I have proved that any uncountable subset of real numbers has the
same
cardinality as that of the set of all of real numbers.

I have proved that any uncountable subset of real numbers has the
same
cardinality as that of the set of all of real numbers.

In fact I have proved that continuum and uncountable are equivalent
mathematical concepts.

In article <95521eae-e92a-483b-a8d0-13629b0d2...@x19g2000prg.googlegroups.com>,
Student of Math  <omar.hosse...@gmail.com> wrote:

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.

Can you please STATE the continuum hypothesis? I have the strong
suspicion that you cannot state it.

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

No, you are incorrect. The Archimedean property of the real numbers is
a THEOREM of ZFC.

What is true is that the Archimedean property is independent of the
field axioms (and of other axiomatic systems for algebra). However,
the fact that the REAL NUMBERS form an Archimedean field is a
perfectly fine theorem of ZF, it is NOT independent from ZF.

--
=====================================================================> "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

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."

On Apr 14, 11:02=A0pm, magi...@math.berkeley.edu (Arturo Magidin) wrote:
In article <95521eae-e92a-483b-a8d0-13629b0d2...@x19g2000prg.googlegroups.=
com>,
Student of Math =A0<omar.hosse...@gmail.com> wrote:

On Apr 14, 6:41=3DA0pm, magi...@math.berkeley.edu (Arturo Magidin) wrote:=


=A0 =A0[...]

it is just =A0archimedian property of real numbers that i assume to
prove continuum hypothesis.

Can you please STATE the continuum hypothesis? I have the strong
suspicion that you cannot state it.

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

No, you are incorrect. The Archimedean property of the real numbers is
a THEOREM of ZFC.

What is true is that the Archimedean property is independent of the
field axioms (and of other axiomatic systems for algebra). However,
the fact that the REAL NUMBERS form an Archimedean field is a
perfectly fine theorem of ZF, it is NOT independent from ZF.

If anyone can prove archimedian property of real numbers?????

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

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

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

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

If anyone can prove archimedian property of real numbers?????

In article <0abe5352-f6f1-441d-ad65-7368132ab...@u36g2000prf.googlegroups.com>,
Student of Math  <omar.hosse...@gmail.com> wrote:





On Apr 14, 11:02=A0pm, magi...@math.berkeley.edu (Arturo Magidin) wrote:
In article <95521eae-e92a-483b-a8d0-13629b0d2...@x19g2000prg.googlegroups.> >com>,
Student of Math =A0<omar.hosse...@gmail.com> wrote:

On Apr 14, 6:41=3DA0pm, magi...@math.berkeley.edu (Arturo Magidin) wrote:
=A0 =A0[...]

it is just =A0archimedian property of real numbers that i assume to
prove continuum hypothesis.

Can you please STATE the continuum hypothesis? I have the strong
suspicion that you cannot state it.

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

No, you are incorrect. The Archimedean property of the real numbers is
a THEOREM of ZFC.

What is true is that the Archimedean property is independent of the
field axioms (and of other axiomatic systems for algebra). However,
the fact that the REAL NUMBERS form an Archimedean field is a
perfectly fine theorem of ZF, it is NOT independent from ZF.
If anyone can prove archimedian property of real numbers?????

It follows from the supremum property, which is provable for the real
numbers.

The Archimedean property for real numbers states that for every e,M>0
there exists a natural number N such that Ne > M.

Fix M>0; let A be the set of all e>0 such that Ne<M for all natural
number N. We claim the set is empty (hence the Archimedean property
holds).

Note that if k>0 is not in A, then any f>k is also not in A: for if
k>0 is such that there exists a natural number N such that Nk>M, then
from f>k>0 we get Nf>Nk>M, hence f is also not in A. And note that M
is not in A.

Next, suppose by way of contradiction that A is not empty. Then A is a
nonempty set of real numbers, and is bounded above (by M, for
example), hence it has a supremum, or least upper bound. Call this
supremum s. Since A is nonempty, there is an e>0 in A, and s>=e, hence
s is positive.

Is s in A? Well, if s is in A, then for every natural number N we have
Ns<M. Then 2s is also in A, because for every natural number K, 2K is
a natural  number, so K(2s) = (2K)s < M. Thus, 2s is in A. But 2>1 and
s>0, so 2s > 1s = s. Hence s, the least upper bound of A, is strictly
smaller than an element of A, which is contradiction. Thus, s cannot
be in A.

But then, consider (1/2)s. This is positive (since s is
positive). Since s is not in A, there exists a natural number N such
that Ns > M. But then 2N is a natural number, and 2N((1/2)s) = Ns>M,
hence (1/2)s is also not in N. Thus, (1/2)s is an upper bound of A,
but it is strictly smaller than the least upper bound of A (which is
s). Thus, we cannot have s not in A.

Thus, the assumption that A is nonempty leads to the absurd
proposition that its least upper bound is both in and not in A. Thus,
A must be empty. Therefore, for every e>0, e is not in A. Hence
there exists a natural number N such that Ne>M.

Thus, the real numbers satisfy the Archimedean Property.

(As for the Supremum Property for the real numbers, it follows for
example from the definition of real numbers as Dedekind Cuts; if A is
a set of real number, each real number x determined by a cut
{L_x|R_x}, and is bounded above by y={L_y|R_y}, then we can construct
the real number {L|R} whose left set L is the union of L_x for every x
in A; the fact that A is bounded above by y guarantees that this gives
you a Dedekind cut, and then it is a trivial matter to show that this
{L|R} is indeed the supremum of A).

--
=====================================================================> "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 -

Where you have used ZF(C) in your proof?

OK!
Where you have used ZF(C) in your proof?

Where you have used ZF(C) in your proof?

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

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.

Best regards,
Omar hosseiny

On Apr 14, 6:43 pm, Aatu Koskensilta <aatu.koskensi...@xortec.fi
wrote:
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

I have proved that any uncountable subset of real numbers has the
same
cardinality as that of the set of all of real numbers.

In fact I have proved that continuum and uncountable are equivalent
mathematical concepts.