On 2008-04-14, in sci.math, Student of Math wrote:

I think i have found a way to answer to Cantor's Continuum
hypothesis by using this assumption : [0,1/3^n]={0} when n tends to
the infinity.

Well, what's your answer? On the face of it the triviality you note
has nothing whatsoever to do with the continuum hypothesis.

If this assumption is from ZFC?

It is provable in ZF that 0 is the only number that is in [0, 1/(3^n)]
for all n.

--
Aatu Koskensilta (aatu.koskensi...@xortec.fi)

"Wovon man nicht sprechen kann, daruber muss man schweigen"
 - Ludwig Wittgenstein, Tractatus Logico-Philosophicus

On 14-04-2008 14:58, Student of Math wrote:





I think i have found a way to answer to Cantor's Continuum
hypothesis by using this assumption : [0,1/3^n]={0} when n tends to
the infinity.
Well, what's your answer? On the face of it the triviality you note
has nothing whatsoever to do with the continuum hypothesis.

If this assumption is from ZFC?
It is provable in ZF that 0 is the only number that is in [0, 1/(3^n)]
for all n.

--
Aatu Koskensilta (aatu.koskensi...@xortec.fi)

"Wovon man nicht sprechen kann, daruber muss man schweigen"
 - Ludwig Wittgenstein, Tractatus Logico-Philosophicus

perhaps it maybe strang or impossible but, I think I have prove
Continuum
Hypothesis with this assumption that :[0,1/3^n]={0} when n tends to
infinity.

The Continuum Hypothesis can neither be proved nor disproved in ZFC.

Can you at least explain the _meaning_ of the assertion "[0,1/3^n]={0}
when n tends to infinity"?

Best regards,

Jose Carlos Santos- Hide quoted text -

- Show quoted text -

perhaps it maybe strang or impossible but, I think I have prove
Continuum
Hypothesis with this assumption that :[0,1/3^n]=3D{0} when n tends to
infinity.

The Continuum Hypothesis can neither be proved nor disproved in ZFC.

Can you at least explain the _meaning_ of the assertion "[0,1/3^n]=3D{0}
when n tends to infinity"?

it means that if A_n=3D[0,1/3^n] be a sequence of intevales then
A_(infinity)=3D{0},

in the other words the only number that can be an
element of any of (A_n)s is 0.

In article <f753d4ac-ad1d-4571-bf40-42941ab44...@x19g2000prg.googlegroups.com>,
Student of Math  <omar.hosse...@gmail.com> wrote:

perhaps it maybe strang or impossible but, I think I have prove
Continuum
Hypothesis with this assumption that :[0,1/3^n]=3D{0} when n tends to
infinity.

The Continuum Hypothesis can neither be proved nor disproved in ZFC.

Can you at least explain the _meaning_ of the assertion "[0,1/3^n]=3D{0}
when n tends to infinity"?

it means that if A_n=3D[0,1/3^n] be a sequence of intevales then
A_(infinity)=3D{0},

You have not defined "A_(infinity)", so this assertion is useless.

Presumably, you want A_(infinty) = /\_{n in N} A_n.

in the other words the only number that can be an
element of any of (A_n)s is 0.

No, "an element of any of the (A_n)s" means the UNION of the A_n's,
and in that case, you get either [0,1] or [0,1/3], depending on
whether your index set starts with n=0 or n=1 (respectively). You mean
"the only number that is an element of ALL of the A_n's is 0."

It is still a mystery why it is you think this proves "the continuum
hypothesis." Can you state the continuum hypothesis, and say how this
trivial fact about certain nested integrals allegedly "proves" the
continuum hypothesis?

--
=====================================================================> "It's not denial. I'm just very selective about
 what I accept as reality."
    --- Calvin ("Calvin and Hobbes" by Bill Watterson)
=====================================================================
Arturo Magidin
magidin-at-member-ams-org

It is still a mystery why it is you think this proves "the continuum
hypothesis." Can you state the continuum hypothesis, and say how this
trivial fact about certain nested integrals allegedly "proves" the
continuum hypothesis?

I use a new classification on sequences of 0s and 1s also,

my quastion is that if this assumption is independent from ZFC:
" Let A_n=3D[0,1/3^n] then the only number that is an element of all
of (A_n)s is 0=3Dzero."

On 2008-04-14, in sci.math, Student of Math wrote:

I use a new classification on sequences of 0s and 1s also,
my quastion is that if this assumption is independent from ZFC:
" Let A_n=[0,1/3^n] then the only number that is an element of all
of (A_n)s is 0=zero."

As said, it is trivially provable in ZF(C); that is, it is provable
that for every positive real r there is an n such that 1/(3^n) < r. It
remains completely obscure why you think this has anything to do with
the continuum hypothesis.

--
Aatu Koskensilta (aatu.koskensi...@xortec.fi)

"Wovon man nicht sprechen kann, daruber muss man schweigen"
 - Ludwig Wittgenstein, Tractatus Logico-Philosophicus

On Apr 14, 6:11=A0pm, Aatu Koskensilta <aatu.koskensi...@xortec.fi
wrote:
On 2008-04-14, in sci.math, Student of Math wrote:

I use a new classification on sequences of 0s and 1s also,
my quastion is that if this assumption is independent from ZFC:
" Let A_n=3D[0,1/3^n] then the only number that is an element of all
of (A_n)s is 0=3Dzero."

As said, it is trivially provable in ZF(C); that is, it is provable
that for every positive real r there is an n such that 1/(3^n) < r. It
remains completely obscure why you think this has anything to do with
the continuum hypothesis.

--
Aatu Koskensilta (aatu.koskensi...@xortec.fi)

"Wovon man nicht sprechen kann, daruber muss man schweigen"
=A0- Ludwig Wittgenstein, Tractatus Logico-Philosophicus

but relation "<" is not defined as an acxiom in ZF.

I think i have found a way to answer to Cantor's Continuum
hypothesis by using this assumption : [0,1/3^n]={0} when n tends to
the infinity.

If this assumption is from ZFC?

I think i have found a way to answer to Cantor's Continuum
hypothesis by using this assumption : [0,1/3^n]={0} when n tends to
the infinity.
Well, what's your answer? On the face of it the triviality you note
has nothing whatsoever to do with the continuum hypothesis.

If this assumption is from ZFC?
It is provable in ZF that 0 is the only number that is in [0, 1/(3^n)]
for all n.

--
Aatu Koskensilta (aatu.koskensi...@xortec.fi)

"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus

perhaps it maybe strang or impossible but, I think I have prove
Continuum
Hypothesis with this assumption that :[0,1/3^n]={0} when n tends to
infinity.

I think i have found a way to answer to Cantor's Continuum
hypothesis by using this assumption : [0,1/3^n]={0} when n tends to
the infinity.
Well, what's your answer? On the face of it the triviality you note
has nothing whatsoever to do with the continuum hypothesis.
If this assumption is from ZFC?
It is provable in ZF that 0 is the only number that is in [0, 1/(3^n)]
for all n.
--
Aatu Koskensilta (aatu.koskensi...@xortec.fi)
"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
perhaps it maybe strang or impossible but, I think I have prove
Continuum
Hypothesis with this assumption that :[0,1/3^n]={0} when n tends to
infinity.
The Continuum Hypothesis can neither be proved nor disproved in ZFC.

Can you at least explain the _meaning_ of the assertion "[0,1/3^n]={0}
when n tends to infinity"?

it means that if A_n=[0,1/3^n] be a sequence of intevales then
A_(infinity)={0}

,in the other words the only number that can be an element
of any of (A_n)s is 0.

Can you at least explain the _meaning_ of the assertion "[0,1/3^n]={0}
when n tends to infinity"?

Best regards,

Jose Carlos Santos- Hide quoted text -

- Show quoted text -

In article <638f7745-efe8-4aae-803d-f198d783d...@b9g2000prh.googlegroups.com>,
Student of Math  <omar.hosse...@gmail.com> wrote:





On Apr 14, 6:11=A0pm, Aatu Koskensilta <aatu.koskensi...@xortec.fi
wrote:
On 2008-04-14, in sci.math, Student of Math wrote:

I use a new classification on sequences of 0s and 1s also,
my quastion is that if this assumption is independent from ZFC:
" Let A_n=3D[0,1/3^n] then the only number that is an element of all
of (A_n)s is 0=3Dzero."

As said, it is trivially provable in ZF(C); that is, it is provable
that for every positive real r there is an n such that 1/(3^n) < r. It
remains completely obscure why you think this has anything to do with
the continuum hypothesis.

--
Aatu Koskensilta (aatu.koskensi...@xortec.fi)

"Wovon man nicht sprechen kann, daruber muss man schweigen"
=A0- Ludwig Wittgenstein, Tractatus Logico-Philosophicus

but relation "<" is not defined as an acxiom in ZF.

It does not have to be an axiom; it just needs to be expressible in
terms of ZF, and it can be.

For example, if you construct real numbers as Dedekind cuts of
rational numbers, then you can compare the real r={A|B} and the real
s={C|D}, by defining

 r  = s   if and only if A=C.
 r <= s   if and only if   A is a subset of C
 r  < s   if and only if A is a subset of C and A=/=C.

Dedekind cuts are constructed from the rationals; the rationals and
their ordering (which is required to define Dedekind cuts) can be
defined in ZF using the integers and their order. The integers and
their order can be defined in ZF using the natural numbers and their
order. The natural numbers and their order can be defined in ZF using
the Axiom of Infinity and the relation "is an elemetn of". Thus, the
reals and their order "<" can be defined in ZF, using the axioms of
ZF.

So it matters not that "<" is not directly an axiom; it can be
expresse in ZF, and that is all you need.

--
=====================================================================> "It's not denial. I'm just very selective about
 what I accept as reality."
    --- Calvin ("Calvin and Hobbes" by Bill Watterson)
=====================================================================
Arturo Magidin
magidin-at-member-ams-org- Hide quoted text -

- Show quoted text -

On 2008-04-14, in sci.math, Student of Math wrote:

but relation "<" is not defined as an acxiom in ZF.

In saying that it is provable in ZF(C) that for every positive real r
there is an n such that 1/(3^n) < r we mean that the set theoretic
formalisation, ultimately in terms of the membership relation and the
usual logical notions, of that statement is provable ZF(C).

But let's set that aside and consider instead your "answer" to the
continuum hypothesis -- how does it go?

--
Aatu Koskensilta (aatu.koskensi...@xortec.fi)

"Wovon man nicht sprechen kann, daruber muss man schweigen"
 - Ludwig Wittgenstein, Tractatus Logico-Philosophicus

On Apr 14, 6:41=A0pm, magi...@math.berkeley.edu (Arturo Magidin) wrote:

it is just archimedian property of real numbers that i assume to
prove continuum hypothesis.

As you must know Archimedian property of real numbers is independent
from ZFC.