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Correct presentation of the diagonal argument
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WM
Nov 21, 2015, 8:31:26 AM11/21/15
to
Assume a list of all infinite digit sequences.
Every finite digit sequence d_1,d_2,d_3,...,d_n of an antidiagonal can be found (infinitely often) in the remainder of the list:
For all n in |N there exists k > n such that:
d_1,d_2,d_3,...,d_n = q_(k1),q_(k2),q_(k3),...,q_(kn)
There is no contradiction with Cantor's diagonal argument, if formalized in the correct way:
For all n in |N and for all k =< n:
d_1,d_2,d_3,...,d_n =/= q_(k1),q_(k2),q_(k3),...,q_(kn)
Regards, WM
Every finite digit sequence d_1,d_2,d_3,...,d_n of an antidiagonal can be found (infinitely often) in the remainder of the list:
For all n in |N there exists k > n such that:
d_1,d_2,d_3,...,d_n = q_(k1),q_(k2),q_(k3),...,q_(kn)
There is no contradiction with Cantor's diagonal argument, if formalized in the correct way:
For all n in |N and for all k =< n:
d_1,d_2,d_3,...,d_n =/= q_(k1),q_(k2),q_(k3),...,q_(kn)
Regards, WM
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Message has been deleted
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Dan Christensen
Nov 21, 2015, 10:41:09 AM11/21/15
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Theorem: The set T of infinite strings of digits (0 through 9) is uncountable.
Proof:
We make use of the following lemma: A set S is uncountable if S is non-empty and there exists no surjection f: N --> S. (See my formal proof in DC Proof format at: http://www.dcproof.com/CountableIfSurjectionFromN.htm)
000000... is an element of T. Therefore, T is non-empty.
Suppose we have a function f: N --> T.
We can construct an infinite string of digits D (the so-called anti-diagonal) as follows:
Let the kth digit in D be 1 if the kth digit in f(k) is 0; 0 otherwise.
D is an infinite string of 0's and 1's. So it is an element of T. It also differs from every string in the range of f. Therefore, f cannot be a surjection.
Generalizing, no function f: N --> T is a surjection. Therefore, T is uncountable.
Dan
Download my DC Proof 2.0 software at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
Proof:
We make use of the following lemma: A set S is uncountable if S is non-empty and there exists no surjection f: N --> S. (See my formal proof in DC Proof format at: http://www.dcproof.com/CountableIfSurjectionFromN.htm)
000000... is an element of T. Therefore, T is non-empty.
Suppose we have a function f: N --> T.
We can construct an infinite string of digits D (the so-called anti-diagonal) as follows:
Let the kth digit in D be 1 if the kth digit in f(k) is 0; 0 otherwise.
D is an infinite string of 0's and 1's. So it is an element of T. It also differs from every string in the range of f. Therefore, f cannot be a surjection.
Generalizing, no function f: N --> T is a surjection. Therefore, T is uncountable.
Dan
Download my DC Proof 2.0 software at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
Ben Bacarisse
Nov 21, 2015, 11:14:22 AM11/21/15
to
exists m in |N such that for all i in |N d_i = q_{m,i}
That would be a real result since it *would* contradict Cantor's
theorem. Coming up with things that don't contradict theorems is a
fool's errand.
(I prefer q_{k,1} and so on to your q_(k1). In TeX, the q_{i,j} would
then like the 2D array they are.)
--
Ben.
Virgil
Nov 21, 2015, 11:36:29 AM11/21/15
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In article <9ba6f90a-6897-4634...@googlegroups.com>,
It has been proved that, at least everywhere outside of WM's witless
worthless wacky world of WMytheology, no such LIST can exist.
If one restricts one's list to only only those infinite digit sequences
which are eventually periodic, one can, at least in principle, list all
of them, but without some such constraint, it is provably not possible!
So WM is
WRONG !
AGAIN ! !
AS USUAL ! ! !
--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)
It has been proved that, at least everywhere outside of WM's witless
worthless wacky world of WMytheology, no such LIST can exist.
If one restricts one's list to only only those infinite digit sequences
which are eventually periodic, one can, at least in principle, list all
of them, but without some such constraint, it is provably not possible!
So WM is
WRONG !
AGAIN ! !
AS USUAL ! ! !
--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)
Virgil
Nov 21, 2015, 11:42:10 AM11/21/15
to
In article <36578baa-1238-445b...@googlegroups.com>,
You know that WM will not understand it however often yo post it!
Dan Christensen <Dan_Chr...@sympatico.ca> wrote:
> Theorem: The set T of infinite strings of digits (0 through 9) is
> uncountable.
>
> Proof:
>
> We make use of the following lemma: A set S is uncountable if S is non-empty
> and there exists no surjection f: N --> S. (See my formal proof in DC Proof
> format at: http://www.dcproof.com/CountableIfSurjectionFromN.htm)
>
> 000000... is an element of T. Therefore, T is non-empty.
>
> Suppose we have a function f: N --> T.
>
> We can construct an infinite string of digits D (the so-called anti-diagonal)
> as follows:
>
> Let the kth digit in D be 1 if the kth element in f(k) is 0; 1 otherwise.
> Theorem: The set T of infinite strings of digits (0 through 9) is
> uncountable.
>
> Proof:
>
> We make use of the following lemma: A set S is uncountable if S is non-empty
> and there exists no surjection f: N --> S. (See my formal proof in DC Proof
> format at: http://www.dcproof.com/CountableIfSurjectionFromN.htm)
>
> 000000... is an element of T. Therefore, T is non-empty.
>
> Suppose we have a function f: N --> T.
>
> We can construct an infinite string of digits D (the so-called anti-diagonal)
> as follows:
>
>
> D is an infinite string of 0's and 1's. So it is an element of T. It also
> differs from every string in the range of f. Therefore, f cannot be a
> surjection, and T is uncountable.
> D is an infinite string of 0's and 1's. So it is an element of T. It also
> differs from every string in the range of f. Therefore, f cannot be a
>
>
> Dan
>
> Download my DC Proof 2.0 software at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com
Is there some reason for posting the same proof multiple times?
>
> Dan
>
> Download my DC Proof 2.0 software at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com
You know that WM will not understand it however often yo post it!
George Greene
Nov 21, 2015, 3:36:00 PM11/21/15
to
On Saturday, November 21, 2015 at 8:31:26 AM UTC-5, WM wrote:
> Assume a list of all infinite digit sequences.
Since no such list exists (if by "list" you mean what everybody else means,
> Assume a list of all infinite digit sequences.
something that is countably infinitely long and in which, despite this, every element on it is only finitely far away from every other element), this assumption is pregnant with any and all contradictions as consequences.
You can prove ANYthing, literally, from that assumption. The truth of what
you have "proved" will NOT be confirmed or even supported by such a proof.
> Every finite digit sequence d_1,d_2,d_3,...,d_n of an antidiagonal
It doesn't *change* anything important but it does *simplify* MANY things -
for starters, the anti-diagonal becomes unique: THE anti-diagonal.
It's a definable function. If you can't go all the way down to 2 then we'll settle for n just being even -- in that case you can define the
anti-diagonal-digit of d_x to be d_(1+n-x), and speaking of simplifications,
if you had STARTED WITH 0 instead of 1, it would just be d_(n-x).
WithOUT loss of generality, PLEASE, *THE* anti-diagonal.
> can be found (infinitely often) in the remainder of the list:
if the list contains EVERY infinite sequence. That does not require
demonstration; that's just obvious.
>
> For all n in |N
kindergarten teacher may have told you.
> there exists k > n such that:
>
> d_1,d_2,d_3,...,d_n = q_(k1),q_(k2),q_(k3),...,q_(kn)
> There is no contradiction with Cantor's diagonal
> argument, if formalized in the correct way:
George Greene
Nov 21, 2015, 3:42:20 PM11/21/15
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On Saturday, November 21, 2015 at 8:31:26 AM UTC-5, WM wrote:
> Assume a list of all infinite digit sequences.
>
> Every finite digit sequence d_1,d_2,d_3,...,d_n of an antidiagonal can be found (infinitely often) in the remainder of the list:
>
> For all n in |N there exists k > n such that:
>
> d_1,d_2,d_3,...,d_n = q_(k1),q_(k2),q_(k3),...,q_(kn)
You have said that there is a k > n but you haven't said what k1, k2, or any other k is. Let alone what q is.
>
> Every finite digit sequence d_1,d_2,d_3,...,d_n of an antidiagonal can be found (infinitely often) in the remainder of the list:
>
> For all n in |N there exists k > n such that:
>
> d_1,d_2,d_3,...,d_n = q_(k1),q_(k2),q_(k3),...,q_(kn)
Ben Bacarisse
Nov 21, 2015, 4:41:46 PM11/21/15
to
list. I'd write it q_{k,n}.
--
Ben.
WM
Nov 21, 2015, 4:58:44 PM11/21/15
to
Regards, WM
WM
Nov 21, 2015, 5:03:06 PM11/21/15
to
Am Samstag, 21. November 2015 16:41:09 UTC+1 schrieb Dan Christensen:
> D is an infinite string of 0's and 1's. So it is an element of T. It also differs from every string in the range of f.
For every n in |N: d_1,d_2,d_3,...,d_n differs from the first n rows.
> D is an infinite string of 0's and 1's. So it is an element of T. It also differs from every string in the range of f.
Every row n is followed by infinitely many rows. Infinitely many contain the sedquence d_1,d_2,d_3,...,d_n. Therefore a statement about every n does not prove anything about all n.
Regards, WM
WM
Nov 21, 2015, 5:10:16 PM11/21/15
to
Am Samstag, 21. November 2015 17:14:22 UTC+1 schrieb Ben Bacarisse:
> WM <wolfgang.m...@hs-augsburg.de> writes:
>
> > Assume a list of all infinite digit sequences.
> >
> > Every finite digit sequence d_1,d_2,d_3,...,d_n of an antidiagonal can
> > be found (infinitely often) in the remainder of the list:
> >
> > For all n in |N there exists k > n such that:
> >
> > d_1,d_2,d_3,...,d_n = q_(k1),q_(k2),q_(k3),...,q_(kn)
> >
> > There is no contradiction with Cantor's diagonal argument, if
> > formalized in the correct way:
> >
> > For all n in |N and for all k =< n:
> >
> > d_1,d_2,d_3,...,d_n =/= q_(k1),q_(k2),q_(k3),...,q_(kn)
>
> Yet you can not prove that
>
> exists m in |N such that for all i in |N d_i = q_{m,i}
That is impossible because there is nothing like all i.
> WM <wolfgang.m...@hs-augsburg.de> writes:
>
> > Assume a list of all infinite digit sequences.
> >
> > Every finite digit sequence d_1,d_2,d_3,...,d_n of an antidiagonal can
> > be found (infinitely often) in the remainder of the list:
> >
> > For all n in |N there exists k > n such that:
> >
> > d_1,d_2,d_3,...,d_n = q_(k1),q_(k2),q_(k3),...,q_(kn)
> >
> > There is no contradiction with Cantor's diagonal argument, if
> > formalized in the correct way:
> >
> > For all n in |N and for all k =< n:
> >
> > d_1,d_2,d_3,...,d_n =/= q_(k1),q_(k2),q_(k3),...,q_(kn)
>
> Yet you can not prove that
>
> exists m in |N such that for all i in |N d_i = q_{m,i}
All we can prove and all Cantor can prove is shown above.
>
> That would be a real result since it *would* contradict Cantor's
> theorem. Coming up with things that don't contradict theorems is a
> fool's errand.
Coming up with claims that are not supported by mathematics is fools errand too. Cantor's claim that every n is the same as all n is of that kind.
> That would be a real result since it *would* contradict Cantor's
> theorem. Coming up with things that don't contradict theorems is a
> fool's errand.
Every n is followed by infinitely many naturals. All n are not followed by naturals. But Cantor's proof concerns only every n (proof: he never shows a last one). Just like my proof concerns every n.
Regards, WM
Message has been deleted
Dan Christensen
Nov 21, 2015, 6:06:28 PM11/21/15
to
(Corrected typo)
On Saturday, November 21, 2015 at 5:03:06 PM UTC-5, WM wrote:
> Am Samstag, 21. November 2015 16:41:09 UTC+1 schrieb Dan Christensen:
>
>
What WM snipped without indicating:
Proof:
We make use of the following lemma: A set S is uncountable if S is non-empty and there exists no surjection f: N --> S. (See my formal proof in DC Proof format at: http://www.dcproof.com/CountableIfSurjectionFromN.htm)
000000... is an element of T. Therefore, T is non-empty.
Suppose we have a function f: N --> T.
We can construct an infinite string of digits D (the so-called anti-diagonal) as follows:
Let the kth digit in D be 1 if the kth digit in f(k) is 0; 0 otherwise.
On Saturday, November 21, 2015 at 5:03:06 PM UTC-5, WM wrote:
> Am Samstag, 21. November 2015 16:41:09 UTC+1 schrieb Dan Christensen:
>
>
Proof:
We make use of the following lemma: A set S is uncountable if S is non-empty and there exists no surjection f: N --> S. (See my formal proof in DC Proof format at: http://www.dcproof.com/CountableIfSurjectionFromN.htm)
000000... is an element of T. Therefore, T is non-empty.
Suppose we have a function f: N --> T.
We can construct an infinite string of digits D (the so-called anti-diagonal) as follows:
Let the kth digit in D be 1 if the kth digit in f(k) is 0; 0 otherwise.
> > D is an infinite string of 0's and 1's. So it is an element of T. It also differs from every string in the range of f.
>
> For every n in |N: d_1,d_2,d_3,...,d_n differs from the first n rows.
If you are talking about the anti-diagonal string D as defined above (in the part you snipped), it will differ from every string in the range of f. In the string given by f(k), D will differ from f(k) at position k. Do you not agree?
>
> For every n in |N: d_1,d_2,d_3,...,d_n differs from the first n rows.
Rupert
Nov 21, 2015, 6:11:17 PM11/21/15
to
FredJeffries
Nov 21, 2015, 6:30:11 PM11/21/15
to
Ross A. Finlayson
Nov 21, 2015, 7:02:34 PM11/21/15
to
Rupert
Nov 21, 2015, 7:13:45 PM11/21/15
to
If you are working in the second-order theory of complete ordered fields it is quite likely that you will take it as an axiom, if on the other hand you are working in ZFC then you can define the field of real numbers and prove that it has the least upper bound property. Just to give two examples.
Ross A. Finlayson
Nov 21, 2015, 7:43:21 PM11/21/15
to
property in modern set-theoretic algebra, how is
this proven?
(Thank you an outline will suffice.)
"Sufficiency" of "completeness" of the "complete
ordered field" is not the same thing, for reals
as roots of polynomials.
https://en.wikipedia.org/wiki/Least-upper-bound_property
"It is possible to prove the least-upper-bound property
using the assumption that every Cauchy sequence of real
numbers converges." Uh - that is a definition of a real
number in ZF, a sequence that is Cauchy, that thus converges,
that of all those there are, the "real number" is those.
Instead it's proven via properties of the image of this
sweep function.
This is for definitions in continuity.
Rupert
Nov 21, 2015, 7:47:21 PM11/21/15
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On Sunday, November 22, 2015 at 1:43:21 AM UTC+1, Ross A. Finlayson wrote:
> On Saturday, November 21, 2015 at 4:13:45 PM UTC-8, Rupert wrote:
> > On Sunday, November 22, 2015 at 1:02:34 AM UTC+1, Ross A. Finlayson wrote:
> > > On Saturday, November 21, 2015 at 3:30:11 PM UTC-8, FredJeffries wrote:
> > > > On Saturday, November 21, 2015 at 3:11:17 PM UTC-8, Rupert wrote:
> > > > > On Saturday, November 21, 2015 at 2:31:26 PM UTC+1, WM wrote:
> > > > > > Assume a list of all infinite digit sequences.
> > > > > >
> > > > > > Every finite digit sequence d_1,d_2,d_3,...,d_n of an antidiagonal can be found (infinitely often) in the remainder of the list:
> > > > > >
> > > > > > For all n in |N there exists k > n such that:
> > > > > >
> > > > > > d_1,d_2,d_3,...,d_n = q_(k1),q_(k2),q_(k3),...,q_(kn)
> > > > > >
> > > > > > There is no contradiction with Cantor's diagonal argument, if formalized in the correct way:
> > > > > >
> > > > > > For all n in |N and for all k =< n:
> > > > > >
> > > > > > d_1,d_2,d_3,...,d_n =/= q_(k1),q_(k2),q_(k3),...,q_(kn)
> > > > > >
> > > > > > Regards, WM
> > > > >
> > > > > The diagonal argument is about the properties of the entire infinite anti-diagonal sequence, not finite initial segments of it.
> > > >
> > > > OOOOOOOOOOOOoooooooooooo..... You said a naughty. The Professor does not allow any talk about sets having properties that cannot be reduced to properties of its elements.
> > >
> > > Is LUB a property or an axiom?
> >
> > Obviously depends on your choice of which axioms to work with, doesn't it.
> >
> > If you are working in the second-order theory of complete ordered fields it is quite likely that you will take it as an axiom, if on the other hand you are working in ZFC then you can define the field of real numbers and prove that it has the least upper bound property. Just to give two examples.
>
> "Gaplessness" property or "Least Upper Bound / LUB"
> property in modern set-theoretic algebra, how is
> this proven?
Well, we need to agree on a definition of the real numbers before I can prove it.
> On Saturday, November 21, 2015 at 4:13:45 PM UTC-8, Rupert wrote:
> > On Sunday, November 22, 2015 at 1:02:34 AM UTC+1, Ross A. Finlayson wrote:
> > > On Saturday, November 21, 2015 at 3:30:11 PM UTC-8, FredJeffries wrote:
> > > > On Saturday, November 21, 2015 at 3:11:17 PM UTC-8, Rupert wrote:
> > > > > On Saturday, November 21, 2015 at 2:31:26 PM UTC+1, WM wrote:
> > > > > > Assume a list of all infinite digit sequences.
> > > > > >
> > > > > > Every finite digit sequence d_1,d_2,d_3,...,d_n of an antidiagonal can be found (infinitely often) in the remainder of the list:
> > > > > >
> > > > > > For all n in |N there exists k > n such that:
> > > > > >
> > > > > > d_1,d_2,d_3,...,d_n = q_(k1),q_(k2),q_(k3),...,q_(kn)
> > > > > >
> > > > > > There is no contradiction with Cantor's diagonal argument, if formalized in the correct way:
> > > > > >
> > > > > > For all n in |N and for all k =< n:
> > > > > >
> > > > > > d_1,d_2,d_3,...,d_n =/= q_(k1),q_(k2),q_(k3),...,q_(kn)
> > > > > >
> > > > > > Regards, WM
> > > > >
> > > > > The diagonal argument is about the properties of the entire infinite anti-diagonal sequence, not finite initial segments of it.
> > > >
> > > > OOOOOOOOOOOOoooooooooooo..... You said a naughty. The Professor does not allow any talk about sets having properties that cannot be reduced to properties of its elements.
> > >
> > > Is LUB a property or an axiom?
> >
> > Obviously depends on your choice of which axioms to work with, doesn't it.
> >
> > If you are working in the second-order theory of complete ordered fields it is quite likely that you will take it as an axiom, if on the other hand you are working in ZFC then you can define the field of real numbers and prove that it has the least upper bound property. Just to give two examples.
>
> "Gaplessness" property or "Least Upper Bound / LUB"
> property in modern set-theoretic algebra, how is
> this proven?
I propose that we define a real number to be a set S of rational numbers which is not empty and not all of Q, with the property that if a is in S and b<a then b is in S, and also if a is in S then there exists a b>a such that b is in S. The order relation is given by inclusion. Are you happy to accept this definition?
Obviously I cannot prove anything unless we first agree on a definition.
Ross A. Finlayson
Nov 21, 2015, 7:57:34 PM11/21/15
to
I have already used in modern
algebra nor as contiguous sequence,
nor Hardy's, and, those are positive
rational numbers.
This is where sometimes
we note or see noted in
class that various definitions
of "continuity" are satisfied
by the rational numbers, into
where we axiomatize the
gaplessness property or LUB.
George Greene
Nov 21, 2015, 8:38:30 PM11/21/15
to
On Saturday, November 21, 2015 at 8:31:26 AM UTC-5, WM wrote:
> There is no contradiction with Cantor's diagonal argument,
> if formalized in the correct way:
There IS SO TOO, dumbass. The anti-diagonal IS NOT ON the list, yet you said there was a list OF ALL the sequences. The anti-diagonal is a sequence and IT IS NOT ON the list. THAT IS a contradiction.
> if formalized in the correct way:
> For all n in |N and for all k =< n:
>
> d_1,d_2,d_3,...,d_n =/= q_(k1),q_(k2),q_(k3),...,q_(kn)
it would still be completely irrelevant, because you are talking about
FINITE PREFIXES of the anti-diagonal, NOT THE WHOLE anti-diagonal -- which
is AN INFINITE digit-sequence -- a kind of thing that you have previously said
does not exist, even though you are talking about an infinite list OF them.
Rupert
Nov 21, 2015, 8:39:12 PM11/21/15
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If you don't like that definition then you give me your definition of a real number, and if it's a correct definition I'll prove that the real numbers so defined have the least upper bound property.
George Greene
Nov 21, 2015, 8:40:09 PM11/21/15
to
On Saturday, November 21, 2015 at 4:41:46 PM UTC-5, Ben Bacarisse wrote:
> q_(kn) is supposed to be the n'th element of the k'th sequence in the
> list. I'd write it q_{k,n}.
Thank you kindly. I sort of though that q was the whole square list but the point is, you have to make HIM say what things are. He never gets it right.
> q_(kn) is supposed to be the n'th element of the k'th sequence in the
> list. I'd write it q_{k,n}.
It has to be HIS error in order for the debate-points to be scored corectly.
George Greene
Nov 21, 2015, 8:42:30 PM11/21/15
to
On Saturday, November 21, 2015 at 5:03:06 PM UTC-5, WM wrote:
> Every row n is followed by infinitely many rows.
> Infinitely many contain the sedquence d_1,d_2,d_3,...,d_n.
> Therefore a statement about every n does not prove anything about all n.
Of course it does. Every MEANS "all" in this context. What you are TRYING to
> Infinitely many contain the sedquence d_1,d_2,d_3,...,d_n.
> Therefore a statement about every n does not prove anything about all n.
say is that it doesn't prove anything about THE WHOLE ANTI-DIAGONAL as a thing.
A statement "about every n" IS a statement "about all n", despite NOT being
a statement about some infinite thing.
If every n has the property that the nth element of the list is not equal to some thing d, then that DOES prove that d is not on the list.
Ross A. Finlayson
Nov 21, 2015, 9:19:45 PM11/21/15
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Jim Burns
Nov 21, 2015, 10:30:19 PM11/21/15
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Rupert, but still it strikes me that you are being too
generous to Ross here.
It seems to me that Ross's question is "Is there a mathematical
object with these properties you all claim for them?"
in particular with the least upper bound property.
In order to show that there is such an object, I don't
think you need Ross's agreement that this object _is_
what we shall call the real numbers -- at least at the
beginning of the argument. That is, of course, where we
want to end.
I think that you do need Ross's agreement on which properties
this mathematical object should have in order to be
considered the real numbers. But I think you already have
Ross's agreement on that. He seems to be disputing (or asking
about, to give it a friendlier spin) that there is such
an object. Or, specifically, an object with the least upper
bound property.
It would also be necessary to prove that the other real-number
properties are true of this object in order to show it is
what we call the real numbers, but since those all are also
true of the rationals, I suspect that they are much easier to
believe in, and Ross may not need much convincing there.
There are several ways to prove that _something we are justified_
_in calling a complete order field_ (AKA the real numbers)
exists and it seems to me that Ross _should_ be satisfied with
any of them.
Ross, do you object, as long as you see how the least upper
bound property is true of the mathematical object?
Ross A. Finlayson
Nov 21, 2015, 10:40:50 PM11/21/15
to
am of the mind that it is axiomatized or
defined so in most constructions of the real
numbers that are so defined via algebra's
complete ordered field, for gaplessness/LUB.
I agree it's consistent: I think it's contrived.
Also I think it's perfectly suitable and a necessary
contrivance there.
What I'm looking for is where the LUB property
is evident or emergent from the construction
of the real numbers (or a reasonable bounded
segment thereof), instead of defined/axiomatized
(as "measure 1.0" is also, again a perfectly
suitable contrivance to keep foundations from
disturbing real analysis foundations).
Virgil
Nov 21, 2015, 10:41:43 PM11/21/15
to
In article <ef1bf8ab-7ff3-47a9...@googlegroups.com>,
Cantor has provided two entirely different but both totally valid
mathematical proofs showing that no surjection from |N to |R is possible.
WM has shown himself totally unable to refute either of those proofs, at
least anywhere in the proper mathematical world free from the many
manifold corruptions of WM's own wild weird wacky worthless world of
WMytheology.
So WM is
WRONG !
AGAIN ! !
AS USUAL ! ! !
WM <wolfgang.m...@hs-augsburg.de> wrote:
> Coming up with claims that are not supported by mathematics is fools
> errand
And one which WM is perpetually engaged in!
> Coming up with claims that are not supported by mathematics is fools
> errand
Cantor has provided two entirely different but both totally valid
mathematical proofs showing that no surjection from |N to |R is possible.
WM has shown himself totally unable to refute either of those proofs, at
least anywhere in the proper mathematical world free from the many
manifold corruptions of WM's own wild weird wacky worthless world of
WMytheology.
So WM is
WRONG !
AGAIN ! !
AS USUAL ! ! !
Virgil
Nov 21, 2015, 10:51:04 PM11/21/15
to
In article <5177bb2b-8695-42f2...@googlegroups.com>,
But Cantor's second , antidiagonal, theorem deos not rely on WM's false
representations of it, nor do WM's phony attacks on Cantor's second
proof vitiate his first proof!
Thus Cantor has two valid proofs of his own claim, while WM has no Valid
proofs of WM's counterclaims.
So WM is
LYING !
(stating that there cannot be any surjection from the set of all
naturals to the set of all reals).
If one assumes every real has been indexed with a different natural, it
will follow below that some real would have to be indexed by a natural
larger than infinitely many other naturals. Which is impossible
everywhere outside of WMytheology.
Lemma; Assume every real has been indexed with a different natural.
Then every open real interval will contain two reals of lowest possible
index for that interval, those points delimiting an open subinterval all
of whose internal points have higher indices than its endpoints.
Proof of lemma : obvious!
Proof of theorem: Iterate the lemma to produce a nested sequence of such
closed intervals each a subset of its predecessor and each interval with
all its points having indices greater than any points outside it.
Such an infinite sequence of nested closed real intervals is known not
to have non-empty intersection. But any point inside all of its
intervals must have a natural number index larger than all the
infinitely many different natural numbers occurring as endpoints of the
infinite sequence of intervals defined above containing it. Which is
impossible in proper mathematics!
Thus the original assumption that one can surject the naturals onto the
reals is false and the theorem is true, and the set of reals is
uncountable!!
representations of it, nor do WM's phony attacks on Cantor's second
proof vitiate his first proof!
Thus Cantor has two valid proofs of his own claim, while WM has no Valid
proofs of WM's counterclaims.
So WM is
LYING !
AGAIN ! !
AS USUAL ! ! !
Cantor's First Proof that the set of reals cannot be enumerated revisited
AS USUAL ! ! !
(stating that there cannot be any surjection from the set of all
naturals to the set of all reals).
If one assumes every real has been indexed with a different natural, it
will follow below that some real would have to be indexed by a natural
larger than infinitely many other naturals. Which is impossible
everywhere outside of WMytheology.
Lemma; Assume every real has been indexed with a different natural.
Then every open real interval will contain two reals of lowest possible
index for that interval, those points delimiting an open subinterval all
of whose internal points have higher indices than its endpoints.
Proof of lemma : obvious!
Proof of theorem: Iterate the lemma to produce a nested sequence of such
closed intervals each a subset of its predecessor and each interval with
all its points having indices greater than any points outside it.
Such an infinite sequence of nested closed real intervals is known not
to have non-empty intersection. But any point inside all of its
intervals must have a natural number index larger than all the
infinitely many different natural numbers occurring as endpoints of the
infinite sequence of intervals defined above containing it. Which is
impossible in proper mathematics!
Thus the original assumption that one can surject the naturals onto the
reals is false and the theorem is true, and the set of reals is
uncountable!!
Virgil
Nov 21, 2015, 10:57:44 PM11/21/15
to
In article <51535547-bd85-445a...@googlegroups.com>,
For any finite initial string of digits, there are uncountably many
reals starting with that finite initial string of digits, so
Mueckenheim's Meaningless Maunderings about finite initial strings of
digits are irrelevant.
reals starting with that finite initial string of digits, so
Mueckenheim's Meaningless Maunderings about finite initial strings of
digits are irrelevant.
Virgil
Nov 21, 2015, 11:00:57 PM11/21/15
to
Jim Burns
Nov 21, 2015, 11:29:30 PM11/21/15
to
Rupert has already offered to prove this to you, and I
don't want to steal any thunder from him.
However, I will offer a quick preview of the proof
(because I find it so beautiful).
Suppose we have the rationals, with the usual operations
on the rationals, '+', '-', '*', '<', those necessary
in order to be an ordered field, such as the rationals or
the reals or some other ordered field.
Suppose we have sets of rationals. In particular, suppose
we have the set D of all sets x of rationals with these four
properties:
Let x e D.
x is not empty
(Ea) a e x
x is not Q
(Ea) ~(a e Q)
x continues all the way down, without holes.
(Aa)(Ab) ( a e x & b < a ) -> b e x
x does not contain a maximum.
(Aa) a e x -> (Eb)( b e x & a < b )
Theorem:
The set of all such x is a complete ordered field.
(This is what Rupert offered to proved to you)
The a and b above here are rationals, and the '<' up there
is the '<' you would expect to use with the rationals.
Another order can be defined on these sets, which
can be proven to have the properties needed for the
order of an ordered field, plus one more property,
the _least upper bound property_ .
(This is a theorem, Ross, not an assertion of an axiom.
The assumption is that these sets in D exist.)
The newly defined order is set inclusion (as Rupert has
already mentioned). if x e D and y e D, then
x is less than y iff x c y and exactly one of
x c y , x = y , y c x is true, and so on.
Here's the beautiful part:
For every bounded, non-empty set A of elements of D
(the things we claim are real numbers because that
is how they behave) the _union of A_ is also in D
(satisfies those four conditions above) and is the
_least upper bound_ of A (bounds A and is less than
or equal to any other bound).
I haven't proven this for you, but it is provable, given no
more assumptions than we have already mentioned.
I think that you would agree that that this is emergent
from the properties of sets and rationals, and not
assumed.
Virgil
Nov 21, 2015, 11:46:11 PM11/21/15
to
In article <5651449A...@att.net>,
A perfectly good model can be formed from equivalence classes of Cauchy
sequences of rationals, where two such sequences are deemed equivalent
if any only if their termwise difference sequence converges to 0.
sequences of rationals, where two such sequences are deemed equivalent
if any only if their termwise difference sequence converges to 0.
Ross A. Finlayson
Nov 22, 2015, 12:23:38 AM11/22/15
to
ordered field with them being their own subsets,
where, as sets they are already equivalence classes
of ratios (reduced fractions) in a usual development
of the rationals from the laws of arithmetic.
That said, besides so that those are rationals
already, and not containing themselves ("x does
not contain a maximum", _itself_), it still is
not so clearly establishing that "the rationals
aren't dense", where, thusly there's a value
(Eb)( b e x & a < b ), Ec a < b < c < x
where, that would both be pretty rough in the
usual US undergraduate abstract algebra curriculum
and also I am not yet finding it compelling.
Rupert
Nov 22, 2015, 1:01:19 AM11/22/15
to
Rupert
Nov 22, 2015, 1:03:07 AM11/22/15
to
Maybe I misunderstood him.
Virgil
Nov 22, 2015, 1:06:56 AM11/22/15
to
In article <745a743a-4b57-4765...@googlegroups.com>,
reals!
Every real is the limit of many Cauchy sequences of rationals, two such
sequences converging to the same real if and only if the sequence of
their termwise differences is a sequence converging to zero.
Having such termwise difference sequence converge to zero is an
equivalence relation on the set of all such convergent sequences of
rationals, and their equivalence classes under that relation is a
standard model for the field of real numbers.
At least among mathematicians!
> > A perfectly good model can be formed from equivalence classes of Cauchy
> > sequences of rationals, where two such sequences are deemed equivalent
> > if any only if their termwise difference sequence converges to 0.
> > --
> > Virgil
> > "Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)
>
> You are conflating elements of D the rationals'
> ordered field with them being their own subsets
You have much yet to learn, including of this standard model for the
> > sequences of rationals, where two such sequences are deemed equivalent
> > if any only if their termwise difference sequence converges to 0.
> > --
> > Virgil
> > "Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)
>
> You are conflating elements of D the rationals'
> ordered field with them being their own subsets
reals!
Every real is the limit of many Cauchy sequences of rationals, two such
sequences converging to the same real if and only if the sequence of
their termwise differences is a sequence converging to zero.
Having such termwise difference sequence converge to zero is an
equivalence relation on the set of all such convergent sequences of
rationals, and their equivalence classes under that relation is a
standard model for the field of real numbers.
At least among mathematicians!
Virgil
Nov 22, 2015, 1:29:17 AM11/22/15
to
In article <b5700358-548a-4c48...@googlegroups.com>,
1. if a in S and b is rational with b < a then b in S, and
2. For every S there is a rational, c, not in S.
because every real is the least upper bound of such an S.
And to keep the representations unique, no S is to have maximal member.
The above will turn out to have for each real number 1 and only 1 set S
as the set of all rationals less than that real number.
> > >
> > > Obviously I cannot prove anything unless we first agree on
> > > a definition.
> >
> > I don't doubt that you know much more about this than I do,
> > Rupert, but still it strikes me that you are being too
> > generous to Ross here.
> >
> > It seems to me that Ross's question is "Is there a mathematical
> > object with these properties you all claim for them?"
> > in particular with the least upper bound property.
> >
> > In order to show that there is such an object, I don't
> > think you need Ross's agreement that this object _is_
> > what we shall call the real numbers -- at least at the
> > beginning of the argument. That is, of course, where we
> > want to end.
> >
> > I think that you do need Ross's agreement on which properties
> > this mathematical object should have in order to be
> > considered the real numbers.
You need the agreement of a knowledgeable mathematician, but Ross is
not, at least not yet, sufficiently knowledgeable.
Rupert <rupertm...@yahoo.com> wrote:
> On Sunday, November 22, 2015 at 4:30:19 AM UTC+1, Jim Burns wrote:
> > On 11/21/2015 7:47 PM, Rupert wrote:
> > > On Sunday, November 22, 2015 at 1:43:21 AM UTC+1,
> > > Ross A. Finlayson wrote:
> >
> > >> "Gaplessness" property or "Least Upper Bound / LUB"
> > >> property in modern set-theoretic algebra, how is
> > >> this proven?
> > >
> > > Well, we need to agree on a definition of the real numbers
> > > before I can prove it.
> > >
> > > I propose that we define a real number to be a set S of
> > > rational numbers which is not empty and not all of Q,
> > > with the property that if a is in S and b<a then b is in S,
> > > and also if a is in S then there exists a b>a such that b
> > > is in S. The order relation is given by inclusion.
> > > Are you happy to accept this definition?
Better would be as the set of sets of rationals S such that
> On Sunday, November 22, 2015 at 4:30:19 AM UTC+1, Jim Burns wrote:
> > On 11/21/2015 7:47 PM, Rupert wrote:
> > > On Sunday, November 22, 2015 at 1:43:21 AM UTC+1,
> > > Ross A. Finlayson wrote:
> >
> > >> "Gaplessness" property or "Least Upper Bound / LUB"
> > >> property in modern set-theoretic algebra, how is
> > >> this proven?
> > >
> > > Well, we need to agree on a definition of the real numbers
> > > before I can prove it.
> > >
> > > I propose that we define a real number to be a set S of
> > > rational numbers which is not empty and not all of Q,
> > > with the property that if a is in S and b<a then b is in S,
> > > and also if a is in S then there exists a b>a such that b
> > > is in S. The order relation is given by inclusion.
> > > Are you happy to accept this definition?
1. if a in S and b is rational with b < a then b in S, and
2. For every S there is a rational, c, not in S.
because every real is the least upper bound of such an S.
And to keep the representations unique, no S is to have maximal member.
The above will turn out to have for each real number 1 and only 1 set S
as the set of all rationals less than that real number.
> > >
> > > Obviously I cannot prove anything unless we first agree on
> > > a definition.
> >
> > I don't doubt that you know much more about this than I do,
> > Rupert, but still it strikes me that you are being too
> > generous to Ross here.
> >
> > It seems to me that Ross's question is "Is there a mathematical
> > object with these properties you all claim for them?"
> > in particular with the least upper bound property.
> >
> > In order to show that there is such an object, I don't
> > think you need Ross's agreement that this object _is_
> > what we shall call the real numbers -- at least at the
> > beginning of the argument. That is, of course, where we
> > want to end.
> >
> > I think that you do need Ross's agreement on which properties
> > this mathematical object should have in order to be
> > considered the real numbers.
not, at least not yet, sufficiently knowledgeable.
Ross A. Finlayson
Nov 22, 2015, 2:05:03 AM11/22/15
to
Ross A. Finlayson
Nov 22, 2015, 2:32:46 AM11/22/15
to
Eg, an irrational number?
2 Assume that there exists an equivalence class of Cauchy sequences of rational numbers, containing one Cauchy sequence s say, which is an upper bound for S.
For example a non-rational real.
3 Then there is a rational number r which is an upper bound for all but finitely many terms of s.
That is not so for rational r and irrational s.
Is that mistaken?
4 Consequencely, given any Cauchy sequence of rational numbers t whose equivalence class belongs to S, the rational number r is an upper bound for all finitely many terms of t.
5 Thus, the rational number r is an upper bound for S, so rational numbers which are upper bounds for S exist.
6 For each positive integer n, let r_n denote the least rational number with denominator dividing n which is an upper bound for S.
7 The sequence (r_1, r_2, ... r_n, ...) is a Cauchy sequence of rational numbers whose equivalence class is a least upper bound for S.
Rupert
Nov 22, 2015, 3:58:48 AM11/22/15
to
For some sufficiently large positive integer N, we have |s_n - s_m| <= 1 for all n, m > N.
Let r = s_(N+1)+1.
This establishes my claim.
Rupert
Nov 22, 2015, 3:59:47 AM11/22/15
to
On Sunday, November 22, 2015 at 7:29:17 AM UTC+1, Virgil wrote:
> In article <b5700358-548a-4c48...@googlegroups.com>,
> Rupert <rupertm...@yahoo.com> wrote:
>
> > On Sunday, November 22, 2015 at 4:30:19 AM UTC+1, Jim Burns wrote:
> > > On 11/21/2015 7:47 PM, Rupert wrote:
> > > > On Sunday, November 22, 2015 at 1:43:21 AM UTC+1,
> > > > Ross A. Finlayson wrote:
> > >
> > > >> "Gaplessness" property or "Least Upper Bound / LUB"
> > > >> property in modern set-theoretic algebra, how is
> > > >> this proven?
> > > >
> > > > Well, we need to agree on a definition of the real numbers
> > > > before I can prove it.
> > > >
> > > > I propose that we define a real number to be a set S of
> > > > rational numbers which is not empty and not all of Q,
> > > > with the property that if a is in S and b<a then b is in S,
> > > > and also if a is in S then there exists a b>a such that b
> > > > is in S. The order relation is given by inclusion.
> > > > Are you happy to accept this definition?
>
>
> Better would be as the set of sets of rationals S such that
> 1. if a in S and b is rational with b < a then b in S, and
> 2. For every S there is a rational, c, not in S.
> because every real is the least upper bound of such an S.
> And to keep the representations unique, no S is to have maximal member.
That's equivalent to the definition I gave.
> In article <b5700358-548a-4c48...@googlegroups.com>,
> Rupert <rupertm...@yahoo.com> wrote:
>
> > On Sunday, November 22, 2015 at 4:30:19 AM UTC+1, Jim Burns wrote:
> > > On 11/21/2015 7:47 PM, Rupert wrote:
> > > > On Sunday, November 22, 2015 at 1:43:21 AM UTC+1,
> > > > Ross A. Finlayson wrote:
> > >
> > > >> "Gaplessness" property or "Least Upper Bound / LUB"
> > > >> property in modern set-theoretic algebra, how is
> > > >> this proven?
> > > >
> > > > Well, we need to agree on a definition of the real numbers
> > > > before I can prove it.
> > > >
> > > > I propose that we define a real number to be a set S of
> > > > rational numbers which is not empty and not all of Q,
> > > > with the property that if a is in S and b<a then b is in S,
> > > > and also if a is in S then there exists a b>a such that b
> > > > is in S. The order relation is given by inclusion.
> > > > Are you happy to accept this definition?
>
>
> Better would be as the set of sets of rationals S such that
> 1. if a in S and b is rational with b < a then b in S, and
> 2. For every S there is a rational, c, not in S.
> because every real is the least upper bound of such an S.
> And to keep the representations unique, no S is to have maximal member.
Virgil
Nov 22, 2015, 5:17:01 AM11/22/15
to
In article <dd815df4-e6fa-4ca6...@googlegroups.com>,
Not quite!
In mine, a rational can be the largest member of a set S,
in yours no S can ever have a largest member.
So in mine, there is a clear distinction between those S's representing
rationals and those S's representing irrationals which is not present
in yours.
In mine, a rational can be the largest member of a set S,
in yours no S can ever have a largest member.
So in mine, there is a clear distinction between those S's representing
rationals and those S's representing irrationals which is not present
in yours.
WM
Nov 22, 2015, 5:23:22 AM11/22/15
to
Am Sonntag, 22. November 2015 00:11:17 UTC+1 schrieb Rupert:
> The diagonal argument is about the properties of the entire infinite anti-diagonal sequence, not finite initial segments of it.
If an infinite sequence cannot be distinguished from all finite sequences then it is interesting to know why its entirety is assumed to exist as an entity different from the set of all finite sequences or how this could be proved.
> The diagonal argument is about the properties of the entire infinite anti-diagonal sequence, not finite initial segments of it.
Regards, WM
WM
Nov 22, 2015, 5:23:23 AM11/22/15
to
Am Sonntag, 22. November 2015 00:06:28 UTC+1 schrieb Dan Christensen:
> > > D is an infinite string of 0's and 1's. So it is an element of T. It also differs from every string in the range of f.
> >
> > For every n in |N: d_1,d_2,d_3,...,d_n differs from the first n rows.
>
> If you are talking about the anti-diagonal string D as defined above
> > > D is an infinite string of 0's and 1's. So it is an element of T. It also differs from every string in the range of f.
> >
> > For every n in |N: d_1,d_2,d_3,...,d_n differs from the first n rows.
>
I would not talk about it unless I find some proof that it differs from all finite initial segments. It's like in the Binary Tree:
If an infinite path cannot be distinguished from all finite paths then it is interesting to know why its entirety is assumed to exist as an entity different from the set of all finite paths or how this could be proved
> (in the part you snipped), it will differ from every string in the range of f. In the string given by f(k), D will differ from f(k) at position k.
And beyond the k-th row there will be infinitely many duplicates of d_1,d_2,d_3,...,d_n. This holds for every k in |N. Do you not agree?
Regards, WM
WM
Nov 22, 2015, 5:29:43 AM11/22/15
to
Am Sonntag, 22. November 2015 02:38:30 UTC+1 schrieb George Greene:
> On Saturday, November 21, 2015 at 8:31:26 AM UTC-5, WM wrote:
>
> > There is no contradiction with Cantor's diagonal argument,
> > if formalized in the correct way:
>
> There IS SO TOO, dumbass. The anti-diagonal IS NOT ON the list, yet you said there was a list OF ALL the sequences. The anti-diagonal is a sequence and IT IS NOT ON the list. THAT IS a contradiction.
Every initial segment is infinitely often on the list. Since zthe infinite sequence is nothing but all its finite initial segments, this is a contraediction
> On Saturday, November 21, 2015 at 8:31:26 AM UTC-5, WM wrote:
>
> > There is no contradiction with Cantor's diagonal argument,
> > if formalized in the correct way:
>
> There IS SO TOO, dumbass. The anti-diagonal IS NOT ON the list, yet you said there was a list OF ALL the sequences. The anti-diagonal is a sequence and IT IS NOT ON the list. THAT IS a contradiction.
In order to resolve it, you must believe that the infinite sequence differs from all (not only from every) finite segment. That is easily shown to be wrong.
>
> > For all n in |N and for all k =< n:
> >
> > d_1,d_2,d_3,...,d_n =/= q_(k1),q_(k2),q_(k3),...,q_(kn)
>
> This is completely unintelligible to begin with, but even if it made sense,
It is obvious.
> > For all n in |N and for all k =< n:
> >
> > d_1,d_2,d_3,...,d_n =/= q_(k1),q_(k2),q_(k3),...,q_(kn)
>
> This is completely unintelligible to begin with, but even if it made sense,
> it would still be completely irrelevant, because you are talking about
> FINITE PREFIXES of the anti-diagonal
>, NOT THE WHOLE anti-diagonal -- which
> is AN INFINITE digit-sequence -- a kind of thing that you have previously said
> does not exist, even though you are talking about an infinite list OF them.
Regards, WM
WM
Nov 22, 2015, 5:31:58 AM11/22/15
to
Am Sonntag, 22. November 2015 02:42:30 UTC+1 schrieb George Greene:
> A statement "about every n" IS a statement "about all n"
Every finite set of rational numbers can be well-ordered by size. So, according to your logic, all rational numbers can be well-ordered by size.
> A statement "about every n" IS a statement "about all n"
Every natural number is followed by infinitely many others. So all natural numbers are followed by infinitely many others?
REgards, WM
Alan Smaill
Nov 22, 2015, 5:35:03 AM11/22/15
to
WM <wolfgang.m...@hs-augsburg.de> writes:
> Am Sonntag, 22. November 2015 00:11:17 UTC+1 schrieb Rupert:
>
>
>> The diagonal argument is about the properties of the entire infinite
>> anti-diagonal sequence, not finite initial segments of it.
>
> If an infinite sequence cannot be distinguished from all finite
> sequences
Can you yourself distinguish the (potentially) infinite sequence
> Am Sonntag, 22. November 2015 00:11:17 UTC+1 schrieb Rupert:
>
>
>> The diagonal argument is about the properties of the entire infinite
>> anti-diagonal sequence, not finite initial segments of it.
>
> If an infinite sequence cannot be distinguished from all finite
> sequences
1,2,3,... from the finite sequence 1,2 ?
>
> Regards, WM
--
Alan Smaill
Virgil
Nov 22, 2015, 5:35:24 AM11/22/15
to
In article <1352c25c-0833-41cc...@googlegroups.com>,
number, so any such nonempty set would have to be a singleton set having
only one Cauchy sequence member.
terms are larger having the same limit, so here cannot be any such thng
as an upper bound for a set S.
>
> For example a non-rational real.
>
> 3 Then there is a rational number r which is an upper bound for all
> but finitely many terms of s.
For every Cauchy sequence there is a rational upper bound on its entire
set of terms!
> 1 Let S be a nonempty set of equivalence classes of Cauchy sequences
> of rational numbers.
Any one equivalence class of Cauchy sequences represents one real
> of rational numbers.
number, so any such nonempty set would have to be a singleton set having
only one Cauchy sequence member.
>
> Eg, an irrational number?
>
> 2 Assume that there exists an equivalence class of Cauchy sequences
> of rational numbers, containing one Cauchy sequence s say, which is
> an upper bound for S.
Given any Cauchy sequence, there is another Cauchy sequence all of whose
> Eg, an irrational number?
>
> 2 Assume that there exists an equivalence class of Cauchy sequences
> of rational numbers, containing one Cauchy sequence s say, which is
> an upper bound for S.
terms are larger having the same limit, so here cannot be any such thng
as an upper bound for a set S.
>
> For example a non-rational real.
>
> 3 Then there is a rational number r which is an upper bound for all
> but finitely many terms of s.
set of terms!
Virgil
Nov 22, 2015, 5:41:02 AM11/22/15
to
In article <0c86f7bf-30a5-4f29...@googlegroups.com>,
The whole point is that sequences of rationals that converge to a real
in |R all are Cauchy sequences in |Q.
in |R all are Cauchy sequences in |Q.
Virgil
Nov 22, 2015, 5:48:49 AM11/22/15
to
In article <8089fb89-94bf-469f...@googlegroups.com>,
worthless wacky world of WMytheology.
WM <wolfgang.m...@hs-augsburg.de> wrote:
> Am Sonntag, 22. November 2015 02:42:30 UTC+1 schrieb George Greene:
>
>
> > A statement "about every n" IS a statement "about all n"
>
> Every finite set of rational numbers can be well-ordered by size. So,
> according to your logic, all rational numbers can be well-ordered by
> size.
Perhaps by WM's corrupt logic, but nowhere outside of WM's witless
> Am Sonntag, 22. November 2015 02:42:30 UTC+1 schrieb George Greene:
>
>
> > A statement "about every n" IS a statement "about all n"
>
> Every finite set of rational numbers can be well-ordered by size. So,
> according to your logic, all rational numbers can be well-ordered by
> size.
worthless wacky world of WMytheology.
Virgil
Nov 22, 2015, 5:52:19 AM11/22/15
to
In article <94cf3fc2-26e5-49d5...@googlegroups.com>,
> wrong.
Any infinite sequence differs from any and every finite sequence
everywhere outside of WM's witless worthless wacky world of WMytheology.
So WM is
LYING !
AGAIN ! !
AS USUAL ! ! !
WM <wolfgang.m...@hs-augsburg.de> wrote:
> Every initial segment is infinitely often on the list. Since zthe infinite
> sequence is nothing but all its finite initial segments, this is a
> contraediction
>
> In order to resolve it, you must believe that the infinite sequence differs
> from all (not only from every) finite segments. That is easily shown to be
> Every initial segment is infinitely often on the list. Since zthe infinite
> sequence is nothing but all its finite initial segments, this is a
> contraediction
>
> In order to resolve it, you must believe that the infinite sequence differs
> wrong.
Any infinite sequence differs from any and every finite sequence
everywhere outside of WM's witless worthless wacky world of WMytheology.
So WM is
LYING !
AGAIN ! !
AS USUAL ! ! !
Virgil
Nov 22, 2015, 5:57:41 AM11/22/15
to
In article <0e0191d9-2a5e-42e6...@googlegroups.com>,
WMytheology.
>
agree, since if they differ properly in even one place they are
different numerals and represent different numbers.
So WM is
WRONG !
WM <wolfgang.m...@hs-augsburg.de> wrote:
> Am Sonntag, 22. November 2015 00:06:28 UTC+1 schrieb Dan Christensen:
>
>
> > > > D is an infinite string of 0's and 1's. So it is an element of T. It
> > > > also differs from every string in the range of f.
> > >
> > > For every n in |N: d_1,d_2,d_3,...,d_n differs from the first n rows.
> >
> > If you are talking about the anti-diagonal string D as defined above
>
> I would not talk about it unless I find some proof that it differs from all
> finite initial segments. It's like in the Binary Tree:
>
> If an infinite path cannot be distinguished from all finite paths
It can be everywhere outside of WM's witless worthless wacky world of
> Am Sonntag, 22. November 2015 00:06:28 UTC+1 schrieb Dan Christensen:
>
>
> > > > D is an infinite string of 0's and 1's. So it is an element of T. It
> > > > also differs from every string in the range of f.
> > >
> > > For every n in |N: d_1,d_2,d_3,...,d_n differs from the first n rows.
> >
> > If you are talking about the anti-diagonal string D as defined above
>
> I would not talk about it unless I find some proof that it differs from all
> finite initial segments. It's like in the Binary Tree:
>
> If an infinite path cannot be distinguished from all finite paths
WMytheology.
>
> > (in the part you snipped), it will differ from every string in the range of
> > f. In the string given by f(k), D will differ from f(k) at position k.
>
> And beyond the k-th row there will be infinitely many duplicates of
> d_1,d_2,d_3,...,d_n. This holds for every k in |N. Do you not agree?
It is irrelevant at how many places two sequences of decimal digits
> > f. In the string given by f(k), D will differ from f(k) at position k.
>
> And beyond the k-th row there will be infinitely many duplicates of
> d_1,d_2,d_3,...,d_n. This holds for every k in |N. Do you not agree?
agree, since if they differ properly in even one place they are
different numerals and represent different numbers.
So WM is
WRONG !
Virgil
Nov 22, 2015, 5:59:40 AM11/22/15
to
In article <346c6c5c-84fc-4b38...@googlegroups.com>,
sequence needs more schooling!
WM <wolfgang.m...@hs-augsburg.de> wrote:
> Am Sonntag, 22. November 2015 00:11:17 UTC+1 schrieb Rupert:
>
>
> > The diagonal argument is about the properties of the entire
> > infinite anti-diagonal sequence, not finite initial segments of it.
>
> If an infinite sequence cannot be distinguished from all finite
> sequences
Anyone like WM who cannot distinguish a finite sequence from an infinite
> Am Sonntag, 22. November 2015 00:11:17 UTC+1 schrieb Rupert:
>
>
> > The diagonal argument is about the properties of the entire
> > infinite anti-diagonal sequence, not finite initial segments of it.
>
> If an infinite sequence cannot be distinguished from all finite
> sequences
sequence needs more schooling!
Alan Smaill
Nov 22, 2015, 6:05:02 AM11/22/15
to
WM <wolfgang.m...@hs-augsburg.de> writes:
> Am Sonntag, 22. November 2015 02:42:30 UTC+1 schrieb George Greene:
>
>> A statement "about every n" IS a statement "about all n"
>
> Every finite set of rational numbers can be well-ordered by size. So,
> according to your logic, all rational numbers can be well-ordered by
> size.
No -- according to the "logic", a statement about every finite set
> Am Sonntag, 22. November 2015 02:42:30 UTC+1 schrieb George Greene:
>
>> A statement "about every n" IS a statement "about all n"
>
> Every finite set of rational numbers can be well-ordered by size. So,
> according to your logic, all rational numbers can be well-ordered by
> size.
is a statement about all finite sets.
Do you think that there are only finitely many rationals?
> Every natural number is followed by infinitely many others. So all
> natural numbers are followed by infinitely many others?
all n (n in |N -> {m in |N: m > n} is infinite)
or whether
{ m in |N : all n in |N. m > n } is infinite ?
>
> REgards, WM
--
Alan Smaill
WM
Nov 22, 2015, 6:34:15 AM11/22/15
to
Am Sonntag, 22. November 2015 12:05:02 UTC+1 schrieb Alan Smaill:
> WM <wolfgang.m...@hs-augsburg.de> writes:
>
> > Am Sonntag, 22. November 2015 02:42:30 UTC+1 schrieb George Greene:
> >
> >> A statement "about every n" IS a statement "about all n"
> >
> > Every finite set of rational numbers can be well-ordered by size. So,
> > according to your logic, all rational numbers can be well-ordered by
> > size.
>
> No -- according to the "logic", a statement about every finite set
> is a statement about all finite sets.
> Do you think that there are only finitely many rationals?
Why did you delete the countability proof for infinitely many rationals?
> WM <wolfgang.m...@hs-augsburg.de> writes:
>
> > Am Sonntag, 22. November 2015 02:42:30 UTC+1 schrieb George Greene:
> >
> >> A statement "about every n" IS a statement "about all n"
> >
> > Every finite set of rational numbers can be well-ordered by size. So,
> > according to your logic, all rational numbers can be well-ordered by
> > size.
>
> No -- according to the "logic", a statement about every finite set
> is a statement about all finite sets.
> Do you think that there are only finitely many rationals?
Regards, WM
WM
Nov 22, 2015, 6:34:45 AM11/22/15
to
Regards, WM
Rupert
Nov 22, 2015, 7:50:23 AM11/22/15
to
Rupert
Nov 22, 2015, 7:51:26 AM11/22/15
to
Ben Bacarisse
Nov 22, 2015, 8:51:32 AM11/22/15
to
WM <wolfgang.m...@hs-augsburg.de> writes:
> Am Samstag, 21. November 2015 17:14:22 UTC+1 schrieb Ben Bacarisse:
>>
>> exists m in |N such that for all i in |N d_i = q_{m,i}
>
> That is impossible because there is nothing like all i.
Odd that both your claims have the same quantifiers. But I'm glad you
can't prove anything that contradicts set theory. Keep posting provable
results about non-existent sequences!
<snip>
--
Ben.
> Am Samstag, 21. November 2015 17:14:22 UTC+1 schrieb Ben Bacarisse:
>> WM <wolfgang.m...@hs-augsburg.de> writes:
>>
>> > Assume a list of all infinite digit sequences.
>> >
>> > Every finite digit sequence d_1,d_2,d_3,...,d_n of an antidiagonal can
>> > be found (infinitely often) in the remainder of the list:
>> >
>> > For all n in |N there exists k > n such that:
>> >
>> > d_1,d_2,d_3,...,d_n = q_(k1),q_(k2),q_(k3),...,q_(kn)
>> >
>>
>> > Assume a list of all infinite digit sequences.
>> >
>> > Every finite digit sequence d_1,d_2,d_3,...,d_n of an antidiagonal can
>> > be found (infinitely often) in the remainder of the list:
>> >
>> > For all n in |N there exists k > n such that:
>> >
>> > d_1,d_2,d_3,...,d_n = q_(k1),q_(k2),q_(k3),...,q_(kn)
>> >
>> > There is no contradiction with Cantor's diagonal argument, if
>> > formalized in the correct way:
>> >
>> > formalized in the correct way:
>> >
>> > For all n in |N and for all k =< n:
>> >
>> > d_1,d_2,d_3,...,d_n =/= q_(k1),q_(k2),q_(k3),...,q_(kn)
>>
>> Yet you can not prove that
>> >
>> > d_1,d_2,d_3,...,d_n =/= q_(k1),q_(k2),q_(k3),...,q_(kn)
>>
>>
>> exists m in |N such that for all i in |N d_i = q_{m,i}
>
> That is impossible because there is nothing like all i.
Odd that both your claims have the same quantifiers. But I'm glad you
can't prove anything that contradicts set theory. Keep posting provable
results about non-existent sequences!
<snip>
--
Ben.
Dan Christensen
Nov 22, 2015, 10:28:54 AM11/22/15
to
On Sunday, November 22, 2015 at 5:23:23 AM UTC-5, WM wrote:
> Am Sonntag, 22. November 2015 00:06:28 UTC+1 schrieb Dan Christensen:
>
>
> > > > D is an infinite string of 0's and 1's. So it is an element of T. It also differs from every string in the range of f.
> > >
> > > For every n in |N: d_1,d_2,d_3,...,d_n differs from the first n rows.
> >
> > If you are talking about the anti-diagonal string D as defined above
>
What WM snipped without indicating:
> Am Sonntag, 22. November 2015 00:06:28 UTC+1 schrieb Dan Christensen:
>
>
> > > > D is an infinite string of 0's and 1's. So it is an element of T. It also differs from every string in the range of f.
> > >
> > > For every n in |N: d_1,d_2,d_3,...,d_n differs from the first n rows.
> >
> > If you are talking about the anti-diagonal string D as defined above
>
If you are talking about the anti-diagonal string D as defined above (in the part you snipped), it will differ from every string in the range of f. In the string given by f(k), D will differ from f(k) at position k. Do you not agree?
> I would not talk about it unless I find some proof that it differs from all finite initial segments. It's like in the Binary Tree:
>
I'm not interested in your hand-waving obfuscations attempting to muddy the waters here, Mucke. It reminds me of a very weak student's failed attempt to solve a problem. As here, it was often easier to simply give the correct solution than to pinpoint all the many silly errors made by that student. EOD
**********
More absurd quotes from Wolfgang Muckenheim (WM):
"In my system, two different numbers can have the same value."
-- sci.math, 2014/10/16
"1+2 and 2+1 are different numbers."
-- sci.math, 2014/10/20
"1/9 has no decimal representation."
-- sci.math, 2015/09/22
"Axioms are rubbish!"
-- sci.math, 2014/11/19
"No set is countable, not even |N."
-- sci.logic, 2015/08/05
"A [natural] number with aleph_0 digits is not less than aleph_0."
-- sci.math, 2015/08/12
"The notion of aleph_0 is not meaningful."
-- sci.math, 2015/08/28
Slipping ever more deeply into madness...
"There is no actually infinite set |N."
-- sci.math, 2015/10/26
"|N is not covered by the set of natural numbers."
-- sci.math, 2015/10/26
A special word of caution to students: Do not attempt to use WM's "system" in any course work in any high school, college or university on the planet. You will fail miserably. His system is certainly no "shortcut" to success in mathematics. It is truly a dead-end.
Dan
Download my DC Proof 2.0 software at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
Virgil
Nov 22, 2015, 1:56:14 PM11/22/15
to
In article <a58b9c03-9c35-4ffb...@googlegroups.com>,
sequences because outside of WM's witless worthless wacky world of
WMytheology no such thing exists.
WM <wolfgang.m...@hs-augsburg.de> wrote:
> > Can you yourself distinguish the (potentially) infinite sequence
> > 1,2,3,... from the finite sequence 1,2 ?
>
> The potentially infinite sequence cannot be distinguished from all finite
> sequences because it is not more than a finite sequence with variable upper
> bound.
A potentially infinite sequence cannot be distinguished from finite
> > Can you yourself distinguish the (potentially) infinite sequence
> > 1,2,3,... from the finite sequence 1,2 ?
>
> The potentially infinite sequence cannot be distinguished from all finite
> sequences because it is not more than a finite sequence with variable upper
> bound.
sequences because outside of WM's witless worthless wacky world of
WMytheology no such thing exists.
WM
Nov 22, 2015, 2:05:57 PM11/22/15
to
Am Sonntag, 22. November 2015 14:51:32 UTC+1 schrieb Ben Bacarisse:
> WM <wolfgang.m...@hs-augsburg.de> writes:
>
> > Am Samstag, 21. November 2015 17:14:22 UTC+1 schrieb Ben Bacarisse:
> >> WM <wolfgang.m...@hs-augsburg.de> writes:
> >>
> >> > Assume a list of all infinite digit sequences.
> >> >
> >> > Every finite digit sequence d_1,d_2,d_3,...,d_n of an antidiagonal can
> >> > be found (infinitely often) in the remainder of the list:
> >> >
> >> > For all n in |N there exists k > n such that:
> >> >
> >> > d_1,d_2,d_3,...,d_n = q_(k1),q_(k2),q_(k3),...,q_(kn)
> >> >
> >> > There is no contradiction with Cantor's diagonal argument, if
> >> > formalized in the correct way:
> >> >
> >> > For all n in |N and for all k =< n:
> >> >
> >> > d_1,d_2,d_3,...,d_n =/= q_(k1),q_(k2),q_(k3),...,q_(kn)
> >>
> >> Yet you can not prove that
> >>
> >> exists m in |N such that for all i in |N d_i = q_{m,i}
> >
> > That is impossible because there is nothing like all i.
>
> Odd that both your claims have the same quantifiers.
This is not a claim but the explanation of the resulting contradiction. It results from the assumption that there are aleph_0 natural numbers n. For each n we find
> WM <wolfgang.m...@hs-augsburg.de> writes:
>
> > Am Samstag, 21. November 2015 17:14:22 UTC+1 schrieb Ben Bacarisse:
> >> WM <wolfgang.m...@hs-augsburg.de> writes:
> >>
> >> > Assume a list of all infinite digit sequences.
> >> >
> >> > Every finite digit sequence d_1,d_2,d_3,...,d_n of an antidiagonal can
> >> > be found (infinitely often) in the remainder of the list:
> >> >
> >> > For all n in |N there exists k > n such that:
> >> >
> >> > d_1,d_2,d_3,...,d_n = q_(k1),q_(k2),q_(k3),...,q_(kn)
> >> >
> >> > There is no contradiction with Cantor's diagonal argument, if
> >> > formalized in the correct way:
> >> >
> >> > For all n in |N and for all k =< n:
> >> >
> >> > d_1,d_2,d_3,...,d_n =/= q_(k1),q_(k2),q_(k3),...,q_(kn)
> >>
> >> Yet you can not prove that
> >>
> >> exists m in |N such that for all i in |N d_i = q_{m,i}
> >
> > That is impossible because there is nothing like all i.
>
> Odd that both your claims have the same quantifiers.
d_1,d_2,d_3,...,d_n = q_(k1),q_(k2),q_(k3),...,q_(kn)
Is there anything wrong?
> But I'm glad you
> can't prove anything that contradicts set theory.
We can assume aleph_0 approximations to pi. That does not make pi being represented by fractions or digits. Therefore no diagonal number will represent an irrational number.
Regards, WM
WM
Nov 22, 2015, 2:06:05 PM11/22/15
to
Am Sonntag, 22. November 2015 13:51:26 UTC+1 schrieb Rupert:
> On Sunday, November 22, 2015 at 11:23:22 AM UTC+1, WM wrote:
> > Am Sonntag, 22. November 2015 00:11:17 UTC+1 schrieb Rupert:
> >
> >
> > > The diagonal argument is about the properties of the entire infinite anti-diagonal sequence, not finite initial segments of it.
> >
> > If an infinite sequence cannot be distinguished from all finite sequences then it is interesting to know why its entirety is assumed to exist as an entity different from the set of all finite sequences or how this could be proved.
>
> Well, are you happy to allow us to assume the Axiom of Infinity?
Of course. Applying it you will find aleph_0 rational approximations but not a representation of an irrational number by fractions (= digits).
> On Sunday, November 22, 2015 at 11:23:22 AM UTC+1, WM wrote:
> > Am Sonntag, 22. November 2015 00:11:17 UTC+1 schrieb Rupert:
> >
> >
> > > The diagonal argument is about the properties of the entire infinite anti-diagonal sequence, not finite initial segments of it.
> >
> > If an infinite sequence cannot be distinguished from all finite sequences then it is interesting to know why its entirety is assumed to exist as an entity different from the set of all finite sequences or how this could be proved.
>
> Well, are you happy to allow us to assume the Axiom of Infinity?
Regards, WM
WM
Nov 22, 2015, 2:06:13 PM11/22/15
to
Am Sonntag, 22. November 2015 16:28:54 UTC+1 schrieb Dan Christensen:
> So, you cannot say whether D differs from every string in the range of f. Oh, well.
You cannot prove that D exists. All you can prove using the AI and the definition of aleph_0 is that there are aleph_0 natural numbers n such that
> So, you cannot say whether D differs from every string in the range of f. Oh, well.
d_1,d_2,d_3,...,d_n = q_(k1),q_(k2),q_(k3),...,q_(kn)
for infinitely many k > n. Everything else is handwaving and belief in matheology.
Regards, WM
Virgil
Nov 22, 2015, 2:13:57 PM11/22/15
to
In article <8188db72-1acd-4959...@googlegroups.com>,
automatically well-ordered as well.
Only actually infinite sets can be totally ordered in any other way!
While the set of all rationals CAN be well-ordered, but is not then
simultaneously densely ordered.
WM <wolfgang.m...@hs-augsburg.de> wrote:
> Every finite set of rational numbers can be well-ordered by size.
Every finite set of any sort whatsoever that is totally ordered is
> Every finite set of rational numbers can be well-ordered by size.
automatically well-ordered as well.
Only actually infinite sets can be totally ordered in any other way!
While the set of all rationals CAN be well-ordered, but is not then
simultaneously densely ordered.
Rupert
Nov 22, 2015, 2:18:28 PM11/22/15
to
Virgil
Nov 22, 2015, 2:57:05 PM11/22/15
to
In article <58930042-c9ac-4ef2...@googlegroups.com>,
WM <wolfgang.m...@hs-augsburg.de> wrote:
> Everything else is handwaving and belief in matheology.
>
> Regards, WM
Everything by WM is handwaving and belief in his own WMytheology.
> Everything else is handwaving and belief in matheology.
>
> Regards, WM
Virgil
Nov 22, 2015, 3:04:22 PM11/22/15
to
In article <fed67080-9745-45d6...@googlegroups.com>,
Since one cannot represent most fractions as integers, in WMytheology
one can, by the same logic claim no such fractions can exist.
And since one cannot represent most naturals by numreal, in WMytheology
one can, by the same logic claim no such naturals can exist.
But those outside of WM's witless worthless wacky world of WMytheology,
agree that even without being able to find names for all of them, such
things as integers, rationals and reals exist.
So WM is
WRONG !
AGAIN ! !
AS USUAL ! ! !
one can, by the same logic claim no such fractions can exist.
And since one cannot represent most naturals by numreal, in WMytheology
one can, by the same logic claim no such naturals can exist.
But those outside of WM's witless worthless wacky world of WMytheology,
agree that even without being able to find names for all of them, such
things as integers, rationals and reals exist.
So WM is
WRONG !
AGAIN ! !
AS USUAL ! ! !
Dan Christensen
Nov 22, 2015, 3:05:35 PM11/22/15
to
On Sunday, November 22, 2015 at 2:06:13 PM UTC-5, WM wrote:
> Am Sonntag, 22. November 2015 16:28:54 UTC+1 schrieb Dan Christensen:
>
>
> > So, you cannot say whether D differs from every string in the range of f. Oh, well.
>
> You cannot prove that D exists.
Yes, you can. The strings in question are just functions mapping N to the set of characters used (at least 2 characters). You can, for example, construct the function D: N --> {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} (for base-10) such that D(k)=1 if the the kth position of the kth string in your list is 0; 0 otherwise. You can then easily prove that D is the required string.
> Am Sonntag, 22. November 2015 16:28:54 UTC+1 schrieb Dan Christensen:
>
>
> > So, you cannot say whether D differs from every string in the range of f. Oh, well.
>
> You cannot prove that D exists.
See a similar construction of an anti-diagonal h in my proof, already posted here at http://www.dcproof.com/CantorDiagonal.htm
Looks like you have failed again, Mucke.
Virgil
Nov 22, 2015, 3:20:56 PM11/22/15
to
In article <0e055822-7fe4-400f...@googlegroups.com>,
Unless that non-terminating anti-diagonal can be proved always and
necessarily to be eventually repeating, it very well can represent an
irrational, as that as what all such non-terminating and never
eventually repeating decimals do!
A nonterminating not-eventually-repeating decimal, by defining a
monotone bounded sequence of rationals, always defines a real number
which by construction cannot be rational.
E.g. SUM(n in |N) 1/10^(n!)
If WM wishes to claim this number to be rational, let him prove it by
presenting its alleged fractional form.
necessarily to be eventually repeating, it very well can represent an
irrational, as that as what all such non-terminating and never
eventually repeating decimals do!
A nonterminating not-eventually-repeating decimal, by defining a
monotone bounded sequence of rationals, always defines a real number
which by construction cannot be rational.
E.g. SUM(n in |N) 1/10^(n!)
If WM wishes to claim this number to be rational, let him prove it by
presenting its alleged fractional form.
Jim Burns
Nov 22, 2015, 3:31:09 PM11/22/15
to
On 11/22/2015 1:03 AM, Rupert wrote:
> On Sunday, November 22, 2015 at 4:30:19 AM UTC+1,
> Jim Burns wrote:
>> I think that you do need Ross's agreement on which properties
> On Sunday, November 22, 2015 at 4:30:19 AM UTC+1,
> Jim Burns wrote:
>> this mathematical object should have in order to be
>> considered the real numbers. But I think you already have
>> Ross's agreement on that.
>
> Yeah, okay, sure, but then I would need to prove all of the
> complete ordered field axioms, not just the least upper bound
> property. I was hoping to avoid the effort of doing that by
> getting an agreement that my definition is correct. I take
> that to be the main point of the dispute here.
>
> Maybe I misunderstood him.
No, you're probably right. I think I didn't think my
post through well enough.
It could also be that I am biased toward the Dedekind
cut model of the reals and I was disappointed that it
got put aside.
The l.u.b. is the union! Who wouldn't fall in love with that?
(I never claimed I was not strange.)
WM
Nov 22, 2015, 3:32:25 PM11/22/15
to
Am Sonntag, 22. November 2015 21:05:35 UTC+1 schrieb Dan Christensen:
> On Sunday, November 22, 2015 at 2:06:13 PM UTC-5, WM wrote:
> > Am Sonntag, 22. November 2015 16:28:54 UTC+1 schrieb Dan Christensen:
> >
> >
> > > So, you cannot say whether D differs from every string in the range of f. Oh, well.
> >
> > You cannot prove that D exists.
>
> Yes, you can. The strings in question are just functions mapping N to the set of characters used (at least 2 characters). You can, for example, construct the function D: N --> {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} (for base-10) such that D(k)=1 if the the kth position of the kth string in your list is 0; 0 otherwise. You can then easily prove that D is the required string.
The 5 lines above are a finite definition. Of course it is possible to define an infinite sequence by a finite word, like "0.111...", or like you can conjure the devil and believe that he is present.
> On Sunday, November 22, 2015 at 2:06:13 PM UTC-5, WM wrote:
> > Am Sonntag, 22. November 2015 16:28:54 UTC+1 schrieb Dan Christensen:
> >
> >
> > > So, you cannot say whether D differs from every string in the range of f. Oh, well.
> >
> > You cannot prove that D exists.
>
> Yes, you can. The strings in question are just functions mapping N to the set of characters used (at least 2 characters). You can, for example, construct the function D: N --> {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} (for base-10) such that D(k)=1 if the the kth position of the kth string in your list is 0; 0 otherwise. You can then easily prove that D is the required string.
It is impossible however to distinguish a really infinite sequence from all its finite segments.
Probably you will not be able to understand the difference. Most mathematicians need more than one go to understand.
Regards, WM
Virgil
Nov 22, 2015, 3:50:23 PM11/22/15
to
In article <a15d2be1-3e49-414d...@googlegroups.com>,
WM <wolfgang.m...@hs-augsburg.de> wrote:
> It is impossible however to distinguish a really infinite sequence
> from all its finite segments.
Not for a mathematician!
> It is impossible however to distinguish a really infinite sequence
> from all its finite segments.
Dan Christensen
Nov 22, 2015, 3:55:46 PM11/22/15
to
Jim Burns
Nov 22, 2015, 4:06:22 PM11/22/15
to
On 11/22/2015 12:23 AM, Ross A. Finlayson wrote:
> On Saturday, November 21, 2015 at 8:46:11 PM UTC-8,
> Virgil wrote:
>> In article <5651449A...@att.net>,
>> Jim Burns <james....@att.net> wrote:
[...]
>>> Suppose we have the rationals, with the usual operations
>>> on the rationals, '+', '-', '*', '<', those necessary
>>> in order to be an ordered field, such as the rationals or
>>> the reals or some other ordered field.
>>>
>>> Suppose we have sets of rationals. In particular, suppose
>>> we have the set D of all sets x of rationals with these four
>>> properties:
>>>
>>> Let x e D.
>>>
>>> x is not empty
>>> (Ea) a e x
>>>
>>> x is not Q
>>> (Ea) ~(a e Q)
>>>
>>> x continues all the way down, without holes.
>>> (Aa)(Ab) ( a e x & b < a ) -> b e x
>>>
>>> x does not contain a maximum.
>>> (Aa) a e x -> (Eb)( b e x & a < b )
>>>
>>> Theorem:
>>> The set of all such x is a complete ordered field.
>>> (This is what Rupert offered to proved to you)
[...]
> You are conflating elements of D the rationals'
> ordered field with them being their own subsets,
> where, as sets they are already equivalence classes
> of ratios (reduced fractions) in a usual development
> of the rationals from the laws of arithmetic.
I am presenting a different model of the reals than
Rupert is. It is probably best if, for the sake of
clarity, you only discuss one model at a time.
It was probably my mistake to bring this up here and now.
In the model I'm presenting, there are
-- the set of rationals, Q
-- members of Q, rational numbers
(Here I have used a and b as variables ranging
over the rational numbers.)
-- subsets of Q, sets containing rational numbers
and satisfying four conditions, above.
(Here I have used x and y as variables ranging
over this kind of _set_ of rational numbers.)
-- the set D of all the sets of rational numbers
_of this kind_ , satisfying the four conditions
This is a subset of P(Q), the powerset of the
rationals.
I'm not using equivalence classes, though Rupert is.
> On Saturday, November 21, 2015 at 8:46:11 PM UTC-8,
> Virgil wrote:
>> In article <5651449A...@att.net>,
>> Jim Burns <james....@att.net> wrote:
[...]
>>> Suppose we have the rationals, with the usual operations
>>> on the rationals, '+', '-', '*', '<', those necessary
>>> in order to be an ordered field, such as the rationals or
>>> the reals or some other ordered field.
>>>
>>> Suppose we have sets of rationals. In particular, suppose
>>> we have the set D of all sets x of rationals with these four
>>> properties:
>>>
>>> Let x e D.
>>>
>>> x is not empty
>>> (Ea) a e x
>>>
>>> x is not Q
>>> (Ea) ~(a e Q)
>>>
>>> x continues all the way down, without holes.
>>> (Aa)(Ab) ( a e x & b < a ) -> b e x
>>>
>>> x does not contain a maximum.
>>> (Aa) a e x -> (Eb)( b e x & a < b )
>>>
>>> Theorem:
>>> The set of all such x is a complete ordered field.
>>> (This is what Rupert offered to proved to you)
[...]
> You are conflating elements of D the rationals'
> ordered field with them being their own subsets,
> where, as sets they are already equivalence classes
> of ratios (reduced fractions) in a usual development
> of the rationals from the laws of arithmetic.
I am presenting a different model of the reals than
Rupert is. It is probably best if, for the sake of
clarity, you only discuss one model at a time.
It was probably my mistake to bring this up here and now.
In the model I'm presenting, there are
-- the set of rationals, Q
-- members of Q, rational numbers
(Here I have used a and b as variables ranging
over the rational numbers.)
-- subsets of Q, sets containing rational numbers
and satisfying four conditions, above.
(Here I have used x and y as variables ranging
over this kind of _set_ of rational numbers.)
-- the set D of all the sets of rational numbers
_of this kind_ , satisfying the four conditions
This is a subset of P(Q), the powerset of the
rationals.
I'm not using equivalence classes, though Rupert is.
Ross A. Finlayson
Nov 22, 2015, 4:28:37 PM11/22/15
to
On Sunday, November 22, 2015 at 2:41:02 AM UTC-8, Virgil wrote:
> In article <0c86f7bf-30a5-4f29...@googlegroups.com>,
> "Ross A. Finlayson" <ross.fi...@gmail.com> wrote:
>
> > On Saturday, November 21, 2015 at 10:06:56 PM UTC-8, Virgil wrote:
> > > In article <745a743a-4b57-4765...@googlegroups.com>,
> > > "Ross A. Finlayson" <ross.fi...@gmail.com> wrote:
> > >
> > > > > A perfectly good model can be formed from equivalence classes of Cauchy
> > > > > sequences of rationals, where two such sequences are deemed equivalent
> > > > > if any only if their termwise difference sequence converges to 0.
> > > reals!
> > >
> > > Every real is the limit of many Cauchy sequences of rationals, two such
> > > sequences converging to the same real if and only if the sequence of
> > > their termwise differences is a sequence converging to zero.
> > >
> > > Having such termwise difference sequence converge to zero is an
> > > equivalence relation on the set of all such convergent sequences of
> > > rationals, and their equivalence classes under that relation is a
> > > standard model for the field of real numbers.
> > > At least among mathematicians!
>
> The whole point is that sequences of rationals that converge to a real
> in |R all are Cauchy sequences in |Q.
sequences of rationals
that converge to a real
in |R all are Cauchy
sequences in |Q. "
But, sequences of rationals
that converge don't necessarily
converge to a normal form of a
sequence as in |Q, i.e., possible
an irrational.
But, that's an irrational. Seems you're
setting out having assumed that which
you're to prove.
(I am having found Least Upper Bound
neatly evident via sweep/EF that also
builds continuity, and also "measure 1.0").
> In article <0c86f7bf-30a5-4f29...@googlegroups.com>,
> "Ross A. Finlayson" <ross.fi...@gmail.com> wrote:
>
> > On Saturday, November 21, 2015 at 10:06:56 PM UTC-8, Virgil wrote:
> > > In article <745a743a-4b57-4765...@googlegroups.com>,
> > > "Ross A. Finlayson" <ross.fi...@gmail.com> wrote:
> > >
> > > > > A perfectly good model can be formed from equivalence classes of Cauchy
> > > > > sequences of rationals, where two such sequences are deemed equivalent
> > > > > if any only if their termwise difference sequence converges to 0.
> > > > > --
> > > > > Virgil
> > > > > "Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)
> > > >
> > > > > Virgil
> > > > > "Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)
> > > >
> > > > You are conflating elements of D the rationals'
> > > > ordered field with them being their own subsets
> > >
> > > You have much yet to learn, including of this standard model for the
> > > > ordered field with them being their own subsets
> > >
> > > reals!
> > >
> > > Every real is the limit of many Cauchy sequences of rationals, two such
> > > sequences converging to the same real if and only if the sequence of
> > > their termwise differences is a sequence converging to zero.
> > >
> > > Having such termwise difference sequence converge to zero is an
> > > equivalence relation on the set of all such convergent sequences of
> > > rationals, and their equivalence classes under that relation is a
> > > standard model for the field of real numbers.
> > > At least among mathematicians!
> > > --
> > > Virgil
> > > "Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)
> >
> > "The equivalence classes of sequences that are Cauchy"
> > > Virgil
> > > "Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)
> >
>
> The whole point is that sequences of rationals that converge to a real
> in |R all are Cauchy sequences in |Q.
> --
> Virgil
> "Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)
"The whole point is that
> Virgil
> "Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)
sequences of rationals
that converge to a real
in |R all are Cauchy
sequences in |Q. "
But, sequences of rationals
that converge don't necessarily
converge to a normal form of a
sequence as in |Q, i.e., possible
an irrational.
But, that's an irrational. Seems you're
setting out having assumed that which
you're to prove.
(I am having found Least Upper Bound
neatly evident via sweep/EF that also
builds continuity, and also "measure 1.0").
Virgil
Nov 22, 2015, 4:48:18 PM11/22/15
to
In article <68145a86-f3e6-4c9d...@googlegroups.com>,
Wrong. As usual!
Every sequence of rationals which is non-decreasong and bounded above
defines real number without any explicit limit or convergence
requirements.
Then any sequence whose termwise differences from such a sequence form a
sequence converging to zero is declared to be equivalent to it.
Then each such equivalence class of sequences "is" a real number and
each real number then "is" such an equivalence class of sequences!
Every sequence of rationals which is non-decreasong and bounded above
defines real number without any explicit limit or convergence
requirements.
Then any sequence whose termwise differences from such a sequence form a
sequence converging to zero is declared to be equivalent to it.
Then each such equivalence class of sequences "is" a real number and
each real number then "is" such an equivalence class of sequences!
Ross A. Finlayson
Nov 22, 2015, 4:57:59 PM11/22/15
to
using density properties that it's
in R?
What here are you using to fill R?
(Answer is convergent sequences of
Q, that aren't in Q.)
Here let's keep in mind usual differences
between (and among): integer ratios, reduced
fractions as normal forms, sequences that
are Cauchy that converge to a rational,
and sequences that are Cauchy that don't (of
rationals).
Ross A. Finlayson
Nov 22, 2015, 5:00:41 PM11/22/15
to
polynomials that aren't rational
before all the transcendentals
from geometry's e and pi before
axiomatizing LUB/gaplessness for
building the complete ordered
field from algebra.
Virgil
Nov 22, 2015, 5:53:24 PM11/22/15
to
In article <2ef62fd7-6fbf-4f2d...@googlegroups.com>,
That some real numbers are easier to identify than others does not make
those others any less real!
those others any less real!
Virgil
Nov 22, 2015, 6:12:20 PM11/22/15
to
In article <4c98b926-8f53-4300...@googlegroups.com>,
On the contrary, I can use sets of those sets of rationals that are
bounded above.
>
> Here let's keep in mind usual differences
> between (and among): integer ratios, reduced
> fractions as normal forms, sequences that
> are Cauchy that converge to a rational,
> and sequences that are Cauchy that don't (of
> rationals).
Every non-empty set of rationals that is bounded above determines a real
number as its "least upper bound" in R, and every real number is
determined by such a set, the only issue is determining which such sets
of rationals will have the same least upper bounds.
If one restricts the set of sets of rationals to
1. those that are bounded above, and
2. do not contain any least upper bound, and
3. do contain every rational less than any member
then the set of such sets bijects with |R.
bounded above.
>
> Here let's keep in mind usual differences
> between (and among): integer ratios, reduced
> fractions as normal forms, sequences that
> are Cauchy that converge to a rational,
> and sequences that are Cauchy that don't (of
> rationals).
number as its "least upper bound" in R, and every real number is
determined by such a set, the only issue is determining which such sets
of rationals will have the same least upper bounds.
If one restricts the set of sets of rationals to
1. those that are bounded above, and
2. do not contain any least upper bound, and
3. do contain every rational less than any member
then the set of such sets bijects with |R.
Ross A. Finlayson
Nov 22, 2015, 7:03:52 PM11/22/15
to
presentation of the diagonal argument"?
"... then the set of such sets bijects with |R",
what are the properties of this function that
maps the sets of rationals to the reals now?
Here again: the set of rationals is not
containing itself or it is. That is,
rationals are (each) bounding themselves above,
and rationals (all) is all you yet have there.
> 1. those that are bounded above, and
> 2. do not contain any least upper bound, and
representation space to get all the value
representation space. (That's actually
part of the effect.)
Jim Burns
Nov 22, 2015, 7:22:20 PM11/22/15
to
On 11/22/2015 6:12 PM, Virgil wrote:
> Every non-empty set of rationals that is bounded above
> determines a real number as its "least upper bound" in R,
> and every real number is determined by such a set, the only
> issue is determining which such sets of rationals will have
> the same least upper bounds.
>
> If one restricts the set of sets of rationals to
> 1. those that are bounded above, and
> 2. do not contain any least upper bound, and
> 3. do contain every rational less than any member
> then the set of such sets bijects with |R.
I think you want to specifically exclude {}, which is
> Every non-empty set of rationals that is bounded above
> determines a real number as its "least upper bound" in R,
> and every real number is determined by such a set, the only
> issue is determining which such sets of rationals will have
> the same least upper bounds.
>
> If one restricts the set of sets of rationals to
> 1. those that are bounded above, and
> 2. do not contain any least upper bound, and
> 3. do contain every rational less than any member
> then the set of such sets bijects with |R.
bounded above because there is a rational larger than all
of its none at all members. And you maybe want to specifically
exclude Q for clarity, at least.
Of course, the set of all these sets still bijects with
the reals if you add in a few extra sets ( {}, Q ),
but I don't think it would be isomorphic, which would
be very important in the current context.
Ben Bacarisse
Nov 22, 2015, 8:02:33 PM11/22/15
to
WM <wolfgang.m...@hs-augsburg.de> writes:
> Am Sonntag, 22. November 2015 14:51:32 UTC+1 schrieb Ben Bacarisse:
>> WM <wolfgang.m...@hs-augsburg.de> writes:
>>
>> > Am Samstag, 21. November 2015 17:14:22 UTC+1 schrieb Ben Bacarisse:
>> >> WM <wolfgang.m...@hs-augsburg.de> writes:
>> >>
>> >> > Assume a list of all infinite digit sequences.
<snip>
> Am Sonntag, 22. November 2015 14:51:32 UTC+1 schrieb Ben Bacarisse:
>> WM <wolfgang.m...@hs-augsburg.de> writes:
>>
>> > Am Samstag, 21. November 2015 17:14:22 UTC+1 schrieb Ben Bacarisse:
>> >> WM <wolfgang.m...@hs-augsburg.de> writes:
>> >>
>> >> > Assume a list of all infinite digit sequences.
>
> d_1,d_2,d_3,...,d_n = q_(k1),q_(k2),q_(k3),...,q_(kn)
>
> Is there anything wrong?
>
>> But I'm glad you
>> can't prove anything that contradicts set theory.
>
> The above equality shows a contradiction.
Then you must reject the assumption you started with: "Assume a list of
> d_1,d_2,d_3,...,d_n = q_(k1),q_(k2),q_(k3),...,q_(kn)
>
> Is there anything wrong?
>
>> But I'm glad you
>> can't prove anything that contradicts set theory.
>
> The above equality shows a contradiction.
all infinite digit sequences". Glad we agree on that.
<snip>
--
Ben.
Virgil
Nov 22, 2015, 8:15:11 PM11/22/15
to
In article <56525C3A...@att.net>,
but Q is already excluded by not being bounded above.
A non-empty set of rationals is bounded above only if there is some
rational larger then every one of its members, and Q is NOT such a set.
Jim Burns <james....@att.net> wrote:
> On 11/22/2015 6:12 PM, Virgil wrote:
>
> > Every non-empty set of rationals that is bounded above
> > determines a real number as its "least upper bound" in R,
> > and every real number is determined by such a set, the only
> > issue is determining which such sets of rationals will have
> > the same least upper bounds.
> >
> > If one restricts the set of sets of rationals to
> > 1. those that are bounded above, and
> > 2. do not contain any least upper bound, and
> > 3. do contain every rational less than any member
> > then the set of such sets bijects with |R.
>
> I think you want to specifically exclude {}, which is
> bounded above because there is a rational larger than all
> of its none at all members. And you maybe want to specifically
> exclude Q for clarity, at least.
Thank you! I do need specifically to exclude the empty set,
> On 11/22/2015 6:12 PM, Virgil wrote:
>
> > Every non-empty set of rationals that is bounded above
> > determines a real number as its "least upper bound" in R,
> > and every real number is determined by such a set, the only
> > issue is determining which such sets of rationals will have
> > the same least upper bounds.
> >
> > If one restricts the set of sets of rationals to
> > 1. those that are bounded above, and
> > 2. do not contain any least upper bound, and
> > 3. do contain every rational less than any member
> > then the set of such sets bijects with |R.
>
> I think you want to specifically exclude {}, which is
> bounded above because there is a rational larger than all
> of its none at all members. And you maybe want to specifically
> exclude Q for clarity, at least.
but Q is already excluded by not being bounded above.
A non-empty set of rationals is bounded above only if there is some
rational larger then every one of its members, and Q is NOT such a set.
>
> Of course, the set of all these sets still bijects with
> the reals if you add in a few extra sets ( {}, Q ),
> but I don't think it would be isomorphic, which would
> be very important in the current context.
> Of course, the set of all these sets still bijects with
> the reals if you add in a few extra sets ( {}, Q ),
> but I don't think it would be isomorphic, which would
> be very important in the current context.
Virgil
Nov 22, 2015, 8:26:41 PM11/22/15
to
In article <30effd1e-681a-4188...@googlegroups.com>,
It is not the set of all sets of rationals, only some of those sets of
rational which are bounded above but contqain every rational less than
any of their own members, which in the set of all reals pairs each real
with the set of all rationals less than it.
> That is, rationals are (each) bounding themselves above,
> and rationals (all) is all you yet have there.
On the contrary, I have a family of sets of rationals with one and only
one such set for each real!
rational which are bounded above but contqain every rational less than
any of their own members, which in the set of all reals pairs each real
with the set of all rationals less than it.
>
> Here again: the set of rationals is not
> containing itself or it is.
Every set "contains itself" as a subset!
> Here again: the set of rationals is not
> containing itself or it is.
> That is, rationals are (each) bounding themselves above,
> and rationals (all) is all you yet have there.
one such set for each real!
>
> > 1. those that are bounded above, and
> > 2. do not contain any least upper bound, and
>
> So, it looks here you are reducing the value
> representation space to get all the value
> representation space. (That's actually
> part of the effect.)
> > 1. those that are bounded above, and
> > 2. do not contain any least upper bound, and
>
> So, it looks here you are reducing the value
> representation space to get all the value
> representation space. (That's actually
> part of the effect.)
Ross A. Finlayson
Nov 22, 2015, 9:29:23 PM11/22/15
to
1 is real, rational, integer,
and etcetera, but as types,
integer ratios are not equivalence
classes of reduced fractions are
not sets of rationals _dense_
and bounded above.
The rationals are dense as the
ordered field, also they're dense
in the complete ordered field.
So, your definition of rationals
augmented as "sets of themselves"
then to point to things between
themselves, should have that they
are dense besides bounded above
(or they'd also be "incomplete"
as of the ordered field).
Then, where any set is its own
subset, your conflation there is
like taking a subset of an
ordinal, calling it an ordinal,
and finding the ordinals "contains"
itself.
So, to get "all the sequences for
a rational that are Cauchy, and
all those less", those are also
_dense_ in the ordered field.
George Greene
Nov 23, 2015, 12:49:37 AM11/23/15
to
On Saturday, November 21, 2015 at 7:02:34 PM UTC-5, Ross A. Finlayson wrote:
> Is LUB a property or an axiom?
It's a FUNCTION. Or functor, or constructor. The thing that it is a function OF is a thing that does not occur in the universe of discourse, at FIRST-order,
anywyay. Lub is going to be a 2nd-order function in the theory and its argument is going to be a 1st-order variable (a collection of 0th-order objects from the domain, i.e., a collection of reals).
You can sort of make Lub an axiom-schema at first-order if you make the set-of-which-the-lub-is-being-taken definable by a first-order schema.
An axiomatization of the reals is going to have all AND ONLY reals in its domain; in particular, it is not going to have pairs or sequences of reals, despite the fact that those are easy to encode.
> Is LUB a property or an axiom?
It's a FUNCTION. Or functor, or constructor. The thing that it is a function OF is a thing that does not occur in the universe of discourse, at FIRST-order,
anywyay. Lub is going to be a 2nd-order function in the theory and its argument is going to be a 1st-order variable (a collection of 0th-order objects from the domain, i.e., a collection of reals).
You can sort of make Lub an axiom-schema at first-order if you make the set-of-which-the-lub-is-being-taken definable by a first-order schema.
An axiomatization of the reals is going to have all AND ONLY reals in its domain; in particular, it is not going to have pairs or sequences of reals, despite the fact that those are easy to encode.
George Greene
Nov 23, 2015, 12:51:44 AM11/23/15
to
On Saturday, November 21, 2015 at 8:31:26 AM UTC-5, WM wrote:
> Assume a list of all infinite digit sequences.
>
> Assume a list of all infinite digit sequences.
>
> Every finite digit sequence d_1,d_2,d_3,...,d_n of an antidiagonal can be found (infinitely often) in the remainder of the list:
I don't see why you feel the need to go to the remainder of the list.
If you define the first element right, every finite digit sequence THAT EXISTS can be found IN THE FIRST element of the list. No need to go looking to any later ones.
WM
Nov 23, 2015, 3:27:09 AM11/23/15
to
Am Montag, 23. November 2015 06:51:44 UTC+1 schrieb George Greene:
> On Saturday, November 21, 2015 at 8:31:26 AM UTC-5, WM wrote:
> > Assume a list of all infinite digit sequences.
> >
> > Every finite digit sequence d_1,d_2,d_3,...,d_n of an antidiagonal can be found (infinitely often) in the remainder of the list:
>
> I don't see why you feel the need to go to the remainder of the list.
That is required because d_1,d_2,d_3,...,d_n is provably not in the first n rows. And it is possible, because the remainder is always infinite.
> On Saturday, November 21, 2015 at 8:31:26 AM UTC-5, WM wrote:
> > Assume a list of all infinite digit sequences.
> >
> > Every finite digit sequence d_1,d_2,d_3,...,d_n of an antidiagonal can be found (infinitely often) in the remainder of the list:
>
> I don't see why you feel the need to go to the remainder of the list.
> If you define the first element right, every finite digit sequence THAT EXISTS can be found IN THE FIRST element of the list. No need to go looking to any later ones.
Regards, WM
WM
Nov 23, 2015, 3:30:42 AM11/23/15
to
Am Montag, 23. November 2015 02:02:33 UTC+1 schrieb Ben Bacarisse:
> >> >> > Assume a list of all infinite digit sequences.
> <snip>
> >
> > d_1,d_2,d_3,...,d_n = q_(k1),q_(k2),q_(k3),...,q_(kn)
> >
> > Is there anything wrong?
> >
> >> But I'm glad you
> >> can't prove anything that contradicts set theory.
> >
> > The above equality shows a contradiction.
>
> Then you must reject the assumption you started with: "Assume a list of
> all infinite digit sequences". Glad we agree on that.
That does not matter because it is sufficient to assume a list of all rationals. The result is the same contradiction.
> >> >> > Assume a list of all infinite digit sequences.
> <snip>
> >
> > d_1,d_2,d_3,...,d_n = q_(k1),q_(k2),q_(k3),...,q_(kn)
> >
> > Is there anything wrong?
> >
> >> But I'm glad you
> >> can't prove anything that contradicts set theory.
> >
> > The above equality shows a contradiction.
>
> Then you must reject the assumption you started with: "Assume a list of
> all infinite digit sequences". Glad we agree on that.
Regards, WM
Virgil
Nov 23, 2015, 5:48:23 AM11/23/15
to
In article <75a35495-7a85-4e52...@googlegroups.com>,
WM seems to be claiming that all digit sequences must represent
rationals but infinite ones which do not ever become eventually periodic
don't.
rationals but infinite ones which do not ever become eventually periodic
don't.
So WM is
WRONG !
AGAIN ! !
AS USUAL ! ! !
WRONG !
AGAIN ! !
AS USUAL ! ! !
Virgil
Nov 23, 2015, 5:53:41 AM11/23/15
to
In article <6278430b-ba6c-4c5b...@googlegroups.com>,
none of the infinite sequences listed will equal the antidiagonal
infinite sequence.
Thus WM is again, as always, entirely missing the point!
And WM is
WM <wolfgang.m...@hs-augsburg.de> wrote:
> Am Montag, 23. November 2015 06:51:44 UTC+1 schrieb George Greene:
> > On Saturday, November 21, 2015 at 8:31:26 AM UTC-5, WM wrote:
> > > Assume a list of all infinite digit sequences.
> > >
> > > Every finite digit sequence d_1,d_2,d_3,...,d_n of an antidiagonal can be
> > > found (infinitely often) in the remainder of the list:
> >
> > I don't see why you feel the need to go to the remainder of the list.
>
> That is required because d_1,d_2,d_3,...,d_n is provably not in the first n
> rows. And it is possible, because the remainder is always infinite.
>
> > If you define the first element right, every finite digit sequence THAT
> > EXISTS can be found IN THE FIRST element of the list.
But it is only infinite digit sequences that count, but by construction,
> Am Montag, 23. November 2015 06:51:44 UTC+1 schrieb George Greene:
> > On Saturday, November 21, 2015 at 8:31:26 AM UTC-5, WM wrote:
> > > Assume a list of all infinite digit sequences.
> > >
> > > Every finite digit sequence d_1,d_2,d_3,...,d_n of an antidiagonal can be
> > > found (infinitely often) in the remainder of the list:
> >
> > I don't see why you feel the need to go to the remainder of the list.
>
> That is required because d_1,d_2,d_3,...,d_n is provably not in the first n
> rows. And it is possible, because the remainder is always infinite.
>
> > If you define the first element right, every finite digit sequence THAT
> > EXISTS can be found IN THE FIRST element of the list.
none of the infinite sequences listed will equal the antidiagonal
infinite sequence.
Thus WM is again, as always, entirely missing the point!
And WM is
WM
Nov 23, 2015, 8:21:19 AM11/23/15
to
Am Montag, 23. November 2015 11:48:23 UTC+1 schrieb Virgil:
> > That does not matter because it is sufficient to assume a list of all
> > rationals. The result is the same contradiction.
> >
> WM seems to be claiming that all digit sequences must represent
> rationals but infinite ones which do not ever become eventually periodic
> don't.
What is the first digit that doe not belong to a rational sequence?
> > That does not matter because it is sufficient to assume a list of all
> > rationals. The result is the same contradiction.
> >
> WM seems to be claiming that all digit sequences must represent
> rationals but infinite ones which do not ever become eventually periodic
> don't.
Regards, WM
Ben Bacarisse
Nov 23, 2015, 8:34:14 AM11/23/15
to
WM <wolfgang.m...@hs-augsburg.de> writes:
> Am Montag, 23. November 2015 02:02:33 UTC+1 schrieb Ben Bacarisse:
>
>
>> >> >> > Assume a list of all infinite digit sequences.
>> <snip>
>> >
>> > d_1,d_2,d_3,...,d_n = q_(k1),q_(k2),q_(k3),...,q_(kn)
>> >
>> > Is there anything wrong?
>> >
>> >> But I'm glad you
>> >> can't prove anything that contradicts set theory.
>> >
>> > The above equality shows a contradiction.
>>
>> Then you must reject the assumption you started with: "Assume a list of
>> all infinite digit sequences". Glad we agree on that.
>
> That does not matter
Really? You don't care that you've just claimed another proof of
> Am Montag, 23. November 2015 02:02:33 UTC+1 schrieb Ben Bacarisse:
>
>
>> >> >> > Assume a list of all infinite digit sequences.
>> <snip>
>> >
>> > d_1,d_2,d_3,...,d_n = q_(k1),q_(k2),q_(k3),...,q_(kn)
>> >
>> > Is there anything wrong?
>> >
>> >> But I'm glad you
>> >> can't prove anything that contradicts set theory.
>> >
>> > The above equality shows a contradiction.
>>
>> Then you must reject the assumption you started with: "Assume a list of
>> all infinite digit sequences". Glad we agree on that.
>
> That does not matter
Cantor's theorem? You really *should* care about presenting correct
arguments.
> because it is sufficient to assume a list of all
> rationals. The result is the same contradiction.
WMaths? (As it happens the argument is flawed in all cases but you
don't really care about that.)
--
Ben.
Dan Christensen
Nov 23, 2015, 8:50:20 AM11/23/15
to
> Cantor has "proved" that the set of finite digit sequences is uncountable.
>
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