,1))→([x]=0)). ∀x([x+1]=[x]+1). Z={[r]| r∈R} exists and leads contradiction by theorem 1.♦
The mathematics treating a continuum isn't consistent. The millenniuum problems except P vs NP problem are
provable nonsense.
Theorem 3. Let M(≠{0}) be a module. If the direct sum M(S) exists for an infinite set S,contradiction is led.
Proof. If M(S) is a set, B={x|∃f((f∈M(S))∧(x={y|(y∈S)∧(f(y)≠0)}))} exists as a set by the axiom of
replacement. B is the set of the finite subsets of S.
Put Y=X∪{n(∅)=0,∀x(n({x})=1,∀x∀y((x≠y)→(n({x}∪{y})=2)),…(infinitely)}.
For Y,n(x) is the number of the elements of x for the finite set x.
By these consistent axioms,{n(x)| x∈B}={0}∪N exists and leads contradiction by theorem 1.♦
Generally,you cannot get the projective resolution of the module M by A(M)→M→0 (exact).
You cannot define Ext and Tor then. If M⊗AM" is a set, ∪M⊗AM"=A(M×M")(or =Z(M×M")) exists as a set.
You must not use M⊗AM" when M or M" is an infinite A-module by theorem 3. Fermat conjecture may be
still unsolved. If A[d] exists, {deg(f)|f∈A[d]}={0}∪N exists and leads contradiction by theorem 1.
The mathematics in 20th century may be a fake. You must reconstitute it.
MSC 2010. Primary 03E30;Secondary 03E20.
Key Words and Phrases. A countable set,a continuum,a direct sum.
The abstract. Existence of a countable set or a continuum leads contradiction in the set theory.
1
I indicte that the set theory treating a countable set isn't consistent.
It may be the reason why Gauss didn't admit Abel's paper.
2
Let's write ∅=0,x∪{x}=x+1. Let X be a set of consistent axioms of the set theory.
Let d be a free individual symbol. Let n(x) be a free function symbol.
The mathematical axioms in X don't include d and n(x).
Put X"=X∪{n(∅)=0, ∀x∀y((x∈d)∧(y⊂d)∧(x∉y)→(n({x}∪y)=n(y)+1)).
Theorem 1. Existence of a countable set C leads contradiction in the set theory.
Proof. Put C={c0}∪{c1}∪…(infinitely) (ci≠ck for i≠k) and C"=C-{c0}.
X"∪{c0∈d,…,cm∈d, c0≠c1,…,cm-1≠cm} has a model Mm for which Mm |= d={c0,…,cm}
and Mm |= n(c)=0 when ¬(c⊂d) and Mm |= n(c)=the number of the elements of c when c⊂d
and is consistent for ∀m∈N. So X"∪{c0∈d,c1∈d,…(infinitely),c0≠c1,c0≠c2,…(infinitely)} is consistent.
By these consistent axioms,
n({c1})=1, n({c1}∪{c2})=1+1,…n({c1}∪{c2}∪…(infinitely))=1+1+…(infinitely)
={0}∪{1}∪{2}∪…(infinitely)={0}∪N. n(C")={0}∪N.
Similarly,n(C)=n({c0}∪{c1}∪…(infinitely))=1+1+…(infinitely)={0}∪N.
n(C")={0}∪N=n(C)=n({c0}∪C")=n(C")+1=n(C")∪{n(C")}∋n(C"). {0}∪N∈{0}∪N
{0}∪N is an infinite set but the elements of {0}∪N are finite sets. This is contradiction.♦
The mathemaatics treating a countable set isn't consistent. You need the countable set Q to express BSD
conjecture and Hodge conjecture. So they are provable nonsense.
Theorem 2. Existence of a continuum leads contradiction in the set theory.
Proof. If a continuum exists,R exists as a set by the axiom of replacement. Let [ ] be the symbol of Gauss.
∀x((x∈[0,1))→([x]=0)). ∀x([x+1]=[x]+1). Z={[r]| r∈R} exists and leads contradiction by theorem 1.♦
The mathematics treating a continuum isn't consistent. The millenniuum problems except P vs NP problem are
provable nonsense.
Theorem 3. Let M(≠{0}) be a module. If the direct sum M(S) exists for an infinite set S,contradiction is led.
Proof. If M(S) is a set, B={x|∃f((f∈M(S))∧(x={y|(y∈S)∧(f(y)≠0)}))} exists as a set by the axiom of
replacement. B is the set of the finite subsets of S.
Put Y=X∪{n(∅)=0,∀x(n({x})=1,∀x∀y((x≠y)→(n({x}∪{y})=2)),…(infinitely)}.
For Y,n(x) is the number of the elements of x for the finite set x.
By these consistent axioms,{n(x)| x∈B}={0}∪N exists and leads contradiction by theorem 1.♦
Generally,you cannot get the projective resolution of the module M by A(M)→M→0 (exact).
You cannot define Ext and Tor then. If M⊗AM" is a set, ∪M⊗AM"=A(M×M")(or =Z(M×M")) exists as a set.
You must not use M⊗AM" when M or M" is an infinite A-module by theorem 3. Fermat conjecture may be
still unsolved. If A[d] exists, {deg(f)|f∈A[d]}={0}∪N exists and leads contradiction by theorem 1.
3
The mathematics in 20th century may be a fake. You must reconstitute it.