Well Ordering the Reals
Tony Orlow
I am told it is impossible to well order the real numbers such that there is a
first and a successor operation which defines each other. However, I believe I
have such an enumeration. I have posted a web page at
http://www.people.cornell.edu/pages/aeo6/WellOrder/ .
Please feel free to comment, either here or by email. Is there a valid
objection to this well-ordering, or does it do something not proviously though
possible?
--
Smiles,
Tony
Gene Ward Smith
What well-ordering? I don't see a definition on that page. Can you tell
us the first twenty reals in your well-ordering?
Ross A. Finlayson
To well-order the reals is one of Hilbert's problems.
http://www.google.com/search?q=Hilbert%27s+problems
0, +-1, +-2, +-3, +-4, +-5, +-6, +-7, +-8, +-9, 10 iota.
That's not standard.
Well-order the reals.
Ross
ste...@nomail.com
The first real in his well-ordering is oo, and the second one is -oo.
So far, he is batting 0 for 2. I doubt it gets any better.
Stephen
David Kastrup
I picture big bad judge Tony swatting two oo flying around, banging
his hammer on the desk and shouting "order in the reals!" to an
assembly of rational spectators filling row after row and staring at
him with a blank look.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
Sean Holman
The Ghost In The Machine
<ae...@cornell.edu>
wrote
on Fri, 28 Oct 2005 15:50:27 -0400
<MPG.1dcc4b0a1...@newsstand.cit.cornell.edu>:
Several problems.
[1] +oo and -oo aren't real. This is easily fixable by enumerating the
extended reals, of course, instead of the real numbers, or by
tacking the reals onto the front of the mapping, as you
apparently have done.
[2] What, precisely, is 'x'?
[3] For any mapping m : N -> R, I can find a number d such that
m(i) != d for all i in N. There are various methods by which
one might accomplish this; the simplest is to assume
m2 : N -> [0,1), (or construct m2(x) = m1(i(x)), where i() is
a mapping from [0,+oo) to [0,1)), explicitly construct a
d2's decimal expansion somewhere in [0,1) (the Carnot Diagonal
Proof is or should be well known to most mathematicians), and
then prove d2 is not in m2's image. Since there's a 1-1
correspondence between [0,1) and [0,+oo) -- there's a few,
in fact; the simplest one might be y=x/(1-x), and finding
the inverse yields:
y+1=1/(1-x); 1/(y+1)=(1-x); x=1-(1/(y+1))=y/(y+1)
one can then compute d = d2/(1-d2) which is not in m's image.
[4] Near as I can figure, the first 5 numbers of your enumeration
are -oo, +oo, 0, -1, and +1. Beyond that, I can't tell what's
going on, especially since x^(-oo) = 1 for any real x > 1;
this isn't exactly a midpoint in any logical sense of the term.
It may depend on what one defines as "midpoint"; one simple
one, for example, might be the harmonic mean m = 2/(1/x+1/y).
The main problem with this midpoint is that one only gets
certain members of Q. There are other midpoint defs one
can try but in light of [3] there's not much point.
--
#191, ewi...@earthlink.net
It's still legal to go .sigless.
Ross A. Finlayson
No, because for any cardinal there exists an ordinal, even if that
cardinal is not comparable, ie violates trichotomy. There exists a
larger cardinal that's an ordinal, or it's not a set.
The difference between an existence proof and an example is like the
difference between a guy claiming that Everest, Chomolungma, is
scalable versus the guy that does it.
So, well-order the reals. Extend Cantor's first, that ordering has
adjacent points that are adjacent in the normal ordering of the reals.
We have been discussing this specifically for months and nie upon a
year and more.
For any well-ordering of the reals, it will have adjacent points in the
normal ordering of the reals.
That's counterintuitive because the reals are dense, and complete, they
form a continuum, of individua, at once a set, of elements, and a line,
of points, with there nothing else to be the real complete ordered
field, those same objects are each and all of those things.
Ross
Proginoskes
Tony Orlow wrote:
> Hi All -
>
> I am told it is impossible to well order the real numbers such that there is a
I suspect you're using the wrong definition.
A well-ordering of a set S is a total ordering < such that every
non-empty subset T of S has a first (minimum) element. A well-ordering
of the positive integers is
1, 2, 3, 4, 5, 6, ...
because if you have a set T which is a nonempty subset of the positive
integers, it contains some positive integer n -- not necessarily the
smallest. There are only a finite number of elements of T less than n,
and you just check cases until you find the smallest one. (If 1 is in
T, you're done; otherwise if 2 is in T, you're done; etc., up to n,
where the checking procedure is guaranteed to stop, after a finite
number of checks.)
A well-ordering of the integers is
0, 1, -1, 2, -2, 3, -3, 4, -4, ...
In fact, for a countable set, any enumeration is also a well-ordering.
But the reals are not countable, so the problem is much harder. (The
standard ordering of R is not a well-ordering, because the set (0,1] --
all positive real numbers less than or equal to 1 -- has no smallest
element.)
--- Christopher Heckman
Peter Webb
"Sean Holman" <holm...@yahoo.com> wrote in message
news:16564229.1130535933...@nitrogen.mathforum.org...
The Axiom of Choice is non-constructive.
This doesn't mean that there isn't a constructable choice function over a
set, including R. As far as I know, nobody has proved that there is no
constructable choice function over R (ie that we need the AoC for well
ordering), its just that nobody has ever constructed such a function for R
(or AFAIK for any uncountable set).
G. A. Edgar
<webbfamily...@optusnet.com.au> wrote:
> [...] its just that nobody has ever constructed such a function for R
> (or AFAIK for any uncountable set).
>
>
A constructible well-ordering exists for certain uncountable sets.
Smallest is the "set of all countable ordinals" omega_1.
This goes back to Cantor.
Goedel constructed a well-ordering for a certain subset of the reals,
then showed that it is consistent with ZF (and follows from V=L) that
his subset is all reals.
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
Ross A. Finlayson
How's that?
I wonder how Goedel constructs a well-ordering of the reals.
I read that Feferman says that it is not possible to exemplify a
well-ordering when V =/= L.
V = L.
Thanks,
Ross
Tony Orlow
--
Smiles,
Tony
Tony Orlow
valid. I have updated the page with a more in-depth explanation of the well
ordering, including the first 20 elements, and what I hope is a clear
description of how one proceeds from there. I trust the enumeration will be
clearer if you take a look. My original description of the bijection through
the binary strings was rather terse. Thanks.
--
Smiles,
Tony Orlow
they are the defining points of the line. There is a way to start with 0 and -
0, having +/-oo in the middle, but that would probably offend you more than
having infinity at the "ends" of the line. There is almost surely a way to
begin with +/-1, but hopefully silly nitpicking won't necessitate that, though
it's worth exploring. In any case, you may feel free to reject oo and -oo from
membership in the set, and use them only for a starting point in the
enumeration. This makes the mapping function in my updated description slightly
simpler, which is fine. Personally, I have no problem with starting a
definition of the reals by establishing the infinite number line.
Tony Orlow
> In sci.math, Tony Orlow
> <ae...@cornell.edu>
> wrote
> on Fri, 28 Oct 2005 15:50:27 -0400
> <MPG.1dcc4b0a1...@newsstand.cit.cornell.edu>:
> > Hi All -
> >
> > I am told it is impossible to well order the real numbers such that there is a
> > first and a successor operation which defines each other. However, I believe I
> > have such an enumeration. I have posted a web page at
> > http://www.people.cornell.edu/pages/aeo6/WellOrder/ .
> >
> > Please feel free to comment, either here or by email. Is there a valid
> > objection to this well-ordering, or does it do something not proviously though
> > possible?
>
> Several problems.
>
> [1] +oo and -oo aren't real. This is easily fixable by enumerating the
> extended reals, of course, instead of the real numbers, or by
> tacking the reals onto the front of the mapping, as you
> apparently have done.
him. They can certainly be eliminated as elements, and used only as a starting
point for the enumeration of the finite reals, if that makes you feel better.
That simplifies the quantitative portion of the mapping to the naturals,
slightly.
>
> [2] What, precisely, is 'x'?
values of the elements beyond -1 and 1 depends on x. So, one can do the
enumeration with any number which is in those ranges. I have updated my page
with a more detailed description of the enumeration. The rightmost column of
the second table displays the values for x=1/2.
>
> [3] For any mapping m : N -> R, I can find a number d such that
> m(i) != d for all i in N. There are various methods by which
> one might accomplish this; the simplest is to assume
> m2 : N -> [0,1), (or construct m2(x) = m1(i(x)), where i() is
> a mapping from [0,+oo) to [0,1)), explicitly construct a
> d2's decimal expansion somewhere in [0,1) (the Carnot Diagonal
> Proof is or should be well known to most mathematicians), and
> then prove d2 is not in m2's image. Since there's a 1-1
> correspondence between [0,1) and [0,+oo) -- there's a few,
> in fact; the simplest one might be y=x/(1-x), and finding
> the inverse yields:
>
> y+1=1/(1-x); 1/(y+1)=(1-x); x=1-(1/(y+1))=y/(y+1)
>
> one can then compute d = d2/(1-d2) which is not in m's image.
are several reasons for this. His proof says nothing about real numbers, but
addresses the enumeration of a digital number system with base greater than 1.
The result essentially boils down to the fact that for any set of symbols of
size S, the set of all unique strings of length L constructed from S symbols
has a size S^L, which is greater than L for any S>1. The list of digital
strings is always longer than it is wide. For n binary digits, we have 2^n
strings. So the diagonal from which he constructs his antidiagonal does not
traverse the entire set of real that it purports to. If there are N digits in
each real, and we traverse 1 digits for each real through the diagonal, then
the diagonal traverse N of the 2^N strings. The antidiagonal is not missing
from the list. It is simply below the diagonal. This general proof of the
"uncountability" of the real numbers, and the impossibility of a well ordering,
is simple conflation of what is a propertiy of digital number systems, and more
generally, symbolic languages.
>
> [4] Near as I can figure, the first 5 numbers of your enumeration
> are -oo, +oo, 0, -1, and +1. Beyond that, I can't tell what's
> going on, especially since x^(-oo) = 1 for any real x > 1;
> this isn't exactly a midpoint in any logical sense of the term.
Similarly, for any real in (0,1), x^oo=0.
>
> It may depend on what one defines as "midpoint"; one simple
> one, for example, might be the harmonic mean m = 2/(1/x+1/y).
> The main problem with this midpoint is that one only gets
> certain members of Q. There are other midpoint defs one
> can try but in light of [3] there's not much point.
Limitations of the digital number systems are not to be conflated with
properties of real numbers. The digital enumeration of the reals is actually
valid, when you take into account N=S^L, as above. However, it does suffer from
base-dependency in its structure, requiring trees with 2 children for every
node in binary, ten in decimal. This enumeration does not suffer that drawback,
and enumerates the entire line at once, large and small elements together, in a
linear order.
You suggestion above, 2/(1/x+1/y), looks like it would generate rational
numbers, not all reals. If the majority of reals are irrational, this won't
work. Of course, what is rational probably ultimately depends on the method of
enumeration. In this system, perhaps sqrt(2) can be considered rational, at
least where x=2. It can even be considered something of an integer, the set
being in linear order. But that is not what is going on here. Here, we are
identifying the relationship that ties {-oo,0,oo} and {0,1,oo} and {1,x oo}. I
think I have identified the salient relationship. Yes, no?
Tony Orlow
order such that every element is either greater than or less than any other
element. The diagonal argument is pure conflation, saying nothing about the
real numbers, but barely revealing a property of digital number systems, and
symbolic languages in general. It proves nothing ultimately about the
"uncountability" of the reals, but only serves as the first, and very
rudimentary, demonstration that infinities may have different sizes. it's time
to move beyond the diagonal argument.
Now, what makes you think this gives a subset of the rationals? Look again. I
have updated the page with a more in-depth description of the enumeration.
Among the elements you will find such irrational numbers as sqrt(2) and -1/sqrt
(2). Some of them get far more "irrational" as one continues the enumeration.
Dik T. Winter
> Gene Ward Smith said:
...
> > What well-ordering? I don't see a definition on that page. Can you tell
> > us the first twenty reals in your well-ordering?
>
> Congratulations on being the first to respond! I suppose your question is
> valid. I have updated the page with a more in-depth explanation of the well
> ordering, including the first 20 elements, and what I hope is a clear
> description of how one proceeds from there. I trust the enumeration will be
> clearer if you take a look. My original description of the bijection through
> the binary strings was rather terse. Thanks.
As far as I can see your method does not generate numbers like 3, but only
0 and powers of 2.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
Tony Orlow
>
> Tony Orlow wrote:
> > Hi All -
> >
> > I am told it is impossible to well order the real numbers such that there is a
> > first and a successor operation which defines each other. [...]
>
> I suspect you're using the wrong definition.
>
> A well-ordering of a set S is a total ordering < such that every
> non-empty subset T of S has a first (minimum) element. A well-ordering
> of the positive integers is
>
> 1, 2, 3, 4, 5, 6, ...
>
> because if you have a set T which is a nonempty subset of the positive
> integers, it contains some positive integer n -- not necessarily the
> smallest. There are only a finite number of elements of T less than n,
> and you just check cases until you find the smallest one. (If 1 is in
> T, you're done; otherwise if 2 is in T, you're done; etc., up to n,
> where the checking procedure is guaranteed to stop, after a finite
> number of checks.)
>
> A well-ordering of the integers is
>
> 0, 1, -1, 2, -2, 3, -3, 4, -4, ...
want to use, is it?
>
> In fact, for a countable set, any enumeration is also a well-ordering.
> But the reals are not countable, so the problem is much harder. (The
> standard ordering of R is not a well-ordering, because the set (0,1] --
> all positive real numbers less than or equal to 1 -- has no smallest
> element.)
"Uncountability" of the reals is the result of a conflation of properties of
digital number systems with some vague notion regarding the real numbers. The
standard quantitative ordering of the reals is not considered a well ordering,
because of situations such as you mention, but that CAN be resolved in a sense
through the use of infinitesimals, as Ross and I have been advocating. In any
case, this is not the standard quantitative ordering of the reals, but a
discrete subdivision of the entire real line such that every real number has a
place in a linear order with a first element.
>
> --- Christopher Heckman
Tony Orlow
perhaps? I'll google.
Robert Low
> "Uncountability" of the reals is the result of a conflation of properties of
> digital number systems with some vague notion regarding the real numbers.
Actually, uncountability of the reals is the result of a *precise*
notion regarding the real numbers. And if you're hung up on the idea
that it has something to do with 'digital number systems', you will
continue writing stuff that is 'not even wrong'.
Tony Orlow
> In article <MPG.1dcfdc0cd...@newsstand.cit.cornell.edu> Tony Orlow <ae...@cornell.edu> writes:
> > Gene Ward Smith said:
> ...
> > > What well-ordering? I don't see a definition on that page. Can you tell
> > > us the first twenty reals in your well-ordering?
> >
> > Congratulations on being the first to respond! I suppose your question is
> > valid. I have updated the page with a more in-depth explanation of the well
> > ordering, including the first 20 elements, and what I hope is a clear
> > description of how one proceeds from there. I trust the enumeration will be
> > clearer if you take a look. My original description of the bijection through
> > the binary strings was rather terse. Thanks.
>
> As far as I can see your method does not generate numbers like 3, but only
> 0 and powers of 2.
>
then 2 is going to be an infintiely long bitstring. Similarly, in digital base
2. 1/3 is infinitely long and in digital base 3, 1/2 is infinitely long. This
isn't a problem. All truly infinite sets, when expressed in symbolic language
of any sort, require an infinitely long string to uniquely identify each
element, even if the "finite" ones only have a finite number of SIGNIFICANT
digits. This shouldn't be considered a problem. We have a linear order that
doesn't end.
Dik T. Winter
> Dik T. Winter said:
...
> > As far as I can see your method does not generate numbers like 3, but only
> > 0 and powers of 2.
> >
> If x is 1/2, then 3 is going to be an infinitely long bit string. If x is
> 1/3, then 2 is going to be an infintiely long bitstring. Similarly, in
> digital base 2. 1/3 is infinitely long and in digital base 3, 1/2 is
> infinitely long. This isn't a problem. All truly infinite sets, when
> expressed in symbolic language of any sort, require an infinitely long
> string to uniquely identify each element, even if the "finite" ones only
> have a finite number of SIGNIFICANT digits. This shouldn't be considered
> a problem. We have a linear order that doesn't end.
Indeed, on the other hand it does not make it an enumeration either.
Because an enumeration maps to *finite* integers.
But as you say this is an ordering, you should be able to tell me (when
we assume that x = 2) in what order the numbers 3, 5, 7 and 1/3 come out.
Ross A. Finlayson
So, well-order the reals.
You get into the problem then of mapping any well-ordered set to the
reals, in extension of Cantor's first.
One possibility of solution in consideration that meets an intuition,
of, eg, Leibniz, is that there are adjacent points in the normal
ordering of the reals.
Otherwise the reals aren't a set, rationals are uncountable, etcetera.
ZF is inconsistent. Infinite sets are equivalent.
The real numbers aren't necessarily a digital system, even though the
definitions via Dedekind/Cauchy are of that sort.
Hilbert requests a well-ordering of the reals. So, what is Goedel's
construction?
Ross
Robert Low
> So, well-order the reals.
> You get into the problem then of mapping any well-ordered set to the
> reals, in extension of Cantor's first.
> One possibility of solution in consideration that meets an intuition,
> of, eg, Leibniz, is that there are adjacent points in the normal
> ordering of the reals.
> Otherwise the reals aren't a set, rationals are uncountable, etcetera.
> ZF is inconsistent. Infinite sets are equivalent.
> The real numbers aren't necessarily a digital system, even though the
> definitions via Dedekind/Cauchy are of that sort.
> Hilbert requests a well-ordering of the reals. So, what is Goedel's
> construction?
I think that the random phrase generator need to be re-initialised.
Tony Orlow
which does NOT employ any digital representation? If not, then I suggest you
reevaulate your response. If so, dish it out.
Robert Low
Sure. Fix e>0. Suppose the reals in [0,1] to be countable,
and consider an enumeration of them.
Round the i'th number, put the interval of length e/2^i.
Then all the reals in [0,1] have been covered by a set whose total
measure is sum_i=1^infinity e/2^i = e. If e is less
than 1, then you have covered the unit interval by
a union of intervals the sum of whose lengths is less
than 1.
This contradiction shows that the reals in [0,1]
cannot be countable.
I await the next helping of gibberish with no
enthusiasm whatever.
Tony Orlow
> In article <MPG.1dcff1d0c...@newsstand.cit.cornell.edu> Tony Orlow <ae...@cornell.edu> writes:
> > Dik T. Winter said:
> ...
> > > As far as I can see your method does not generate numbers like 3, but only
> > > 0 and powers of 2.
> > >
> > If x is 1/2, then 3 is going to be an infinitely long bit string. If x is
> > 1/3, then 2 is going to be an infintiely long bitstring. Similarly, in
> > digital base 2. 1/3 is infinitely long and in digital base 3, 1/2 is
> > infinitely long. This isn't a problem. All truly infinite sets, when
> > expressed in symbolic language of any sort, require an infinitely long
> > string to uniquely identify each element, even if the "finite" ones only
> > have a finite number of SIGNIFICANT digits. This shouldn't be considered
> > a problem. We have a linear order that doesn't end.
>
> Indeed, on the other hand it does not make it an enumeration either.
> Because an enumeration maps to *finite* integers.
infinite set, since that will require an infinite number of naturals to map to,
and will require infinite bits to the binary representations by N=S^L.
Wolfram is waiting for a contributed definition of "enumeration", but under
enumerate, he has the following definition, which does not mention finiteness:
"To enumerate a set of objects satisfying some set of properties means to
explicitly produce a listing of all such objects."
>
> But as you say this is an ordering, you should be able to tell me (when
> we assume that x = 2) in what order the numbers 3, 5, 7 and 1/3 come out.
>
That would require deeper knowledge of tetration on my part. Perhaps that will
come with Herman Bosch's brother's work, which I hope to see soon. When I
discussed this with Rusin months ago, he wanted his number 3. We determined
that it did exist as an infinite string, something like ...010101, in whch case
1/3 would be ....010111. I don't know yet how to exactly express 5 and 7, but
they are undoubtedly infinite strings, being mutually prime with 2.
Tony Orlow
> Robert Low wrote:
> > Tony Orlow wrote:
> > > "Uncountability" of the reals is the result of a conflation of properties of
> > > digital number systems with some vague notion regarding the real numbers.
> >
> > Actually, uncountability of the reals is the result of a *precise*
> > notion regarding the real numbers. And if you're hung up on the idea
> > that it has something to do with 'digital number systems', you will
> > continue writing stuff that is 'not even wrong'.
>
> So, well-order the reals.
>
> You get into the problem then of mapping any well-ordered set to the
> reals, in extension of Cantor's first.
>
> One possibility of solution in consideration that meets an intuition,
> of, eg, Leibniz, is that there are adjacent points in the normal
> ordering of the reals.
really consistent with the intermediate value theorem.
>
> Otherwise the reals aren't a set, rationals are uncountable, etcetera.
little sleight of hand.
>
> ZF is inconsistent. Infinite sets are equivalent.
>
> The real numbers aren't necessarily a digital system, even though the
> definitions via Dedekind/Cauchy are of that sort.
>
> Hilbert requests a well-ordering of the reals. So, what is Goedel's
> construction?
than I could afford this morning.
>
> Ross
Tony Orlow
> Tony Orlow wrote:
> > Robert Low said:
> >
> >>Tony Orlow wrote:
> >>
> >>>"Uncountability" of the reals is the result of a conflation of properties of
> >>>digital number systems with some vague notion regarding the real numbers.
> >>
> >>Actually, uncountability of the reals is the result of a *precise*
> >>notion regarding the real numbers. And if you're hung up on the idea
> >>that it has something to do with 'digital number systems', you will
> >>continue writing stuff that is 'not even wrong'.
> > Excuse me, Robert, but can you offer a proof of the uncountability of the reals
> > which does NOT employ any digital representation? If not, then I suggest you
> > reevaulate your response. If so, dish it out.
>
> Sure. Fix e>0. Suppose the reals in [0,1] to be countable,
> and consider an enumeration of them.
>
> Round the i'th number, put the interval of length e/2^i.
this many times? Where are you putting this interval?
> Then all the reals in [0,1] have been covered by a set whose total
> measure is sum_i=1^infinity e/2^i = e. If e is less
> than 1, then you have covered the unit interval by
> a union of intervals the sum of whose lengths is less
> than 1.
Did you just bring in measure theory here? I thought that was entirely separate
from set theory. I am afraid your proof requires a little more explanation. I
await it with bated flatutence.
>
> This contradiction shows that the reals in [0,1]
> cannot be countable.
I don't understand the contradiction.
>
> I await the next helping of gibberish with no
> enthusiasm whatever.
Oh don't be such a sour apple. Be sweet like mine!
Daryl McCullough
>
>Dik T. Winter said:
>> Indeed, on the other hand it does not make it an enumeration either.
>> Because an enumeration maps to *finite* integers.
>Well, gee, that makes it impossible by definition to enumerate ANY actually
>infinite set, since that will require an infinite number of naturals to map to,
>and will require infinite bits to the binary representations by N=S^L.
What you are calling an "actually infinite set" is what everyone
else would call an "uncountably infinite set". With that interpretation,
your claim is correct: It is impossible to enumerate all the elements
of an uncountable set. That's why it's called "uncountable".
On the other hand, what you call a "finite but unbounded set" is what
everyone else would call a "countably infinite set".
--
Daryl McCullough
Ithaca, NY
Robert Low
> Robert Low said:
>>This contradiction shows that the reals in [0,1]
>>cannot be countable.
> I don't understand the contradiction.
You astonish me.
'Sir, I have given you an argument. I cannot
provide you an understanding.'
Dave Seaman
> Tony Orlow wrote:
>> Robert Low said:
>>>This contradiction shows that the reals in [0,1]
>>>cannot be countable.
>> I don't understand the contradiction.
> You astonish me.
Actually, the argument you gave should at least mention the Heine-Borel
theorem for, er, completeness.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>
Robert Low
> On Mon, 31 Oct 2005 17:53:13 +0000, Robert Low wrote:
>>Tony Orlow wrote:
>>>Robert Low said:
>>>>This contradiction shows that the reals in [0,1]
>>>>cannot be countable.
>>>I don't understand the contradiction.
>>You astonish me.
> Actually, the argument you gave should at least mention the Heine-Borel
> theorem for, er, completeness.
But it wouldn't have been nearly as compact then.
Anyway, I know it was just an outline, but what do I need
Heine-Borel for in particular?
Robert Low
The penny just dropped. I could reduce the countable cover
to a finite one, and then I have only a finite number of
intervals to add up. Nice: I hadn't thought of that
at all.
Virgil
Robert Low <mtx...@coventry.ac.uk> wrote:
TO is particularly wrong to invoke dependence on any 'digital number
system', since the *first* Cantor proof was totally independent of a
need for any 'digital number system' to represent the reals
Dave Seaman
How do you show that the outer measure of an interval is its length?
Or, to ask it another way, how do you show that [0,1] can't be covered by
a countable collection of intervals whose total length is less than 1/2?
I can prove those things by invoking Heine-Borel (hence, if there is a
countable collection that works, there is a finite collection that also
works), but perhaps you have a different proof in mind.
Dave Seaman
So it's not so astonishing that someone might not see the contradiction.
Virgil
Tony Orlow <ae...@cornell.edu> wrote:
Cantor's FIRST proof of the uncountability of the reals is just such a
proof, not employing ANY digital representation of the reals, and
totally independent of the existence of any such representations.
So we suggest that TO reevaluate his objections to Robert Low's
evaluation!
Robert Low
> On Mon, 31 Oct 2005 18:11:01 +0000, Robert Low wrote:
>>The penny just dropped. I could reduce the countable cover
>>to a finite one, and then I have only a finite number of
>>intervals to add up. Nice: I hadn't thought of that
>>at all.
> So it's not so astonishing that someone might not see the contradiction.
I admit, I was using a fairly naive approach to 'length'.
And being able to cover the whole interval [0,1]
with a family of intervals the sum of whose lengths
is arbitrarily small is certainly a sign that *something*
is broken.
In any case, do you really think that was the root of Tony's objection?
In any case, I guess that'll teach me to post a 'proof' I hadn't
thought hard about in public...
I'm sure there are much nicer proofs of the uncountability
of R about.
Robert Low
> Cantor's FIRST proof of the uncountability of the reals is just such a
> proof, not employing ANY digital representation of the reals, and
> totally independent of the existence of any such representations.
And it's here...
http://encyclopedia.thefreedictionary.com/Cantor's+first+uncountability+proof
Much more convincing than my effort.
Virgil
Tony Orlow <ae...@cornell.edu> wrote:
> Dik T. Winter said:
> > In article <MPG.1dcff1d0c...@newsstand.cit.cornell.edu> Tony
> > Orlow <ae...@cornell.edu> writes:
> > > Dik T. Winter said:
> > ...
> > > > As far as I can see your method does not generate numbers like 3, but
> > > > only
> > > > 0 and powers of 2.
> > > >
> > > If x is 1/2, then 3 is going to be an infinitely long bit string. If x
> > > is
> > > 1/3, then 2 is going to be an infintiely long bitstring. Similarly, in
> > > digital base 2. 1/3 is infinitely long and in digital base 3, 1/2 is
> > > infinitely long. This isn't a problem. All truly infinite sets, when
> > > expressed in symbolic language of any sort, require an infinitely long
> > > string to uniquely identify each element, even if the "finite" ones only
> > > have a finite number of SIGNIFICANT digits. This shouldn't be considered
> > > a problem. We have a linear order that doesn't end.
> >
> > Indeed, on the other hand it does not make it an enumeration either.
> > Because an enumeration maps to *finite* integers.
> Well, gee, that makes it impossible by definition to enumerate ANY
> actually infinite set, since that will require an infinite number of
> naturals to map to, and will require infinite bits to the binary
> representations by N=S^L.
No! It only makes impossible the enumeration of TOinfinite sets, but
does not impede the ennumeration of Dedekind infinite sets which TO
mislabels as finite.
>
> Wolfram is waiting for a contributed definition of "enumeration", but under
> enumerate, he has the following definition, which does not mention
> finiteness:
>
> "To enumerate a set of objects satisfying some set of properties means to
> explicitly produce a listing of all such objects."
"Listings" are mappings from the Dedekind infinite set of finite
naturals, or some subset, to the set being listed.
> >
> > But as you say this is an ordering, you should be able to tell me (when
> > we assume that x = 2) in what order the numbers 3, 5, 7 and 1/3 come out.
> >
> That would require deeper knowledge of tetration on my part.
In other words, TO doesn't know how his own alleged system works.
However, since TO's "ennumeration" of the reals involves infinite
strings of binaries, and the set of such infinite strings of binaries is
itself uncountable, at least in the standard mathematical meaning of
that word, TO's counting is not counting in any mathematical sense,but
only holds in TOmatics, which, as we all know, is mathematically
irrelevant.
Virgil
Tony Orlow <ae...@cornell.edu> wrote:
> Ross A. Finlayson said:
> > Robert Low wrote:
> > > Tony Orlow wrote:
> > > > "Uncountability" of the reals is the result of a conflation of
> > > > properties of
> > > > digital number systems with some vague notion regarding the real
> > > > numbers.
> > >
> > > Actually, uncountability of the reals is the result of a *precise*
> > > notion regarding the real numbers. And if you're hung up on the idea
> > > that it has something to do with 'digital number systems', you will
> > > continue writing stuff that is 'not even wrong'.
> >
> > So, well-order the reals.
> Consider it done! :D
Not outside of TOmatics.
In TO's infinite binary strings, I presume that those strings with only
finitely many 1's are ordered like the binary integers, so that leaving
out zeros to the left of all non-zero digits we can assume
1 < 10 < 11 < 100 < 101 < 110 < ... in his system.
The one can ask which is larger ....101010 or ...010101 ? And by what
rule does one decide?
Or are these to be TO's mythical infinite strings with two ends?
> >
> > You get into the problem then of mapping any well-ordered set to the
> > reals, in extension of Cantor's first.
> How so? Do you think it holds water?
Cantor's first proof is quite valid in standard mathematics. It is only
in systems like TOmatics, where everything can be proved false, that it
is at all questionable.
> > Hilbert requests a well-ordering of the reals. So, what is Goedel's
> > construction
?
> I was wondering the same thing. I looked it up, but it required more thought
> than I could afford this morning.
It requires more logical thought than TO is capable of even on his best
day.
Dave Seaman
> Dave Seaman wrote:
>> On Mon, 31 Oct 2005 18:11:01 +0000, Robert Low wrote:
>>>The penny just dropped. I could reduce the countable cover
>>>to a finite one, and then I have only a finite number of
>>>intervals to add up. Nice: I hadn't thought of that
>>>at all.
>> So it's not so astonishing that someone might not see the contradiction.
> I admit, I was using a fairly naive approach to 'length'.
> And being able to cover the whole interval [0,1]
> with a family of intervals the sum of whose lengths
> is arbitrarily small is certainly a sign that *something*
> is broken.
> In any case, do you really think that was the root of Tony's objection?
Not at all.
> In any case, I guess that'll teach me to post a 'proof' I hadn't
> thought hard about in public...
> I'm sure there are much nicer proofs of the uncountability
> of R about.
Cantor's first proof is more elementary than the measure-theoretic one,
and it doesn't depend on decimal representations.
Robert Low
> Cantor's first proof is more elementary than the measure-theoretic one,
> and it doesn't depend on decimal representations.
Yes. I didn't know it, but googled for it a few minutes ago. It's
cute.
A N Niel
> >
> Excuse me, Robert, but can you offer a proof of the uncountability of the
> reals
> which does NOT employ any digital representation? If not, then I suggest you
> reevaulate your response. If so, dish it out.
Cantor's original proof of uncountability did not use digital
representation.
Tony Orlow
Wolfram definition of "enumerate" mentions nothing about finiteness. Got a
reference which does? The reals are countable. Was Hilbert a crackpot for
suggesting that this may be the case? It's in his program, and it hasn't been
achieved...until now. So, get with the program.
Tony Orlow
again.
Tony Orlow
and tell you the mistake tomorrow. How's that?
Robert Low
> You can explain your vague terms. Or, maybe you can't. Too bad for you. Try
> again.
OK, if you don't like that verion, go read Cantor's first proof.
It's at
http://en.wikipedia.org/wiki/Cantor's_first_uncountability_proof
Daryl McCullough
You don't really think that Tony could understand that proof, do
you? He has spent literally years struggling with elementary
concepts of infinite sets. He's not really capable of reading
and following a proof on his own (or even with help).
Daryl McCullough
>
>Daryl McCullough said:
>> What you are calling an "actually infinite set" is what everyone
>> else would call an "uncountably infinite set". With that interpretation,
>> your claim is correct: It is impossible to enumerate all the elements
>> of an uncountable set. That's why it's called "uncountable".
>>
>> On the other hand, what you call a "finite but unbounded set" is what
>> everyone else would call a "countably infinite set".
>As I pointed out in the post you responded to here (which part was
>snipped) the Wolfram definition of "enumerate" mentions nothing
>about finiteness.
What you are calling "finite naturals" is what everyone else
calls "naturals". To enumerate a set means to set up a correspondence
between that set and the naturals. An enumeration means the same
thing as a sequence, which is a function whose domain is the
naturals (what you call the "finite naturals").
>Got a reference which does? The reals are countable.
In standard terminology, a set S is countable if there is
a surjection f from the naturals (what you'd call the "finite
naturals") to S. A surjection f from N to S is a function such
that every element of S is in the image of f.
What that means is that a set is countable if it is possible
to order the elements of the set such that every element has
finitely many predecessors (elements that come "earlier" in
the ordering). As I said, standard terminology uses the term
"countably infinite set" to mean approximately what you mean by
"finite unbounded set".
The reals are not countable, by this definition.
>Was Hilbert a crackpot for suggesting that this may be the case?
You are getting confused. Hilbert was not talking about enumerations,
he was talking about well-orderings. A well-ordering on a set S is a
binary ordering relation r that is total, transitive, and well-founded.
A relation r is said to be well-founded if every subset S' of S has
a smallest element, according to ordering r. For every well-ordering
there is a corresponding ordinal, which means that there is a bijection
m from S to a set of ordinals such that
r(x,y) -> m(x) < m(y)
Hilbert wondered whether the reals could be well-ordered. He definitely
didn't wonder if the reals could be countable. He certainly knew that
they weren't countable (according to the standard definition of "countable").
Robert Low
> In article <3snennF...@individual.net>, Robert Low says...
>>Tony Orlow wrote:
>>>You can explain your vague terms. Or, maybe you can't. Too bad for you. Try
>>>again.
>>OK, if you don't like that verion, go read Cantor's first proof.
>>It's at
>>http://en.wikipedia.org/wiki/Cantor's_first_uncountability_proof
> You don't really think that Tony could understand that proof, do
> you?
Just giving him alternatives. In any case, the point wasn't
whether Tony could understand that particular proof, but
to show him that his assumption that uncountability proofs
relied on a particular way of representing reals was
unfounded.
Of course he'll shortly provide us with a 'refutation'
of that proof which makes as much sense as anything else
he writes.
Russell
[snip]
> Hilbert wondered whether the reals could be well-ordered.
Wouldn't such a well-ordering have to be complicated enough
that it couldn't be used to decide CH? (Since CH is undecidable,
right?) It's hard for me to imagine how a well-ordering of the
reals could be specified in any book of finite length while still
leaving CH undecided.
Proginoskes
Peter Webb wrote:
> "Sean Holman" <holm...@yahoo.com> wrote in message
> news:16564229.1130535933...@nitrogen.mathforum.org...
> > You should note that to well order a set and to find an enumeration are
> > not the same thing. Your specific attempt to find an enumeration of the
> > reals (which by a diagonal argument is impossible) seems to only gets a
> > subset of the rationals (and +/- infinity). The axiom that any set has a
> > well-ordering (so called well ordering principle) is equivalent to the
> > axiom of choice, and thus entirely nonconstructive.
>
> The Axiom of Choice is non-constructive.
>
> This doesn't mean that there isn't a constructable choice function over a
> set, including R. As far as I know, nobody has proved that there is no
> constructable choice function over R (ie that we need the AoC for well
> ordering), its just that nobody has ever constructed such a function for R
> (or AFAIK for any uncountable set).
Last night I dreamt that I did. I used the subsets of N, putting the
ones of size 0, then size 1 (ordered lexigraphically), then size 2,
etc., of all finite sizes. But then I realized (still during the dream)
that this was a countable set; I hadn't included the infinte sets.
Really.
--- Christopher Heckman
David McAnally
>"Sean Holman" <holm...@yahoo.com> wrote in message
>news:16564229.1130535933...@nitrogen.mathforum.org...
>> You should note that to well order a set and to find an enumeration are
>> not the same thing. Your specific attempt to find an enumeration of the
>> reals (which by a diagonal argument is impossible) seems to only gets a
>> subset of the rationals (and +/- infinity). The axiom that any set has a
>> well-ordering (so called well ordering principle) is equivalent to the
>> axiom of choice, and thus entirely nonconstructive.
>The Axiom of Choice is non-constructive.
>This doesn't mean that there isn't a constructable choice function over a
>set, including R. As far as I know, nobody has proved that there is no
>constructable choice function over R (ie that we need the AoC for well
>ordering), its just that nobody has ever constructed such a function for R
>(or AFAIK for any uncountable set).
There exist models of ZF in which R cannot be well-ordered (so the
existence of a well-ordering of R is not a theorem of ZF). This
automatically implies that there is no constructible choice function on R,
since the existence of such a constructible function would imply that the
existence of a well-ordering of R is a theorem of ZF.
-----
Ross A. Finlayson
> Ross A. Finlayson wrote:
> > So, well-order the reals.
> > You get into the problem then of mapping any well-ordered set to the
> > reals, in extension of Cantor's first.
> > of, eg, Leibniz, is that there are adjacent points in the normal
> > ordering of the reals.
> > Otherwise the reals aren't a set, rationals are uncountable, etcetera.
> > ZF is inconsistent. Infinite sets are equivalent.
> > The real numbers aren't necessarily a digital system, even though the
> > definitions via Dedekind/Cauchy are of that sort.
> > Hilbert requests a well-ordering of the reals. So, what is Goedel's
> > construction?
No quarter.
>
> I think that the random phrase generator need to be re-initialised.
No dice.
Ross
Robert Low
> Robert Low wrote:
>>I think that the random phrase generator need to be re-initialised.
> No dice.
Ah, there's your problem.
Robert Low
>>>Tony Orlow wrote
>>>>"Uncountability" of the reals is the result of a conflation of properties
>>>>of
>>>>digital number systems with some vague notion regarding the real numbers.
and, rather later
> I was referring to the diagonal proof. I'll take a look at the first tonight
> and tell you the mistake tomorrow. How's that?
You have no shame whatever, do you?
Ross A. Finlayson
N_o problem.
For each cardinal there is an ordinal. Some sets are
ordering-sensitive.
d100: 09, 71, 48, 78, 82, 45, 04, 98, 06, 28, 24, 58, 99
d20; 3, 15, 4, 18
3d6: 7, 13, 9, 17, 12, 16, 12, 13
2d12: hard 12
2d8: 4
d4: 3
Well-order the reals.
Ross
Virgil
Tony Orlow <ae...@cornell.edu> wrote:
> As I pointed out in the post you responded to here (which part was
> snipped) the Wolfram definition of "enumerate" mentions nothing about
> finiteness. Got a reference which does? The reals are countable.
The reals are not countable using the standard definition of the
Dedekind infinite set of finite naturals.
Using TOmatics anything is anything, so that doesn't count!
From Websters Concise Electronic Dictionary
<quote>
1. enumerate
(verb) [enumerated; enumerated; enumerating; enumerates]
count
<\quote>
http://mathworld.wolfram.com/EnumerationProblem.html
<quote>
Enumeration Problem
The problem of determining (or counting) the set of all solutions to a
given problem.
<\quote>
Virgil
Tony Orlow <ae...@cornell.edu> wrote:
> Robert Low said:
> > Tony Orlow wrote:
> > > Robert Low said:
> > >>This contradiction shows that the reals in [0,1]
> > >>cannot be countable.
> > > I don't understand the contradiction.
> >
> > You astonish me.
> >
> > 'Sir, I have given you an argument. I cannot
> > provide you an understanding.'
> >
> You can explain your vague terms. Or, maybe you can't. Too bad for you. Try
> again.
Not needed until TO can "ennumerate" the reals using a set of naturals
that is no more than Dedekind infinite.
That is what "countable" means to everyone but TO.
Virgil
Tony Orlow <ae...@cornell.edu> wrote:
As, after all TO's study, TO has still not found any valid faults with
the diagonal proof, there is not much point in his looking further until
he does find some error in the diagonal proof.
Which will be long after he has explicitly enumerated all the members of
his *N.
Dik T. Winter
> Dik T. Winter said:
...
> > Because an enumeration maps to *finite* integers.
>
> Well, gee, that makes it impossible by definition to enumerate ANY actually
> infinite set, since that will require an infinite number of naturals to map
> to,
Upto here it is right.
> and will require infinite bits to the binary representations by N=S^L.
And this is still only your unfounded opinion.
> Wolfram is waiting for a contributed definition of "enumeration", but under
> enumerate, he has the following definition, which does not mention
> finiteness:
>
> "To enumerate a set of objects satisfying some set of properties means to
> explicitly produce a listing of all such objects."
Yes, so what? In mathematics a list is an ordered set indexed by the
finite integers.
> > But as you say this is an ordering, you should be able to tell me (when
> > we assume that x = 2) in what order the numbers 3, 5, 7 and 1/3 come out.
>
> That would require deeper knowledge of tetration on my part. Perhaps that
> will come with Herman Bosch's brother's work, which I hope to see soon.
> When I discussed this with Rusin months ago, he wanted his number 3. We
> determined that it did exist as an infinite string, something like
> ...010101, in whch case 1/3 would be ....010111. I don't know yet how to
> exactly express 5 and 7, but they are undoubtedly infinite strings, being
> mutually prime with 2.
Something need not be co-prime with 2 to get an infinite string, it is
sufficient to not be a power of 2. 6 also will require an infinite string.
But you have representations (apparently) of 1/3 and 3. Which of those
comes first in your ordering, and why?
More general, take the set of odd integers. As you claim that your ordering
is a well-ordering, you should be able to tell us what odd integer comes
first in your ordering. Or, alternatively, you have to prove that there
*is* an odd integer that comes first.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
Dik T. Winter
> Robert Low said:
...
> > Sure. Fix e>0. Suppose the reals in [0,1] to be countable,
> > and consider an enumeration of them.
>
> So e is any positive number.....
Yes. However, you better think about it as a positive number less than 1.
> > Round the i'th number, put the interval of length e/2^i.
>
> Round the ith number? What do you mean? Round to what? And then divide e
> by 2 this many times? Where are you putting this interval?
Let me reformulate: "put an interval of length e/(2^i) around the i-th
number".
> > Then all the reals in [0,1] have been covered by a set whose total
> > measure is sum_i=1^infinity e/2^i = e. If e is less
> > than 1, then you have covered the unit interval by
> > a union of intervals the sum of whose lengths is less
> > than 1.
>
> Did you just bring in measure theory here? I thought that was entirely
> separate from set theory. I am afraid your proof requires a little more
> explanation. I await it with bated flatutence.
Make that the sum of the lengths of theintervals...
> > This contradiction shows that the reals in [0,1]
> > cannot be countable.
>
> I don't understand the contradiction.
That the interval [0, 1] with length 1 can be covered by a collection of
intervals with a total length less than 1.
Sean Holman
> You should note that to well order a set and to find
> d an enumeration are not
> the same thing. Your specific attempt to find an
> enumeration of the reals
> (which by a diagonal argument is impossible) seems to
> only gets a subset of the
> rationals (and +/- infinity). The axiom that any set
> has a well-ordering (so
> called well ordering principle) is equivalent to the
> axiom of choice, and thus
> entirely nonconstructive.
> >
> successor function, and an
> order such that every element is either greater than
> or less than any other
> element. The diagonal argument is pure conflation,
> saying nothing about the
> real numbers, but barely revealing a property of
> digital number systems, and
> symbolic languages in general. It proves nothing
> ultimately about the
> "uncountability" of the reals, but only serves as the
> first, and very
> rudimentary, demonstration that infinities may have
> different sizes. it's time
> to move beyond the diagonal argument.
>
> Now, what makes you think this gives a subset of the
> rationals? Look again. I
> have updated the page with a more in-depth
> description of the enumeration.
> Among the elements you will find such irrational
> numbers as sqrt(2) and -1/sqrt
> (2). Some of them get far more "irrational" as one
> continues the enumeration.
I tried reading through the webpage again, and I gotta admit, I have no idea how you're generating the numbers, and I don't really see any effort to prove that all real numbers are in this enumeration (or in fact any definition of what you mean by "real numbers').
Perhaps we have different ideas of what "enumeration" means, but I thought an enumeration was a bijection with the natural numbers. I'm fairly certain that this is not the same as a well-ordering, although I also have to admit, I'm no expert in set theory.
I don't know what you mean when you say that, "The diagonal argument is pure conflation." It can be shown that there is a bijection between a subset of the reals (interval (0,1)) and a subset of the set of all infinite strings of digital symbols. The diagonal arument shows that this subset of strings of symbols is uncountable, thereby showing that the cardinality of the reals is uncountable. What's wrong with this?
Ross A. Finlayson
> Ross A. Finlayson said:
> > > Tony Orlow wrote:
> > > > "Uncountability" of the reals is the result of a conflation of properties of
> > > > digital number systems with some vague notion regarding the real numbers.
> > >
> > > Actually, uncountability of the reals is the result of a *precise*
> > > notion regarding the real numbers. And if you're hung up on the idea
> > > that it has something to do with 'digital number systems', you will
> > > continue writing stuff that is 'not even wrong'.
> >
> Consider it done! :D
> >
> > reals, in extension of Cantor's first.
> >
> > of, eg, Leibniz, is that there are adjacent points in the normal
> > ordering of the reals.
> really consistent with the intermediate value theorem.
> >
> > Otherwise the reals aren't a set, rationals are uncountable, etcetera.
> little sleight of hand.
> >
> > ZF is inconsistent. Infinite sets are equivalent.
> >
> > The real numbers aren't necessarily a digital system, even though the
> > definitions via Dedekind/Cauchy are of that sort.
> >
> > Hilbert requests a well-ordering of the reals. So, what is Goedel's
> > construction?
> than I could afford this morning.
> >
Yeah, the intermediate value theorem, that's basically the problem in
serializing the real numbers, because the real numbers are the complete
ordered field, except division by zero is undefined.
They're complete, but they're _ordered_.
While that's so, the rationals are a field, and ring, referring to
being its own closure with respect to division, and subtraction, and
their inverse operations. The irrationals are not, the sum or product
of two irrationals is not necessarily an irrational. The sum or
product of infinitely many rationals may be irrational.
Someone was saying that Goedel well-orders some subset, not necessarily
a proper subset, of the reals, basically the constructible reals as it
refers to that being a well-ordering following from V = L.
So, in ZFC with V = L, in extension of Cantor's first, as the reals are
the complete ordered field, in that they are complete, there are
adjacent points in the well-ordering that are adjacent in the normal
ordering, or the reals are not a set, because, and the rationals are
uncountable, because of a disjoint for each nested interval, and
various other problems.
If you would, please explain or provide reference to how Goedel claims
to well-order the reals.
If _you_ provide an example of the well-ordering of the reals, then
you've solved one of Hilbert's problems.
I already did.
Ross
imagin...@despammed.com
Daryl McCullough wrote:
> Tony Orlow says...
> >
> >Daryl McCullough said:
>
> >> What you are calling an "actually infinite set" is what everyone
> >> else would call an "uncountably infinite set". With that interpretation,
> >> your claim is correct: It is impossible to enumerate all the elements
> >> of an uncountable set. That's why it's called "uncountable".
> >>
> >> On the other hand, what you call a "finite but unbounded set" is what
> >> everyone else would call a "countably infinite set".
>
> >As I pointed out in the post you responded to here (which part was
> >snipped) the Wolfram definition of "enumerate" mentions nothing
> >about finiteness.
>
> What you are calling "finite naturals" is what everyone else
> calls "naturals".
I don't think this is right. Everything Tony has said about "his"
numbers suggests that his "infinite" numbers are actually finite. He
has fairly explicitly said that (for example) doubling every element in
an "infinite" set of naturals produces a set that is "twice as long".
In other words, he explicity rejects any of the normal implications of
being an infinite set. I think it is a vastly better fit if you regard
his "infinite" as referring to something "finite, but imponderably
large". This makes all the "range" stuff more or less work, and
explains the "twilight zone" - just as we can't say there is an
absolute boundary between "interesting" and "uninteresting" numbers, we
can't say there is ever a ponderable number to which adding one yields
an imponderable number.
> In standard terminology, a set S is countable if there is
> a surjection f from the naturals (what you'd call the "finite
> naturals") to S.
Again, this translation into Orlowspeak won't work, because even though
we can't identify the largest ponderable number, we do know they will
run out eventually. So TO will protest that such a set is bound to be
finite.
Brian Chandler
http://imaginatorium.org
Tony Orlow
> Dave Seaman wrote:
> >>Tony Orlow wrote:
> >>>>This contradiction shows that the reals in [0,1]
> >>>>cannot be countable.
> >>>I don't understand the contradiction.
> > Actually, the argument you gave should at least mention the Heine-Borel
> > theorem for, er, completeness.
>
> But it wouldn't have been nearly as compact then.
>
> Anyway, I know it was just an outline, but what do I need
> Heine-Borel for in particular?
>
If it was just an outline, then why are you astonished that it requires a
little more clarification, and why do you refuse to provide it? One might think
you were unable to explain yourself properly.
Tony Orlow
> In article <MPG.1dd0180a7...@newsstand.cit.cornell.edu>,
> Tony Orlow <ae...@cornell.edu> wrote:
>
> > Dik T. Winter said:
> > > In article <MPG.1dcff1d0c...@newsstand.cit.cornell.edu> Tony
> > > > > As far as I can see your method does not generate numbers like 3, but
> > > > > only
> > > > > 0 and powers of 2.
> > > > >
> > > > If x is 1/2, then 3 is going to be an infinitely long bit string. If x
> > > > is
> > > > 1/3, then 2 is going to be an infintiely long bitstring. Similarly, in
> > > > digital base 2. 1/3 is infinitely long and in digital base 3, 1/2 is
> > > > infinitely long. This isn't a problem. All truly infinite sets, when
> > > > expressed in symbolic language of any sort, require an infinitely long
> > > > string to uniquely identify each element, even if the "finite" ones only
> > > > have a finite number of SIGNIFICANT digits. This shouldn't be considered
> > > > a problem. We have a linear order that doesn't end.
> > >
> > > Indeed, on the other hand it does not make it an enumeration either.
> > > Because an enumeration maps to *finite* integers.
>
> > Well, gee, that makes it impossible by definition to enumerate ANY
> > actually infinite set, since that will require an infinite number of
> > representations by N=S^L.
>
> No! It only makes impossible the enumeration of TOinfinite sets, but
> does not impede the ennumeration of Dedekind infinite sets which TO
> mislabels as finite.
The reason it is possible to map any element in your version of enumeration to
a finite natural is because there are a finite number of them. You seem to
think that infinite positions in the sequence are not allowed, but without
them, you have a finite sequence, even if you can't identify where the finite
positions end.
>
> >
> > Wolfram is waiting for a contributed definition of "enumeration", but under
> > enumerate, he has the following definition, which does not mention
> > finiteness:
> >
> > "To enumerate a set of objects satisfying some set of properties means to
> > explicitly produce a listing of all such objects."
>
> naturals, or some subset, to the set being listed.
Got a reference which supports that definition? Wolfram has nothing to say
about the definition of a "listing" as far as I can tell.
> > >
> > > But as you say this is an ordering, you should be able to tell me (when
> > > we assume that x = 2) in what order the numbers 3, 5, 7 and 1/3 come out.
> > >
> > That would require deeper knowledge of tetration on my part.
>
In other words, there is always more to study. Studying any system which opens
up new avenues of thought is worthwhile. You might want to try that sometime.
>
> However, since TO's "ennumeration" of the reals involves infinite
> strings of binaries, and the set of such infinite strings of binaries is
> itself uncountable, at least in the standard mathematical meaning of
> that word, TO's counting is not counting in any mathematical sense,but
> only holds in TOmatics, which, as we all know, is mathematically
> irrelevant.
Irrelevant to your closed loop meanderings perhaps, but pertinent to many
aspects of set theory.
Robert Low
> Robert Low said:
>>>Actually, the argument you gave should at least mention the Heine-Borel
>>>theorem for, er, completeness.
>>
>>But it wouldn't have been nearly as compact then.
>>
>>Anyway, I know it was just an outline, but what do I need
>>Heine-Borel for in particular?
> If it was just an outline, then why are you astonished that it requires a
> little more clarification, and why do you refuse to provide it? One might think
> you were unable to explain yourself properly.
1/ Anybody competent reading my outline can see what is going on.
2/ There is no point in giving more detail for somebody
who won't understand the 'more detail'.
3/ The Heine-Borel theorem tells us that any open cover has
a finite subcover, so we can strengthen the previous statement
from 'for any e>o there is a countable family of intervals of total
length less than e covering the interval [0,1]' to ' for any e>0
there is a finite family of intervals of total length less than e
covering the interval [0,1]'. In the latter statement you
don't even have to know any measure theory to see that there
is a contradiction.
But you didn't raise any substantive issue at all: you simply
made it clear that you don't even grasp the language of
discourse.
Tony Orlow
> In article <MPG.1dd01969...@newsstand.cit.cornell.edu>,
> Tony Orlow <ae...@cornell.edu> wrote:
>
> > Ross A. Finlayson said:
> > > Robert Low wrote:
> > > > Tony Orlow wrote:
> > > > > properties of
> > > > > digital number systems with some vague notion regarding the real
> > > > > numbers.
> > > >
> > > > Actually, uncountability of the reals is the result of a *precise*
> > > > notion regarding the real numbers. And if you're hung up on the idea
> > > > that it has something to do with 'digital number systems', you will
> > > > continue writing stuff that is 'not even wrong'.
> > >
> > > So, well-order the reals.
> > Consider it done! :D
>
>
> In TO's infinite binary strings, I presume that those strings with only
> finitely many 1's are ordered like the binary integers, so that leaving
> out zeros to the left of all non-zero digits we can assume
> 1 < 10 < 11 < 100 < 101 < 110 < ... in his system.
>
> The one can ask which is larger ....101010 or ...010101 ? And by what
> rule does one decide?
Well, for one thing, the first is negative and the second positive, so the
first is obviously smaller. Say we compare ....101011 and ...1010101, they are
positive being x^x^-x^x^-x^x^-x... and x^-x^x^-x^x^-x... The first is x^(0101)^
(0101)^(0101)...., which is x^1^1^1^1..., which is x. The second is x^(1010)^
(1010)^(1010)... which is x^-x^-x^-x^-x..., which is infinite, and abviously
bigger than finite x. Happy? Did you like the ability to break the bit string
into terms like that, which you can recognize? Have a nice chew.
>
> Or are these to be TO's mythical infinite strings with two ends?
Those types of expressions are used for comparing two sets for the purpose of
measurement and expression of infinities and infinitesimals. Here, you are
asking what values these infinite bit strings map to. Given the repeating
pattern in these numbers, it's not hard to boil them down to infinite sequences
of x's in some form or another. Longer cycles of repetition will require more
complex expressions, of course.
> > >
> > > You get into the problem then of mapping any well-ordered set to the
> > > reals, in extension of Cantor's first.
>
> > How so? Do you think it holds water?
>
> in systems like TOmatics, where everything can be proved false, that it
> is at all questionable.
Heh! Well, I wasn't really familiar with Cantor's First. I am sure I saw it
years ago, but didn't really remember it, so I printed it out yesterday, and
looked it over last night. (sigh) Countability is absolutely dependent
according to the standard theory on the finiteness of the indexes on all
elements, apparently. It's really difficult to understand, then, why it claims
to talk about infinity regarding those sets. When one stops requiring that all
element indexes be finite, then one can have a truly infinite set size. Let me
summarize the proof and comment.
Essentially, we assume an enumeration of the reals with a first element and a
sequence thereafter, not in quantitative order. The proof is put in terms of
two sequences, such that sequence A begins with the first element, sequence B
begins with the second, and we alternate between adding an element to A which
is quantitatively between the last element of A and the last element of B, and
adding an element to B which is quantitatively between the last element of B
and the last element of A. In this way, we are narrowing the gap between that
first element of A and the first element of B. We assume some real c lies
between A_n and B_n, which are always different. But then, c can never be in
the reals, since then it would be in the original enumeration, and if so, would
end up in either A or B, the process would stop, and there would no longer be a
real between any two reals.
Hmmm... This is really no different from a single sequence where each element
after the first two is the first enumerated element which lies between the
previous two, quantitatively. So, let's see what sequence this is in The Well
Ordering of the Reals, and get a handle on this element, c. Here is a table of
the elements in the sequence, to illustrate what is going on:
00 -^-oo=-oo
01 x^-oo=oo
10 -x^oo=-0
101 x^-x^oo=1
1011 x^x^-x^oo=x
10111 x^x^x^-x^oo=x^x
101111 x^x^x^x^-x^oo=x^x^x
etc.
Interestingly, the sequence is just that sequence of tetrations (see
http://mathworld.wolfram.com/PowerTower.html ) of the form x^^k, for k in *N.
We can see that this sequence tends toward a bit string of the form
10111...111, which is always the last bit string of any given length. We can
see that 101 is 1, so this becomes the same as ...1111. In my well ordering, I
eliminated 0 as an element and kept -0, eliminating the top half of the set of
bit strings with a most significant bit in any given position, but I could
easily have eliminated -0, and kept the top half of each set, in which case the
sequence proposed by the proof would be:
00 -oo
01 x^-oo=oo
11 x^oo=0
111 x^x^oo=1
1111 x^x^x^oo=x
11111 x^x^x^x^oo=x^x
111111 x^x^x^x^x^oo=x^x^x
...
...111 x^x^x^x^x^x^x^x....
I have to admit that, despite my experience telling me that every contradiction
you produce in your proofs comes from the assumption of some last element of an
unending set, this one blind-sided me. But, given the well ordering I have
proposed, this real number c which is in question as an element in the
enumeration is, amazingly, yet another veiled form of the last element of the
set, since it is precisely that very same element that maps to the entire set,
......1111111111111, the very last element of the unending sequence!!! Amazing.
I am astounded. Set theory has outdone itself yet again, conflating the last
element with some kind of non existence of something unrelated. Any assumption
of a last element will result in a contradiction. This doesn't mean the well
ordering doesn't exist. There is no last element in the naturals, and yet they
are considered well ordered. So, what's the difference?
In short, this proof simply proves that there will be elements between any two
elements which require an infinite number of bits. If this makes the set
"uncountable" (I hate that term), all that really means is it's actually
infinite, requiring infinitely long symbolic representations (given a finite
alphabet) to enumerate all elements. This has no bearing on whether one can
establish a well ordering on the set. It simply means it's infinite.
>
> > > Hilbert requests a well-ordering of the reals. So, what is Goedel's
> > > construction
> ?
> > I was wondering the same thing. I looked it up, but it required more thought
> > than I could afford this morning.
>
> It requires more logical thought than TO is capable of even on his best
> day.
Sure, Virgil. See if you can glean the facts above, and if you're so
knowledgeable, why don't YOU explain Goedel's well ordering of the reals?
Tony Orlow
> On Mon, 31 Oct 2005 18:06:01 +0000, Robert Low wrote:
> > Dave Seaman wrote:
> >>>Tony Orlow wrote:
> >>>>>This contradiction shows that the reals in [0,1]
> >>>>>cannot be countable.
> >>>>I don't understand the contradiction.
> >>>You astonish me.
> >> theorem for, er, completeness.
>
> > But it wouldn't have been nearly as compact then.
>
> > Anyway, I know it was just an outline, but what do I need
> > Heine-Borel for in particular?
>
>
> Or, to ask it another way, how do you show that [0,1] can't be covered by
> a countable collection of intervals whose total length is less than 1/2?
>
> I can prove those things by invoking Heine-Borel (hence, if there is a
> countable collection that works, there is a finite collection that also
> works), but perhaps you have a different proof in mind.
>
>
>
There's really no telling WHAT Robert has in mind.
Tony Orlow
> Dave Seaman wrote:
> >>The penny just dropped. I could reduce the countable cover
> >>to a finite one, and then I have only a finite number of
> >>intervals to add up. Nice: I hadn't thought of that
> >>at all.
> > So it's not so astonishing that someone might not see the contradiction.
>
> I admit, I was using a fairly naive approach to 'length'.
> And being able to cover the whole interval [0,1]
> with a family of intervals the sum of whose lengths
> is arbitrarily small is certainly a sign that *something*
> is broken.
>
> In any case, do you really think that was the root of Tony's objection?
I didn't have enough information to object. I was asking for clarification, but
obviously your proof is not clear to you, so I won't hold my breath.
>
> In any case, I guess that'll teach me to post a 'proof' I hadn't
> thought hard about in public...
We all make mistakes.
>
> I'm sure there are much nicer proofs of the uncountability
> of R about.
Did Goedel construct a well ordering or not? Why is it still included in
Hilbert's Program as unsolved? Obviously these proofs are not generally
accepted as proving conclusively that it can't be done. So, it is my opinion
that any such proof is going to be flawed, almost certainly with some hidden
assumption of a last element. See my reply to Virgil regarding Cantor's First.
denis feldmann
No, he grasps perfectly well the language of trolling. See how cleverly
he used the only counter Dave Seaman so kindly provided him? (btw, I
was very dispointed with that, I was really eagerly waiting for TO
answering using only his own intellectual powers)
Tony Orlow
> Tony Orlow says...
> >
> >Daryl McCullough said:
>
> >> What you are calling an "actually infinite set" is what everyone
> >> else would call an "uncountably infinite set". With that interpretation,
> >> your claim is correct: It is impossible to enumerate all the elements
> >> of an uncountable set. That's why it's called "uncountable".
> >>
> >> On the other hand, what you call a "finite but unbounded set" is what
> >> everyone else would call a "countably infinite set".
>
> >As I pointed out in the post you responded to here (which part was
> >snipped) the Wolfram definition of "enumerate" mentions nothing
> >about finiteness.
>
> What you are calling "finite naturals" is what everyone else
> calls "naturals". To enumerate a set means to set up a correspondence
> between that set and the naturals. An enumeration means the same
> thing as a sequence, which is a function whose domain is the
> naturals (what you call the "finite naturals").
say that we have an infinite enumeration with a domain of *N? What does that
break? Nothing.
>
> >Got a reference which does? The reals are countable.
>
> In standard terminology, a set S is countable if there is
> a surjection f from the naturals (what you'd call the "finite
> naturals") to S. A surjection f from N to S is a function such
> that every element of S is in the image of f.
about actually infinite sets. Set theory shouldn't claim to deal with infinity
when it does everything possible not to.
>
> What that means is that a set is countable if it is possible
> to order the elements of the set such that every element has
> finitely many predecessors (elements that come "earlier" in
> the ordering). As I said, standard terminology uses the term
> "countably infinite set" to mean approximately what you mean by
> "finite unbounded set".
>
> The reals are not countable, by this definition.
numbers. That's why your set of naturals is finite.
>
> >Was Hilbert a crackpot for suggesting that this may be the case?
>
> You are getting confused. Hilbert was not talking about enumerations,
> he was talking about well-orderings. A well-ordering on a set S is a
> binary ordering relation r that is total, transitive, and well-founded.
> A relation r is said to be well-founded if every subset S' of S has
> a smallest element, according to ordering r. For every well-ordering
> there is a corresponding ordinal, which means that there is a bijection
> m from S to a set of ordinals such that
>
> r(x,y) -> m(x) < m(y)
>
> Hilbert wondered whether the reals could be well-ordered. He definitely
> didn't wonder if the reals could be countable. He certainly knew that
> they weren't countable (according to the standard definition of "countable").
No, of course they aren't, because the set is actually infinite. So, what
criterion of a well ordering do you think I have not satisfied?
>
> --
> Daryl McCullough
> Ithaca, NY
>
>
--
Smiles,
Tony Orlow
> In article <3snennF...@individual.net>, Robert Low says...
> >
> >Tony Orlow wrote:
> >> again.
> >
> >
> >It's at
> >
> >http://en.wikipedia.org/wiki/Cantor's_first_uncountability_proof
>
> You don't really think that Tony could understand that proof, do
> concepts of infinite sets. He's not really capable of reading
> and following a proof on his own (or even with help).
>
> --
> Daryl McCullough
> Ithaca, NY
>
>
Read my comments on that "proof", and think again. See where your precious
contradiction comes from, dumbass.
Tony Orlow
Robert Low
> No, he grasps perfectly well the language of trolling.
I really can't decide if he's a troll. If he is, though,
he's just as crazy as if he were sincere, considering the
sheer amount of time and effort he expends here.
Unless he's getting somebody to pay him for every
response he elicits...
Tony Orlow
advice. :D
Tony Orlow
himself put forth. Read my comments on the First, and behave yourself or you'll
be bringing a note home at the end of the day. And, spit out that gum!
Tony Orlow
> Robert Low wrote:
> > Ross A. Finlayson wrote:
> > > Robert Low wrote:
> > >>I think that the random phrase generator need to be re-initialised.
> > > No dice.
> >
> > Ah, there's your problem.
>
> N_o problem.
>
> For each cardinal there is an ordinal. Some sets are
> ordering-sensitive.
>
> d100: 09, 71, 48, 78, 82, 45, 04, 98, 06, 28, 24, 58, 99
> d20; 3, 15, 4, 18
> 3d6: 7, 13, 9, 17, 12, 16, 12, 13
> 2d12: hard 12
> 2d8: 4
> d4: 3
notation.
>
> Well-order the reals.
Okay, just a minute.....okay, they're all in order, which would you like? :D
>
> Ross
Robert Low
> Robert Low said:
>>Tony Orlow wrote:
>>>>>Tony Orlow wrote
>>>>>>"Uncountability" of the reals is the result of a conflation of properties
>>>>>>of
>>>>>>digital number systems with some vague notion regarding the real numbers.
>>and, rather later
>>>I was referring to the diagonal proof. I'll take a look at the first tonight
>>>and tell you the mistake tomorrow. How's that?
>>You have no shame whatever, do you?
> Why should I be ashamed?
Because without batting an eyelid you go from the (incorrect)
claim that proving the reals uncountable depends on a particular
representation of the reals to claiming that you were only
talking about one particular proof.
I am not the one who is unable to clarify the proof he
> himself put forth.
I admit that I can't make it clear to *you*. Nobody
else seemed to have a problem with it except to
point out that I assumed some measure theory that
I didn't need to because I could invoke the Heine-Borel
theorem to avoid it.
But then, nothing else seems to make sense to you except
your own writings, and they don't make sense to anybody
apart from you, so I'm not too concerned that you don't
understand it.
Tony Orlow
> In article <MPG.1dd0180a7...@newsstand.cit.cornell.edu> Tony Orlow <ae...@cornell.edu> writes:
> > Dik T. Winter said:
> ...
> > > Indeed, on the other hand it does not make it an enumeration either.
> > > Because an enumeration maps to *finite* integers.
> >
> > Well, gee, that makes it impossible by definition to enumerate ANY actually
> > infinite set, since that will require an infinite number of naturals to map
> > to,
>
> Upto here it is right.
>
> > and will require infinite bits to the binary representations by N=S^L.
>
> And this is still only your unfounded opinion.
>
> > Wolfram is waiting for a contributed definition of "enumeration", but under
> > enumerate, he has the following definition, which does not mention
> > finiteness:
> >
> > "To enumerate a set of objects satisfying some set of properties means to
> > explicitly produce a listing of all such objects."
>
> Yes, so what? In mathematics a list is an ordered set indexed by the
> finite integers.
"An data structure consisting of an ordered set of elements, each of which may
be a number, another list, etc. A list is usually denoted (, , ..., ) or , and
may also be interpreted as a vector (specifically, an n-vector) or an n-tuple.
Multiplicity matters in a list, so (1, 1, 2) and (1, 2) are not equivalent. "
Hmm... Still no mention of naturals or finiteness. Got a reference?
>
> > > But as you say this is an ordering, you should be able to tell me (when
> > > we assume that x = 2) in what order the numbers 3, 5, 7 and 1/3 come out.
> >
> > That would require deeper knowledge of tetration on my part. Perhaps that
> > will come with Herman Bosch's brother's work, which I hope to see soon.
> > When I discussed this with Rusin months ago, he wanted his number 3. We
> > determined that it did exist as an infinite string, something like
> > ...010101, in whch case 1/3 would be ....010111. I don't know yet how to
> > exactly express 5 and 7, but they are undoubtedly infinite strings, being
> > mutually prime with 2.
>
> Something need not be co-prime with 2 to get an infinite string, it is
> sufficient to not be a power of 2. 6 also will require an infinite string.
Yes, co-prime was not exactly the right word. Let's say one gets an infinite
number of bits if there is a non-common prime denominator. Is that good?
> But you have representations (apparently) of 1/3 and 3. Which of those
> comes first in your ordering, and why?
In each set of strings of a given length (same position for most significant
bit) an integral value comes first, then its negative, then its inverse, so
wherever 3 resides, 1/3 is its successor's successor, being the x^(...) to 3's
x^-(...). I think Dave Rusin confirmed that 3 was something like
...000100010001, in which case 1/3 would be ...00010001011.
>
> More general, take the set of odd integers. As you claim that your ordering
> is a well-ordering, you should be able to tell us what odd integer comes
> first in your ordering. Or, alternatively, you have to prove that there
> *is* an odd integer that comes first.
If I use odd x that is an easy matter, but then you will want me to find a
multiple of 2. I am not yet an expert in calculating the values of tetrations,
but I think the fact that this process divides the interval infinitely is
sufficient to say that it enumerates any real given an infinite number of
iterations.
Daryl McCullough
>The splinter you perceive in my eye is but a reflection of the log in yours.
I'm sure that there is a log in my own eye. I'm not claiming to
be a genius, or even particularly good at math, just to be minimally
competent. You, on the other hand, are a complete idiot.
David Kastrup
> Daryl McCullough said:
>> Tony Orlow says...
>> >
>> >Daryl McCullough said:
>>
>> >> What you are calling an "actually infinite set" is what everyone
>> >> else would call an "uncountably infinite set". With that
>> >> interpretation, your claim is correct: It is impossible to
>> >> enumerate all the elements of an uncountable set. That's why
>> >> it's called "uncountable".
>> >>
>> >> On the other hand, what you call a "finite but unbounded set" is
>> >> what everyone else would call a "countably infinite set".
>>
>> >As I pointed out in the post you responded to here (which part was
>> >snipped) the Wolfram definition of "enumerate" mentions nothing
>> >about finiteness.
>>
>> What you are calling "finite naturals" is what everyone else calls
>> "naturals". To enumerate a set means to set up a correspondence
>> between that set and the naturals. An enumeration means the same
>> thing as a sequence, which is a function whose domain is the
>> naturals (what you call the "finite naturals").
>
> So, what if I consider infinite naturals, despite your fear of them?
> Can I not say that we have an infinite enumeration with a domain of
> *N? What does that break? Nothing.
It just has nothing to do whatsoever with the naturals defined by the
Peano axioms, and thus is irrelevant.
>> >Got a reference which does? The reals are countable.
>>
>> In standard terminology, a set S is countable if there is a
>> surjection f from the naturals (what you'd call the "finite
>> naturals") to S. A surjection f from N to S is a function such that
>> every element of S is in the image of f.
>
> So, countable means finite but unbounded, as I've said for months.
No. It does not mean finite either which way. It does not mean
finite for the member values (in fact, it means nothing whatsoever for
the member values except that they are distinguishable), and it
certainly does not mean finite for the set.
And for a set, there is no such thing as "finite but unbounded". If a
set is unbounded (or even if it just open), then no member is a bound,
and thus for every member there exists at least one larger member.
And that means that if there is at least one element in the set, there
is at least one non-ending chain of strictly increasing members. Now
if you surject all elements in one such chain onto their successor in
the chain, you have bijected the set onto a proper subset of itself.
And that means that _any_ unbounded (and any non-empty open) ordered
set is infinite.
> I am talking about actually infinite sets. Set theory shouldn't
> claim to deal with infinity when it does everything possible not to.
And you shouldn't babble nonsense without a clue. Not after all this
time.
> Well, of course not. No actually infinite set is indexable by only
> finite numbers. That's why your set of naturals is finite.
Nonsense again. The naturals are an infinite set (as explained above)
and is "indexable" by only finite numbers. But its size is _not_ a
finite number. There is no last finite number, something which you
still do not understand.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
Tony Orlow
do you not consider the set of reals in [0,1] to be equal to the set in [0,2]?
What does measure of the continuum have to do with being able to well order the
set of values within it? I mean, the way I see it, fully dense intervals of
reals should be considered sets with size equal to their measure, but that's
not standard set theory.
Tony Orlow
> In article <MPG.1dd04c0c6...@newsstand.cit.cornell.edu>,
> > > >>This contradiction shows that the reals in [0,1]
> > > >>cannot be countable.
> > > > I don't understand the contradiction.
> > >
> > >
> > > 'Sir, I have given you an argument. I cannot
> > > provide you an understanding.'
> > >
> > You can explain your vague terms. Or, maybe you can't. Too bad for you. Try
> > again.
>
> that is no more than Dedekind infinite.
>
> That is what "countable" means to everyone but TO.
>
But that has little to do with a well ordering. Show me where a well order is
necessarily countable. If it were, then Why would Goedel and Hilbert even
consider the possibility of a well-ordering? Stick to topic.
Tony Orlow
> In article <MPG.1dd04cb16...@newsstand.cit.cornell.edu>,
> Tony Orlow <ae...@cornell.edu> wrote:
>
> > Virgil said:
> > > In article <3smn5fF...@individual.net>,
> > > Robert Low <mtx...@coventry.ac.uk> wrote:
> > >
> > > > Tony Orlow wrote:
> > > > > "Uncountability" of the reals is the result of a conflation of
> > > > > properties
> > > > > of
> > > > > digital number systems with some vague notion regarding the real
> > > > > numbers.
> > > >
> > > > notion regarding the real numbers. And if you're hung up on the idea
> > > > that it has something to do with 'digital number systems', you will
> > > > continue writing stuff that is 'not even wrong'.
> > >
> > > system', since the *first* Cantor proof was totally independent of a
> > > need for any 'digital number system' to represent the reals
> > >
> > I was referring to the diagonal proof. I'll take a look at the first tonight
> > and tell you the mistake tomorrow. How's that?
>
> the diagonal proof, there is not much point in his looking further until
> he does find some error in the diagonal proof.
>
> Which will be long after he has explicitly enumerated all the members of
> his *N.
>
Been there, done that, multiple times. Virgil's memory issues are not my
problem.
Daryl McCullough
>
>Daryl McCullough said:
>So, what if I consider infinite naturals, despite your fear of them?
Who said that I *fear* them. There are perfectly coherent theories
of nonstandard integers, and I have passing familiarity with them.
But your particular theory is nonsense.
>Can I not say that we have an infinite enumeration with a domain of
>*N? What does that break? Nothing.
Sure. Given any set A, you can form "bit strings over A", which
are functions from A to {0,1}. But what you can't have is a
surjection from A to the set of bit strings over A. Yes, I
know you think otherwise, but you are an idiot.
>> What that means is that a set is countable if it is possible
>> to order the elements of the set such that every element has
>> finitely many predecessors (elements that come "earlier" in
>> the ordering). As I said, standard terminology uses the term
>> "countably infinite set" to mean approximately what you mean by
>> "finite unbounded set".
>Yes, exactly.
>>
>> The reals are not countable, by this definition.
>
>Well, of course not. No actually infinite set
You mean "uncountably infinite set"
>is indexable by only finite numbers. That's why your set of naturals
>is finite.
You mean "countably infinite". After translation to the
correct terminology, that's right. No uncountably infinite
set can be indexed by a countably infinite set.
>> You are getting confused. Hilbert was not talking about enumerations,
>> he was talking about well-orderings. A well-ordering on a set S is a
>> binary ordering relation r that is total, transitive, and well-founded.
>
>Which of these criteria does this well ordering not meet?
What ordering are you talking about? Are you talking about
the ordering on infinite bit strings
000...0
000...1
000..10
000..11
etc
That ordering is not well-founded. To say that an ordering
is well-founded is to say that there is no infinite sequence
of decreasing elements. For that particular ordering, the
sequence
11000...0
101000...0
1001000...0
10001000...0
...
is an infinite sequence of decreasing elements. So your
ordering is not well-founded, and so is not a well-ordering.
Daryl McCullough
Ithaca, NY
Tony Orlow
> Tony Orlow wrote:
> > Robert Low said:
> >>Dave Seaman wrote:
> >>>Actually, the argument you gave should at least mention the Heine-Borel
> >>>theorem for, er, completeness.
> >>
> >>But it wouldn't have been nearly as compact then.
> >>
> >>Anyway, I know it was just an outline, but what do I need
> >>Heine-Borel for in particular?
> > If it was just an outline, then why are you astonished that it requires a
> > little more clarification, and why do you refuse to provide it? One might think
> > you were unable to explain yourself properly.
>
> 1/ Anybody competent reading my outline can see what is going on.
> 2/ There is no point in giving more detail for somebody
> who won't understand the 'more detail'.
> 3/ The Heine-Borel theorem tells us that any open cover has
> a finite subcover, so we can strengthen the previous statement
> from 'for any e>o there is a countable family of intervals of total
> length less than e covering the interval [0,1]' to ' for any e>0
> there is a finite family of intervals of total length less than e
> covering the interval [0,1]'. In the latter statement you
> don't even have to know any measure theory to see that there
> is a contradiction.
yields a finite number of subintervals", aka, countable set of subintervals.
So, sure, given a finite (countable) number of subdivisions, you will not
enumerate every real, since it is actually an infinite set. What does this say
about the well ordering? Nothing.
>
> But you didn't raise any substantive issue at all: you simply
> made it clear that you don't even grasp the language of
> discourse.
clarify what I am saying, and repeat it ad nauseum, all the time around here.
If your language is unclear, and you are giving an outline of a proof with
which I am not familiar, then you should be prepared to provide clarificatrion
if needed. As it stands, your proof says nothing about the Well Ordering of the
Reals.
Tony Orlow
astonishment at the request for clarification. Then it comes out that he
doesn't know what he's talking about, admits it was vague, and can't clarify
his "proof". I don't think any of you even know what it is you're proving half
the time.
Daryl McCullough
>Did Goedel construct a well ordering or not? Why is it still included in
>Hilbert's Program as unsolved?
Goedel constructed a well ordering of a particular set of reals,
the so-called "constructible reals". It is an unsolved problem
whether there are reals that are not constructible. The axioms
of set theory don't prove that all reals are constructible, and
they also don't prove that there exists any reals that are not
constructible.
If all reals are constructible, then Godel's construction is
a well-ordering of the reals.
--
Daryl McCullough
Ithaca, NY
Tony Orlow
> Tony Orlow wrote:
> > Robert Low said:
> >>Tony Orlow wrote:
> >>>>>Tony Orlow wrote
> >>>>>>"Uncountability" of the reals is the result of a conflation of properties
> >>>>>>of
> >>>>>>digital number systems with some vague notion regarding the real numbers.
> >>and, rather later
> >>>I was referring to the diagonal proof. I'll take a look at the first tonight
> >>>and tell you the mistake tomorrow. How's that?
> >>You have no shame whatever, do you?
> > Why should I be ashamed?
>
> Because without batting an eyelid you go from the (incorrect)
> claim that proving the reals uncountable depends on a particular
> representation of the reals to claiming that you were only
> talking about one particular proof.
as usual, of the unending set. besides, it seems to have been put forth that
well ordering depends on "countability", but that is obviously not the case.
So, what do I care if they aren't countable. They are still enumerable in a
well ordered sequence. So, poopoo to your little countable sets. They are
irrelevant. Hilbert knew the reals were uncountable when he called for a well
ordering, so can it.
>
> I am not the one who is unable to clarify the proof he
> > himself put forth.
>
> I admit that I can't make it clear to *you*. Nobody
> else seemed to have a problem with it except to
> point out that I assumed some measure theory that
> I didn't need to because I could invoke the Heine-Borel
> theorem to avoid it.
professional mathematician, as you can see, and may require clarification
sometimes. If I can dish it out all the time I can expect some back. Perhaps I
will just flip off any questions you have in astonishment at your ignorance of
Bigulosity Theory, and see how you react.
>
> But then, nothing else seems to make sense to you except
> your own writings, and they don't make sense to anybody
> apart from you, so I'm not too concerned that you don't
> understand it.
ordering. Sure, the reals aren't "countable", since the set is actually
infinite, but they are well orderable.
Tony Orlow
Daryl McCullough
>> Yes, so what? In mathematics a list is an ordered set indexed by the
>> finite integers.
>Oh, here it is, Wolfram's "List", under Data Structures:
>
>"An data structure consisting of an ordered set of elements, each of
>which may be a number, another list, etc. A list is usually denoted
>(, , ..., ) or , and may also be interpreted as a vector (specifically,
>an n-vector) or an n-tuple. >Multiplicity matters in a list, so (1, 1, 2)
>and (1, 2) are not equivalent. "
>
>Hmm... Still no mention of naturals or finiteness. Got a reference?
What do you think 'n' means in 'n-tuple'? It's a natural number.
An n-tuple is a structure with n components.
Tony Orlow
"A totally ordered set S is said to be well ordered (or have a well-founded
order) iff every nonempty subset of S has a least element (Ciesielski 1997, p.
38"
Your set would appear to have a least element, in terms of order, of 100...001.
Sorry.
Of course, it does warm my heart to see you using my notation for infinite
naturals and be reminded that, given enough repetition and examples, even a
total moron can learn something new, even if it a sort of horse-foot-stamping
kind of affair.
>
> is an infinite sequence of decreasing elements. So your
> ordering is not well-founded, and so is not a well-ordering.
It has a least member, so it is well ordered. Try again, pith-brain.
Tony Orlow
irrelevant to whether this is a well ordering of the reals.
>
> >> >Got a reference which does? The reals are countable.
> >>
> >> In standard terminology, a set S is countable if there is a
> >> surjection f from the naturals (what you'd call the "finite
> >> naturals") to S. A surjection f from N to S is a function such that
> >> every element of S is in the image of f.
> >
> > So, countable means finite but unbounded, as I've said for months.
>
> No. It does not mean finite either which way. It does not mean
> finite for the member values (in fact, it means nothing whatsoever for
> the member values except that they are distinguishable), and it
> certainly does not mean finite for the set.
number of predecessors.
>
> And for a set, there is no such thing as "finite but unbounded". If a
> set is unbounded (or even if it just open), then no member is a bound,
> and thus for every member there exists at least one larger member.
> And that means that if there is at least one element in the set, there
> is at least one non-ending chain of strictly increasing members. Now
> if you surject all elements in one such chain onto their successor in
> the chain, you have bijected the set onto a proper subset of itself.
call the set of finite naturals infinite, countable or otherwise.
>
> And that means that _any_ unbounded (and any non-empty open) ordered
> set is infinite.
> > I am talking about actually infinite sets. Set theory shouldn't
> > claim to deal with infinity when it does everything possible not to.
>
> And you shouldn't babble nonsense without a clue. Not after all this
> time.
>
> > Well, of course not. No actually infinite set is indexable by only
> > finite numbers. That's why your set of naturals is finite.
>
> Nonsense again. The naturals are an infinite set (as explained above)
> and is "indexable" by only finite numbers. But its size is _not_ a
> finite number. There is no last finite number, something which you
> still do not understand.
is based on a contradiction derived from the assumption of such a beast, in one
veiled form or another.
Tony Orlow
qualifies as a valid reference supporting your claim that a well ordering must
only apply to a countable set. If that were the case, then a well ordering of
the reals would be impossible, and it wouldn't be part of Hilbert's Program.
Perhaps you know more about sets than Hilbert? Gee, I guess it's time for you
to apply for that Fields medal, eh, Daryl?
You know, one CAN talk about infinite-dimensional space, within which each
point would be identified by an n-tuple of coordinates, where n=oo. So, n is
not necessarily finite in that context. Sorry, it's time for you to wake up
from your dream, Daryl. The more you project your idiocy on me, the less
capable I think you are. Keep stamping that foot, Silver! Have an apple.
David Kastrup
> David Kastrup said:
>> Tony Orlow <ae...@cornell.edu> writes:
>>
>> > So, countable means finite but unbounded, as I've said for months.
>>
>> No. It does not mean finite either which way. It does not mean
>> finite for the member values (in fact, it means nothing whatsoever
>> for the member values except that they are distinguishable), and it
>> certainly does not mean finite for the set.
>
> Essentially it does. A countable set consists only of elements with
> a finite number of predecessors.
An infinite number of them, yes. According to which definition does
that make the set finite?
>> And for a set, there is no such thing as "finite but unbounded".
>> If a set is unbounded (or even if it just open), then no member is
>> a bound, and thus for every member there exists at least one larger
>> member. And that means that if there is at least one element in
>> the set, there is at least one non-ending chain of strictly
>> increasing members. Now if you surject all elements in one such
>> chain onto their successor in the chain, you have bijected the set
>> onto a proper subset of itself.
>
> Yeah, yeah, I know your terminology, but it's misleading, and I am
> not going to call the set of finite naturals infinite, countable or
> otherwise.
Fine. So you agree that the set of (finite) naturals as defined by
the Peano axioms is infinite as defined by Dedekind.
>> And that means that _any_ unbounded (and any non-empty open)
>> ordered set is infinite.
>
> In the Dedekind sense, yes, I know. I don't need this repeated.
But that's what mathematicians are talking about when talking about
"infinite sets", and you have not come up with any different
definition. So you have no business complaining.
>> > Well, of course not. No actually infinite set is indexable by only
>> > finite numbers. That's why your set of naturals is finite.
>>
>> Nonsense again. The naturals are an infinite set (as explained
>> above) and is "indexable" by only finite numbers. But its size is
>> _not_ a finite number. There is no last finite number, something
>> which you still do not understand.
>
> I understand it very well. You do not understand that every one of
> your proofs is based on a contradiction derived from the assumption
> of such a beast, in one veiled form or another.
Guffaw. That's really cute coming, from all people in the world, you.