おすすめブログ
@goo_blog
Naoto Meguro. Amateur.
MSC 2010. Primary 03E20;Secondary 08A65.
Key Words and Phrases. Countable sets, infinite sum.
The abstract. The standard mathematics treating the countable sums isn't consistent.
I indicate that the countable set or the continuum leads contradiction in the standard mathematics.
In the module of the countable sums of the rational numbers, v=1+1+…(infinitely)⇒ v=1+v
⇒ 0=v-v=1+v-v=1.
Theorem 1. The standard mathematics treating the set N isn't consistent.
Proof. Assume existence of N. By the axiom of replacement the countable sets exist.
Let F(x) be a function. You may define ∑r∈∅F(r)=0 and ∀x∀y((x∉y)→(∑r∈y∪{x} F(r)=(∑r∈y F(r))+F(x))).
Let c={c1,c2,…} be an arbitrary countable set.
∑r∈∅F(r)=0, ∑r∈{c1}F(r)=F(c1), ∑r∈{c1,c2} F(r)=F(c1)+F(c2),…,∑r∈{c1,c2,…(infinitely)} F(r)=F(c1)+F(c2)+…(infinitely).
The last definition is intuitive and uses infinite symbols for definition of the countable set is intuitive and
uses infinite symbols. When F(r)=u for ∀r∈c, ∑r∈c u=u+u+…(infinitely).
You may set up that Z∋n=n/1 ∈Q and Q∋ 0/2r=0/1=0 ∈Z (r∈N).
You may define R={(∑r∈N c(r)/2r)+n |(n∈Z)∧∀r((c(r)=0)∨(c(r)=1))} (0,1 ∈Z).
R is gotten by the mapping Z×({0,1}N)∋(n,c(r))→(∑r∈N c(r)/2r)+n .
Put 0"=(∑r∈N 0/2r)+0=(∑r∈N 0)+0=∑r∈N∪{0} 0=0+0+…(infinitely) (∈R).
Let G(x) be a function. You may define (∑r∈I F(r))+(∑r∈I G(r))=∑r∈I(F(r)+G(r)).
M={∑r∈N b(r) | ∀r((r∈N)→(b(r)∈Q))} becomes a module whose zero is ∑r∈N 0=0+0+…=0".
R∋(∑r∈N c(r)/2r)+n=∑r∈N c(r-1)/2r-1 ∈M (c(0)=n). So R⊂M.
Put C=∑r∈N 1 and C"=∑r∈N (-1) . C"+C=∑r∈N (-1+1)=∑r∈N 0=0".
C=…((1+1)+1)+1)+…(infinitely)=…((2+1)+1)+1+)+…(infinitely)=2+1+1+…(infinitely).
Put H(1)=1,H(r)=0 (r∈N-{1}), H=∑r∈N H(r)=(∑r∈N-{1} 0)+1=(0+0+…)+1=0"+1.
C+H=(1+1)+(1+0)+(1+0)+…(infinitely)=2+1+1+…(infinitely)=C
0"=C"+C=C"+(C+H)=(C"+C)+H=0"+H=H.
In the standard mathematics, h: Z∋n→(∑r∈N 0/2r)+n=0"+n∈R (⊂M) is an injection.
h(1)=0"+1=H=0"=0"+0=h(0). So 1=0 in Z. This is contradiction.♦
Theorem 2. Existence of the continuum leads contradiction in the standard mathematics.
Proof. Let 2x be the power set of x. If 2N exists,∪ 2N=N exists and leads contradiction by theorem 1.
By the axiom of replacement, existence of the continuum leads contradiction.♦
The standard mathematics treating N,Z,Q,R or C isn't consistent.
The above may be the reason why Gauss didn't admit infinite sets and Abel's paper.
If you agree with Gauss,you must not use the countable sets and the continua. If you admit Abel's
paper,you must admit 0"=0"+1 and cannot evolve the real number theory.
Anyhow R which you know doesn't exist. Couldn't you do something? I couldn't do that.
MSC 2010. Primary 03E20;Secondary 08A65.
Key Words and Phrases. Countable sets, infinite sum.
The abstract. The standard mathematics treating the countable sums isn't consistent.
1
I indicate that the countable set or the continuum leads contradiction in the standard mathematics.
In the module of the countable sums of the rational numbers, v=1+1+…(infinitely)⇒ v=1+v
⇒ 0=v-v=1+v-v=1.
2
Theorem 1. The standard mathematics treating the set N isn't consistent.
Proof. Assume existence of N. By the axiom of replacement the countable sets exist.
Let F(x) be a function. You may define ∑r∈∅F(r)=0 and ∀x∀y((x∉y)→(∑r∈y∪{x} F(r)=(∑r∈y F(r))+F(x))).
Let c={c1,c2,…} be an arbitrary countable set.
∑r∈∅F(r)=0, ∑r∈{c1}F(r)=F(c1), ∑r∈{c1,c2} F(r)=F(c1)+F(c2),…,∑r∈{c1,c2,…(infinitely)} F(r)=F(c1)+F(c2)+…(infinitely).
The last definition is intuitive and uses infinite symbols for definition of the countable set is intuitive and
uses infinite symbols. When F(r)=u for ∀r∈c, ∑r∈c u=u+u+…(infinitely).
You may set up that Z∋n=n/1 ∈Q and Q∋ 0/2r=0/1=0 ∈Z (r∈N).
You may define R={(∑r∈N c(r)/2r)+n |(n∈Z)∧∀r((c(r)=0)∨(c(r)=1))} (0,1 ∈Z).
R is gotten by the mapping Z×({0,1}N)∋(n,c(r))→(∑r∈N c(r)/2r)+n .
Put 0"=(∑r∈N 0/2r)+0=(∑r∈N 0)+0=∑r∈N∪{0} 0=0+0+…(infinitely) (∈R).
Let G(x) be a function. You may define (∑r∈I F(r))+(∑r∈I G(r))=∑r∈I(F(r)+G(r)).
M={∑r∈N b(r) | ∀r((r∈N)→(b(r)∈Q))} becomes a module whose zero is ∑r∈N 0=0+0+…=0".
R∋(∑r∈N c(r)/2r)+n=∑r∈N c(r-1)/2r-1 ∈M (c(0)=n). So R⊂M.
Put C=∑r∈N 1 and C"=∑r∈N (-1) . C"+C=∑r∈N (-1+1)=∑r∈N 0=0".
C=…((1+1)+1)+1)+…(infinitely)=…((2+1)+1)+1+)+…(infinitely)=2+1+1+…(infinitely).
Put H(1)=1,H(r)=0 (r∈N-{1}), H=∑r∈N H(r)=(∑r∈N-{1} 0)+1=(0+0+…)+1=0"+1.
C+H=(1+1)+(1+0)+(1+0)+…(infinitely)=2+1+1+…(infinitely)=C
0"=C"+C=C"+(C+H)=(C"+C)+H=0"+H=H.
In the standard mathematics, h: Z∋n→(∑r∈N 0/2r)+n=0"+n∈R (⊂M) is an injection.
h(1)=0"+1=H=0"=0"+0=h(0). So 1=0 in Z. This is contradiction.♦
Theorem 2. Existence of the continuum leads contradiction in the standard mathematics.
Proof. Let 2x be the power set of x. If 2N exists,∪ 2N=N exists and leads contradiction by theorem 1.
By the axiom of replacement, existence of the continuum leads contradiction.♦
The standard mathematics treating N,Z,Q,R or C isn't consistent.
3
The above may be the reason why Gauss didn't admit infinite sets and Abel's paper.
If you agree with Gauss,you must not use the countable sets and the continua. If you admit Abel's
paper,you must admit 0"=0"+1 and cannot evolve the real number theory.
Anyhow R which you know doesn't exist. Couldn't you do something? I couldn't do that.