goo

Defect of formalism 2

2011-09-30 19:46:35 | Mathematics
A well-ordered set may include decreasing sequence e1>e2>…
for a countable one {e1,e2,…} may not be a set.
Other proofs of theorem1 and theorem2 are written in the article "Defect of ZFC" in
this blog.
Please read it too.
Theorem 3. Let A be a countable axiom system of the standard mathematics.
If A is consistent,there is a model of A for which the general continuum hypothesis
(GCH) isn't formed.
Proof. Assume consistency of A.
Put Dm=A∪{m∈d,Card(d)< Card(d+1)< Card(P(d))}
(m&isin;<b>N,d+1=d∪{d},P(d) is the power set of d.)
Like the proof of theorem 1,
D=∪ Dm is consistent and has a model M for which
  m∈N

M |= m∈d for ∀m∈N, M |= Card(d)< Card(d+1)< Card(P(d)).

d expresses an infinite set. GCH isn't formed.&diams;
The above D is consistent in formalism if A is consistent. D |- Card(d)&ne;Card(d+1)
Card(d)=Card(d+1) is proved from D in the naive set theory by logic which you
cannot write by logical formulas. Such logic leads contradiction in this case.
(D isn't consistent in the naive set theory.)
<pre> 3
If you think that there is no proof of a theorem if there aren't logical formulas
which express a proof of the theorem as Gödel insisted,you lose many
theorems.
If you admit proofs which you cannot write by logical formulas,you are to abandon
formalism and Gödel's incompleteness theorem etc..
Which do you choose?
goo | コメント ( 0 ) | トラックバック ( 0 )