Sneaky *Intractable Diagonal* DisProof of Cantor
Graham Cooper
random digit generator.
LIST(row, digit) = RANDOM(0,9)
For the diagonal proof to be valid, it must provide a MISSING REAL
NUMBER for each of the INFINITELY MANY PERMUTATIONS of the same LIST.
Assume the DIAGONAL is 1 2 3 4 9 9 9 9 9 8 8 8 3 6 8 ...
where
DIAGONAL(LIST) = LIST(1,1) . LIST(2,2) . LIST(3,3) . ...
Suppose we wanted to change digit 1 by substituting LIST' for LIST.
We search for a row that has a matching digit with row 1 at a higher
digit place on the diagonal.
Say row 1 is 1 4 3 4 4 4 4 4 ...
and row 3 is 6 6 3 6 6 6 6 6 ...
We reorder the LIST to LIST' by swapping rows 1 and 3 and the diagonal
becomes
6 2 3 4 9 9 9 9 9 8 8 8 3 6 8 ...
^
Similarly we can change the 2nd digit to
6 9 3 4 9 9 9 9 9 8 8 8 3 6 8 ....
^ ^
Therfore, for all n, the first n digits of the diagonal are arbitrary.
By induction, the diagonal is intractable and not specified at all,
for certain classes of infinite lists.
QED
Ray Vickson
> Start with an infinite list of reals generated by a normal run of a
> random digit generator.
>
> LIST(row, digit) = RANDOM(0,9)
>
> For the diagonal proof to be valid, it must provide a MISSING REAL
> NUMBER for each of the INFINITELY MANY PERMUTATIONS of the same LIST.
No, No, No! It must provide a missing real number for any *given*
list. A permutation of a list is a new list, and so will have a
different missing real number. Where did you ever get the idea that a
single missing number must, necessarily, apply to the diagonals of all
permutations? Where in the Cantor, or later, arguments, was this ever
stated? Hint: it wasn't.
R.G. Vickson
LudovicoVan
On Feb 27, 6:27 pm, Ray Vickson <RGVick...@shaw.ca> wrote:
> On Feb 27, 5:59 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > For the diagonal proof to be valid, it must provide a MISSING REAL
> > NUMBER for each of the INFINITELY MANY PERMUTATIONS of the same LIST.
>
> No, No, No! It must provide a missing real number for any *given*
> list. A permutation of a list is a new list, and so will have a
> different missing real number. Where did you ever get the idea that a
> single missing number must, necessarily, apply to the diagonals of all
> permutations?
I'd expect the permuted list to contain *exactly* the same entries as
the original list. Hence, the missing entry should be a missing entry
in the permuted list as well; conversely, an entry in the permuted
list should be an entry in the original list as well. Where is the
mistake?
-LV
fishfry
<7a2e4367-5e54-4d61...@a11g2000pro.googlegroups.com>,
Graham Cooper <graham...@gmail.com> wrote:
> Start with an infinite list of reals generated by a normal run of a
> random digit generator.
>
> LIST(row, digit) = RANDOM(0,9)
>
> For the diagonal proof to be valid, it must provide a MISSING REAL
> NUMBER for each of the INFINITELY MANY PERMUTATIONS of the same LIST.
>
It would help if you use the same terminology as the rest of us.
A "list" by definition is a function from the natural numbers N to the
reals R. Notationally, a list L is a function L:N -> R.
If you permute a list, you get a different list.
For example:
1
2
3
and
3
2
1
are two different lists.
What you are doing is considering a countable *set* of numbers and
permuting them to get various lists.
Nevertheless, we understand what you mean. But the list is arbitrary.
You already agree that ANY list of reals is necessarily incomplete.
Therefore there is no list of all the reals.
I don't get why you either don't understand, or pretend to not
understand, this fairly simply point.
LudovicoVan
On Feb 27, 7:34 pm, fishfry <BLOCKSPAMfish...@your-mailbox.com> wrote:
> Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > Start with an infinite list of reals generated by a normal run of a
> > random digit generator.
>
> > LIST(row, digit) = RANDOM(0,9)
>
> > For the diagonal proof to be valid, it must provide a MISSING REAL
> > NUMBER for each of the INFINITELY MANY PERMUTATIONS of the same LIST.
>
> It would help if you use the same terminology as the rest of us.
>
> A "list" by definition is a function from the natural numbers N to the
> reals R. Notationally, a list L is a function L:N -> R.
>
> If you permute a list, you get a different list.
>
> For example:
>
> 1
> 2
> 3
>
> and
>
> 3
> 2
> 1
>
> are two different lists.
>
> What you are doing is considering a countable *set* of numbers and
> permuting them to get various lists.
>
> Nevertheless, we understand what you mean.
Do you?
> But the list is arbitrary.
> You already agree that ANY list of reals is necessarily incomplete.
> Therefore there is no list of all the reals.
>
> I don't get why you either don't understand, or pretend to not
> understand, this fairly simply point.
Your fairly simple point adds nothing: we all already know what the
diagonal argument states and concludes. Don't you understand, or is it
you just pretend?
-LV
fishfry
<c6e41a7b-645d-4106...@t15g2000prt.googlegroups.com>,
LudovicoVan <ju...@diegidio.name> wrote:
I genuinely do not understand Herc's continued unwillingness to accept
Cantor's proof. Can you explain Herc's point to me?
LudovicoVan
> I genuinely do not understand Herc's continued unwillingness to accept
> Cantor's proof. Can you explain Herc's point to me?
Herc's point is not about unwillingness, yours rather is. As to what
the OP might mean, just read my first post in this thread.
-LV
Arturo Magidin
The *reasoning* showing that the particular number (obtained by the
diagonal argument) is not on the permuted list would be different from
the reasoning for why it is not in the original list. The original
argument ("different from first number in first decimal digit,
different from second number in second decimal", etc) may be false for
the permuted list. Of course, it is trivial to fix it: if pi is the
permutation in question of the list, then the constructed number is
not on the list, because it is different from the pi(n)-th number on
the list on the nth digit. (it's nth decimal may very well be equal to
the nth decimal of the pi^{-1](n)th number on the list). Of course, we
can construct a new number with the permuted list to which the
original argument will apply with no change, but if you change the
list, even if you just permute it, there is no reason to expect the
rest of the proof to go through without changing a single word. Proofs
are not mad-libs, after all.
--
Arturo Magidin
DonH
news:7a2e4367-5e54-4d61...@a11g2000pro.googlegroups.com...
> Start with an infinite list of reals generated by a normal run of a
> random digit generator.
# 2 * (1+5+12+14)+14 = LYN DEKEL
fishfry
<e5ac014c-499f-493f...@o18g2000prh.googlegroups.com>,
LudovicoVan <ju...@diegidio.name> wrote:
Your first post in this particular thread is your first response to me.
Can you just tell me what Herc's point is in all this? He's very
persistent, and he seems to have something in mind.
LudovicoVan
> In article
> <e5ac014c-499f-493f-8943-0eb1475f7...@o18g2000prh.googlegroups.com>,
> LudovicoVan <ju...@diegidio.name> wrote:
> > On Feb 27, 7:54 pm, fishfry <BLOCKSPAMfish...@your-mailbox.com> wrote:
>
> > > I genuinely do not understand Herc's continued unwillingness to accept
> > > Cantor's proof. Can you explain Herc's point to me?
>
> > Herc's point is not about unwillingness, yours rather is. As to what
> > the OP might mean, just read my first post in this thread.
>
No, it isn't:
https://groups.google.com/group/sci.math/msg/9b4d440cc29861c1
Rather funny that you manage to get lost in a thread with some 10
posts in it.
-LV
LudovicoVan
> On Feb 27, 1:06 pm, LudovicoVan <ju...@diegidio.name> wrote:
> > [Only to sci.math, sci.logic.]
> > On Feb 27, 6:27 pm, Ray Vickson <RGVick...@shaw.ca> wrote:
> > > On Feb 27, 5:59 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > > > For the diagonal proof to be valid, it must provide a MISSING REAL
> > > > NUMBER for each of the INFINITELY MANY PERMUTATIONS of the same LIST.
>
> > > No, No, No! It must provide a missing real number for any *given*
> > > list. A permutation of a list is a new list, and so will have a
> > > different missing real number. Where did you ever get the idea that a
> > > single missing number must, necessarily, apply to the diagonals of all
> > > permutations?
>
> > I'd expect the permuted list to contain *exactly* the same entries as
> > the original list. Hence, the missing entry should be a missing entry
> > in the permuted list as well; conversely, an entry in the permuted
> > list should be an entry in the original list as well. Where is the
> > mistake?
>
> The *reasoning* showing that the particular number (obtained by the
> diagonal argument) is not on the permuted list would be different from
> the reasoning for why it is not in the original list.
Well, certainly different in the details, but not fundamentally
different in the notions involved. Namely, I am still of the advice
that Cooper's statement quoted above is actually true, at least for
the reason I have myself provided. OTOH, upon further reflection, I
agree that it is in fact not a disproof of the diagonal argument: i.e.
the anti-diagonal is not on the list, nor it can be on any of its
permutations! -- I've got definitely convinced after playing with few
tentative constructions, but your argument below about "the
constructed number [not being] on the list because it is different
from the pi(n)-th number on the list on the nth digit" is of course
compelling enough.
Thanks for the feedback,
-LV
Virgil
<8348f24d-d1df-47bd...@m7g2000vbq.googlegroups.com>,
LudovicoVan <ju...@diegidio.name> wrote:
Each such list of reals is also a countable set of reals, and if , for
one ordering of that set, there is shown to be a nonmember of the set,
that non-member is a nonmember for all orderings.
But for EACH listing of the set, of which there are also uncountably
many, there will be an "anti-diagonal" for that listing, showing that
there are uncountably many nonmembers of that countable set of reals.
--
Creationism is code for "Here, we want you to believe this bullshit"
Humans have been successfully applying the principles of Evolution in
agriculture and animal husbandry for over 10,000 years now.
Ray Vickson
Permuting the list produces a different diagonal (that is, a number
with different digits in first, second, third, .... place). Of course,
since the reals are uncountable and there are countably may numbers in
the list, there will be some real that is truly not in any of the
permutations of the list. However, that number need not be the
diagonal of any of the permutations---at least, not obviously so.
(Maybe it is, but that would need proof, not just assertion.)
R.G. Vickson
>
> -LV
Graham Cooper
Let's define JIGGLEDIAG loosely as
LIST(x,1) . LIST(y,2) . LIST(z,3) . ...
where x y z ... is 1to1 with N.
Does the ANTI-DIAG proof still hold using JIGGLEDIAG?
I give you some value of JIGGLEDIAG but do not reveal the list or the
1to1 function for the order of the rows provided of my bona-fide
ordinary single ole LIST.
LudovicoVan
Indeed, it's not about diagonals, it's about anti-diagonals. And A.
Magidin, Virgil and even I have provided, all in all, 6 different
arguments that show that Cooper's assertion is simply true: the anti-
diagonal is not on the list nor it is on any of its possible
permutations.
-LV
Ross A. Finlayson
There's only one antidiagonal in binary. Subject to dual
representation, with the equivalency function it's the item at the end
of the list. The antidiagonal argument in base three and higher
resolves to the argument in base two.
The natural/unit equivalency function is the function between naturals
in order and the unit interval of reals in order, also well-ordering
the reals of the unit interval. It exists (for example constructively
in ZF, modeled by standard functions) and is applied in each of the
presented proofs of the uncountability of the reals, not necessarily
in uniqueness. (It is "a" function where it is not "the" function.)
EF: N <-> R[0,1]
I found it of some interest. There it is. (Domain: counting numbers,
range: line segment.)
Steve, well about Herc or Cooper's argument there, his construction,
well we already discussed some aspects of said construction, as I
recall, about how many ways to make the anti-diagonal there are (and
how it works out). Of course this was some years ago.
Regards,
Ross Finlayson
LudovicoVan
> In article
> <8348f24d-d1df-47bd-b55b-1c237d85f...@m7g2000vbq.googlegroups.com>,
> LudovicoVan <ju...@diegidio.name> wrote:
> > [Only to sci.math, sci.logic.]
> > On Feb 27, 6:27 pm, Ray Vickson <RGVick...@shaw.ca> wrote:
> > > On Feb 27, 5:59 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > > > For the diagonal proof to be valid, it must provide a MISSING REAL
> > > > NUMBER for each of the INFINITELY MANY PERMUTATIONS of the same LIST.
>
> > > No, No, No! It must provide a missing real number for any *given*
> > > list. A permutation of a list is a new list, and so will have a
> > > different missing real number. Where did you ever get the idea that a
> > > single missing number must, necessarily, apply to the diagonals of all
> > > permutations?
>
> > I'd expect the permuted list to contain *exactly* the same entries as
> > the original list. Hence, the missing entry should be a missing entry
> > in the permuted list as well; conversely, an entry in the permuted
> > list should be an entry in the original list as well. Where is the
> > mistake?
>
> Each such list of reals is also a countable set of reals, and if , for
> one ordering of that set, there is shown to be a nonmember of the set,
> that non-member is a nonmember for all orderings.
Yes, of course.
> But for EACH listing of the set, of which there are also uncountably
> many, there will be an "anti-diagonal" for that listing, showing that
> there are uncountably many nonmembers of that countable set of reals.
That is true only as long as there is such a thing as uncountable
sets.
-LV
LudovicoVan
wrote:
> There's only one antidiagonal in binary. Subject to dual
> representation, with the equivalency function it's the item at the end
> of the list.
I sort of "perfectly" understand that:
0: (0)
1: 1(0)
2: 01(0)
3: 11(0)
4: 001(0)
5: 101(0)
6: 011(0)
7: 111(0)
...
w: (1)
But I still wouldn't know how to make that rigorous enough: I'd guess
it requires some notion of "hyper-countability", with "hyper-lists" as
mappings of the kind f : N* -> <0,1>*, or something similar...
-LV
Ray Vickson
It's too bad that Cooper himself could not write as clearly as you do.
His starting message for this thread talked about "diagonals", not
anti-diagonals. However, I am going to give up on trying to understand
any of his messages.
R.G. Vickson
>
> -LV
Graham Cooper
>
> It's too bad that Cooper himself could not write as clearly as you do.
> His starting message for this thread talked about "diagonals", not
> anti-diagonals. However, I am going to give up on trying to understand
> any of his messages.
>
> R.G. Vickson
Coherence is always an issue for opposing stances.
Does anyone commprehend this following question?
-----8<-----------------------------------------
OK forget about LIST'.
Let's define JIGGLEDIAG loosely as
LIST(x,1) . LIST(y,2) . LIST(z,3) . ...
where x y z ... is 1to1 with N.
Does the ANTI-DIAG proof still hold using JIGGLEDIAG?
I give you some value of JIGGLEDIAG but do not reveal the list or the
1to1 function for the order of the rows provided of the single LIST.
----------------------------------------8<---------------
The answer is trivially YES.
The anti-diag proof as you know it should still hold if you are given
a JIGGLEDIAG.
This trivially nullifies your objections to a permutation attack in
the OP.
It stands that, for certain infinite lists:
Mike Terry
news:7a2e4367-5e54-4d61...@a11g2000pro.googlegroups.com...
> random digit generator.
>
> LIST(row, digit) = RANDOM(0,9)
>
> For the diagonal proof to be valid, it must provide a MISSING REAL
> NUMBER for each of the INFINITELY MANY PERMUTATIONS of the same LIST.
>
> Assume the DIAGONAL is 1 2 3 4 9 9 9 9 9 8 8 8 3 6 8 ...
> where
> DIAGONAL(LIST) = LIST(1,1) . LIST(2,2) . LIST(3,3) . ...
>
> Suppose we wanted to change digit 1 by substituting LIST' for LIST.
>
> We search for a row that has a matching digit with row 1 at a higher
> digit place on the diagonal.
>
> Say row 1 is 1 4 3 4 4 4 4 4 ...
> and row 3 is 6 6 3 6 6 6 6 6 ...
>
> We reorder the LIST to LIST' by swapping rows 1 and 3 and the diagonal
> becomes
>
> 6 2 3 4 9 9 9 9 9 8 8 8 3 6 8 ...
> ^
>
> Similarly we can change the 2nd digit to
>
> 6 9 3 4 9 9 9 9 9 8 8 8 3 6 8 ....
> ^ ^
Yes, it's true that with a suitable starting list it can be permuted so that
the diagonal can be made to start with any FINITE sequence of digits, which
is what you're doing with your successive swaps.
>
> Therfore, for all n, the first n digits of the diagonal are arbitrary.
>
> By induction, the diagonal is intractable and not specified at all,
> for certain classes of infinite lists.
It's NOT true that if you attempt to reproduce a given INFINITE diagonal
there will be a permutation of the list that gives that infinite sequence of
digits. If you try to spell out your "induction" proof more thoroughly
(haha) rather than just using it as some kind of magic wand you will find it
doesn't work. The reason your row swapping argument doesn't work for the
infinite case is that you cannot define the meaning of an infinite sequence
of swaps - or at least if you assign it some meaning, then it will not
necessarily be a "permutation" (1-1 mapping) of the original list.
I explained all this to you in considerable detail a few years ago, and I
recall someone else patiently exlained exactly the same to you a few months
back, and yet you're still here spouting the same nonsense. Realistically I
should expect even when I explained it, you had probably already had it
explained by several others over the preceding years when I wasn't around,
as you seem to be stuck in some kind of loop thats repeating over and over
without you ever learning anything.
Well, that's your right I suppose, but anyone who responds to you is
obviously completely wasting their time...
Regards,
Mike.
>
> QED
( - Not!)
Gus Gassmann
It is true as soon as the collection of all real numbers forms a set.
Denying that forces you to deny the powerset axiom, hence you are no
longer talking about ZF.
Graham Cooper
<news.dead.person.sto...@darjeeling.plus.com> wrote:
> "Graham Cooper" <grahamcoop...@gmail.com> wrote in message
Mike, just for the records can you confirm that the following result
is true.
[For certain infinite lists: for all n, the first n digits of the
diagonal are arbitrary.]
That would be super for everyone to see I've cut off the head and only
the tail remains.
Ross A. Finlayson
It just happens that the constructed antidiagonal would follow all the
others in their natural ordering.
The general argument of that form you write is that each integer is
finite so each sequence is finite. As you describe, to see through
that infinite sequences have in consequence infinite sequences, in
them and of them, a concordance with constructible results in for
example ordinals, like 2^X of them, has then all the consequences of
there being infinite integers (infinitely grand besides infinitely
varied). Also nested intervals or "Cantor's first" applies, the other
and various arguments about uncountability of the reals apply,
etcetera.
EF
0: .0
1: .000000000000000000000000000000000.......
2: .000000000000000000000000000000000.......
3: .000000000000000000000000000000000.......
4: .000000000000000000000000000000000.......
5: .000000000000000000000000000000000.......
contrast Reverse Equivalency Function, REF:
0. .(1)
1: .111111111111111111111111111111111......
2: .111111111111111111111111111111111......
3: .111111111111111111111111111111111......
4: .111111111111111111111111111111111......
5: .111111111111111111111111111111111......
In an expanded representation of that form it wouldn't appear that
constant monotonic infinitesimal difference would ever see items of
the ranges being anything but 0 or 1, yet it is a simple consequence
of the construction (or modeling by standard functions), that they
are. Writing them in expansions is simply insufficient to express
them. (Eudoxus/Dedekind/Cauchy is insufficient to exhaust
representations, but is sufficient to model bounds, them being the
continuum of real numbers of the unit line segment or ray, unit ray,
of those real numbers from zero through one.) Yet, it's still a
function, between only natural integers, and only real numbers (from
zero through one, or zero to one).
EF(0) = 0, REF(0) = 1 (EF = 1 - REF)
EF(n) < EF(n+1), REF(n) > REF(n+1) (constant monotonicity)
lim n->oo EF(n) = 1, REF(n) = 0 (EF(oo) =/= 0)
Simple, noticeable. AS expansions, there's only one antidiagonal, and
it is at the end of the list. Then it's as to where the range is
[0,1] or [0,1). Topologically it's one or the other of those. Either
is quite satisfactory.
In as to whether and where and when that suitably represents a
function, throughout the framework of real analysis as so used in the
constructions of proofs of uncountability of sets of the reals, even
without trans-finite induction: in-finite induction suffices to model
it with real functions. (The EF exists.)
This is all quite standard, thank you. Of course its results follow
in the nonstandard along the lines of verisimilitude of results
established standardly and nonstandardly (in for example Internal Set
Theory, co-consistent with ZFC), a non-real function with standard
context.
Regards.
Ross Finlayson
Jim Burns
> Mike, just for the records can you confirm that the
> following result is true.
>
> [For certain infinite lists: for all n, the first n digits
> of the diagonal are arbitrary.]
>
> That would be super for everyone to see I've cut off
> the head and only the tail remains.
The result is true (or could be, for some lists), but
it does not give you what you think it does.
Do you remember the Hydra of Greek mythology, which
Hercules slew as one of his Twelve Labors? It was
a many-headed serpent that grew two heads back
whenever one was cut off.
Hercules had it easy. Each "head", each individual
finite initial digit sequence, has a "tail", all
real numbers that the "head" could belong to,
as large as all the real numbers together.
With your "list the finite initial segments" plan,
each natural number does not correspond to an
individual real number; it corresponds to a
set as large as all the real numbers together.
This will remain true no matter how long the
initial segment gets, as long as it remains
finite. Infinity remains infinitely far away.
Graham Cooper
Just for some persepective on 'rambling', 'coherence' etc....
Does anyone follow Jim's 'argument' here?
PS I just watched my first Hercules Legendary Journey's in 10 years
and yes, they had a scene with Hercules throwing a knife and splitting
the 'weapon' in 2.
I was expecting a big snake, but it was a self firing configured cross
bow spanning between the King and his daughter.
Jim Burns
You complain that I don't answer your questions,
then you complain that I do.
You want an argument, instead? Fine.
Executive summary: A countably infinite digit sequence
has enough degrees of freedom that it can dodge each of
countably infinite fixed digit sequences with at least
one position each.
To prove: Size(Naturals) < Size(Reals).
(By the Cantorian definition of 'Size', this is
the same as "There are no functions from the
naturals to the reals that hit every real.
<proof>
Let myList be an arbitrary function
from the naturals to the reals.
Define the real number Missed_by(myList) as
For all n in N,
Missed_by(myList)[n] =def= Different_from(myList(n)[n]).
Therefore, for all n in N,
Missed_by(myList) not= myList(n).
myList misses Missed_by(myList),
so myList does not hit every real.
Because myList could be any function,
the argument works for /every/ function N -> R,
so /no/ function hits every real, and so ...
Size(Naturals) < Size(reals).
</proof>
Graham Cooper
OK you were just coming down to my level, can't blame you for that!
>
> You want an argument, instead? Fine.
>
> Executive summary: A countably infinite digit sequence
> has enough degrees of freedom that it can dodge each of
> countably infinite fixed digit sequences with at least
> one position each.
>
> To prove: Size(Naturals) < Size(Reals).
>
> (By the Cantorian definition of 'Size', this is
> the same as "There are no functions from the
> naturals to the reals that hit every real.
>
> <proof>
> Let myList be an arbitrary function
> from the naturals to the reals.
>
> Define the real number Missed_by(myList) as
> For all n in N,
> Missed_by(myList)[n] =def= Different_from(myList(n)[n]).
>
> Therefore, for all n in N,
> Missed_by(myList) not= myList(n).
>
> myList misses Missed_by(myList),
> so myList does not hit every real.
>
> Because myList could be any function,
> the argument works for /every/ function N -> R,
> so /no/ function hits every real, and so ...
>
> Size(Naturals) < Size(reals).
> </proof>-
so does your proof work for:
Define the real number Missed_by(myList) as
For all n in N,
Missed_by(myList)[n] =def= Different_from(myList(ONETOONE(n))[n]).
where ONETOONE(n) is a bijection to N.
e.g.
N ONETOONE(n)
1 5
2 4
3 1
4 9
5 10001
6 2
...
LudovicoVan
wrote:
> On Feb 27, 8:03 pm, LudovicoVan <ju...@diegidio.name> wrote:
> > On Feb 27, 10:27 pm, Virgil <vir...@ligriv.com> wrote:
<snip>
> > > But for EACH listing of the set, of which there are also uncountably
> > > many, there will be an "anti-diagonal" for that listing, showing that
> > > there are uncountably many nonmembers of that countable set of reals.
>
> > That is true only as long as there is such a thing as uncountable
> > sets.
>
> It is true as soon as the collection of all real numbers forms a set.
> Denying that forces you to deny the powerset axiom, hence you are no
> longer talking about ZF.
No denial, really, but AFAIK what you are saying is simply not true.
E.g. in a constructive setting the real numbers form a *subcountable*
set. Not that I yet understand how subcountability actually works, but
that is me...
-LV
LudovicoVan
<snip>
> Therfore, for all n, the first n digits of the diagonal are arbitrary.
OK, I can see that: what diagonal we get depends on the specific list
given. But, where is that supposed to lead? (What kind of "permutation
attack" are you envisioning?)
-LV
LudovicoVan
wrote:
> On Feb 27, 4:11 pm, LudovicoVan <ju...@diegidio.name> wrote:
> > On Feb 28, 12:03 am, "Ross A. Finlayson" <ross.finlay...@gmail.com>
> > wrote:
>
> > > There's only one antidiagonal in binary. Subject to dual
> > > representation, with the equivalency function it's the item at the end
> > > of the list.
>
> > I sort of "perfectly" understand that:
>
> > 0: (0)
> > 1: 1(0)
> > 2: 01(0)
> > 3: 11(0)
> > 4: 001(0)
> > 5: 101(0)
> > 6: 011(0)
> > 7: 111(0)
> > ...
> > w: (1)
>
> > But I still wouldn't know how to make that rigorous enough: I'd guess
> > it requires some notion of "hyper-countability", with "hyper-lists" as
> > mappings of the kind f : N* -> <0,1>*, or something similar...
>
> others in their natural ordering.
Although, if we define the list inductively in the natural order, we
get:
0
1
-
1 = 1_10 (anti-diag)
00
01
10
11
--
10 = 2_10
000
001
010
...
---
111 = 7_10
0000
0001
0010
0011
...
----
1100 = 12_10
00000
00001
00010
00011
00100
...
-----
11101 = 29_10
000000
000001
000010
000011
000100
000101
...
------
111110 = 62_10
Etc.
That happens to be sequence http://oeis.org/A102371
Then it seems the antidiagonal is a sequence always at an index >= w
rather than necessarily the last one (which is also consistent with
the fact that there can be more than one single "anti-sequence" for
the list), where the list is made of w columns times 2^w rows... (I'd
object to any argument that treats the list as square.)
I am under the impression that they are not both satisfactory: I'd
insist on it being a mapping to [0,1], in analogy to the balls and
vase problem, where the seeming paradox to me in fact arises from the
ad-hoc mixing of N and N* within the same argument...
> In as to whether and where and when that suitably represents a
> function, throughout the framework of real analysis as so used in the
> constructions of proofs of uncountability of sets of the reals, even
> without trans-finite induction: in-finite induction suffices to model
> it with real functions. (The EF exists.)
>
> This is all quite standard, thank you. Of course its results follow
> in the nonstandard along the lines of verisimilitude of results
> established standardly and nonstandardly (in for example Internal Set
> Theory, co-consistent with ZFC), a non-real function with standard
> context.
This stuff is giving me head aches! I wish I knew more about ordinal
arithmetic...
-LV
Graham Cooper
The purported antidiag proof claims to only need to see the diagonal
to create it's extra element.
The proof claims to work on every permutation of the one set or list.
Then the proof should claim to work with a 1to1 function changing the
order of the rows, as that gives an identical diagonal to a list
permutation.
Missed_by(myList)[n] =def= Different_from(myList(ONETOONE(n))[n]).
where ONETOONE(n) is a bijection to N
Then the *given* diagonal has leading arbitrary digits.
PRIVATE
LIST(y, x)
ONETOONE(n)
PUBLIC
ANYDIAG(n) = LIST(ONETOONE(n))[n]
Now we can role play the Cantor dialog.
CONST: I have an infinite list of reals
CARD: Show me the diagonal and I will form a new real.
CONST: Here is the diagonal of one of the permutations.
SENDS: ANYDIAG
CARD: That will do, the new real is ANTI(ANYDIAG(n)) n=1,2,3...
CONST: I tricked you, ANYDIAG(n) is effectively any digit sequence, I
didn't even (need to) look at my list.
CARD: blah blah blah
CONST: for all n, the first n digits of the diagonal are arbitrary in
regards to the anti-diag proof.
WILL: that means all the digits of the diagonal, for the effect of the
anti-diag proof, are arbitrary.
SOMEONE: that means Cantor's claim of a missing real is not supported.
COOPS: that means there is probably no such thing as uncountable or
cardinality, and INFINITY is just the ... to the right of the X-axis
----> ...
Maybe we need to draw in the ... so this never happens again!
LudovicoVan
> On Feb 28, 8:28 pm, LudovicoVan <ju...@diegidio.name> wrote:
> > On Feb 28, 1:53 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > <snip>
>
> > > Therfore, for all n, the first n digits of the diagonal are arbitrary.
>
> > OK, I can see that: what diagonal we get depends on the specific list
> > given. But, where is that supposed to lead? (What kind of "permutation
> > attack" are you envisioning?)
>
> to create it's extra element.
Yep.
> The proof claims to work on every permutation of the one set or list.
Yep, as long as the proof works, it surely works on any list and on
any permutation thereof: namely, the anti-diagonal will be missing
from the list and from any of its permutations.
> Then the proof should claim to work with a 1to1 function changing the
> order of the rows, as that gives an identical diagonal to a list
> permutation.
No: if you permute the list, the diagonal will be different, and
different will be the corresponding anti-diagonal.
> Missed_by(myList)[n] =def= Different_from(myList(ONETOONE(n))[n]).
> where ONETOONE(n) is a bijection to N
>
> Then the *given* diagonal has leading arbitrary digits.
>
> PRIVATE
> LIST(y, x)
> ONETOONE(n)
>
> PUBLIC
> ANYDIAG(n) = LIST(ONETOONE(n))[n]
>
> Now we can role play the Cantor dialog.
>
> CONST: I have an infinite list of reals
>
> CARD: Show me the diagonal and I will form a new real.
>
> CONST: Here is the diagonal of one of the permutations.
> SENDS: ANYDIAG
>
> CARD: That will do, the new real is ANTI(ANYDIAG(n)) n=1,2,3...
>
> CONST: I tricked you, ANYDIAG(n) is effectively any digit sequence, I
> didn't even (need to) look at my list.
But the trick *is* invalid: the proof works as long as the diagonal is
indeed the diagonal of an underlying list. I.e., to be precise, the
list is arbitrary, not the diagonal itself.
> CARD: blah blah blah
Yep. ;)
> CONST: for all n, the first n digits of the diagonal are arbitrary in
> regards to the anti-diag proof.
But the trick *is* invalid...
> WILL: that means all the digits of the diagonal, for the effect of the
> anti-diag proof, are arbitrary.
Well, I doubt William really said that: OTOH, I too have problems with
a completed anti-diagonal against an ever incomplete list of ever
incomplete sequences...
> SOMEONE: that means Cantor's claim of a missing real is not supported.
>
> COOPS: that means there is probably no such thing as uncountable or
> cardinality, and INFINITY is just the ... to the right of the X-axis
> ----> ...
>
> Maybe we need to draw in the ... so this never happens again!
My informal take:
Counting numbers := { 1, 2, 3, ... too many! }
-LV
Graham Cooper
> On Feb 28, 11:01 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > On Feb 28, 8:28 pm, LudovicoVan <ju...@diegidio.name> wrote:
> > > On Feb 28, 1:53 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > > <snip>
>
> > > > Therfore, for all n, the first n digits of the diagonal are arbitrary.
>
> > > OK, I can see that: what diagonal we get depends on the specific list
> > > given. But, where is that supposed to lead? (What kind of "permutation
> > > attack" are you envisioning?)
>
> > The purported antidiag proof claims to only need to see the diagonal
> > to create it's extra element.
>
> Yep.
>
> > The proof claims to work on every permutation of the one set or list.
>
> Yep, as long as the proof works, it surely works on any list and on
> any permutation thereof: namely, the anti-diagonal will be missing
> from the list and from any of its permutations.
You misinterpreted. The point is:
all (anti)diagonals of all the permutations must be provably missing
from the 1 original list.
>
> > Then the proof should claim to work with a 1to1 function changing the
> > order of the rows, as that gives an identical diagonal to a list
> > permutation.
>
> No: if you permute the list, the diagonal will be different, and
> different will be the corresponding anti-diagonal.
>
Think again.
Constructivist: If I give you the diagonal of one of the permutations
of my list can you tell me a missing real from my list?
Cardinalitist: Yes!
Constructivist: Then equivalently, I can use a remapping function on
the rows of my list and give you the diagonal of that list?
Cardinalitist: Yes! If it's a bijection to N.
Constructivist: Then for my one list, I can give you any of a set of
diagonals?
Cardinalitist: Yes!
Constructivist: Then the 1st digit is arbitrary by using a suitable
list and mapping some row n with row 1 and vice versa so as to only
change the 1st digit of the diagonal of the original list.
Alan Smaill
It is worth remembering that a set can be both subcountable, and
uncountable in the constructive terminology (uncountable
in that there is no effective bijection with the natural numbers).
It is a standard constructive result that the reals are
not countable, in this constructive sense.
> -LV
--
Alan Smaill
Graham Cooper
I thought everyone had agreed with this, but you are all taking it to
mean FOR A DIFFERENT LIST.
I mean the digits are ARBITRARY FROM THE ONE LIST. INTRACTABLE.
----------------------
Put it this way. I have a countable SET OF REALS.
THE ORDER DOESNT MATTER!
I can give you JUST THE DIAGONAL FROM ANY ORDER I WANT.
SAME SET. INFINITE POSSIBLE DIAGONALS
LudovicoVan
> On Feb 28, 9:36 pm, LudovicoVan <ju...@diegidio.name> wrote:
> > On Feb 28, 11:01 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
> > > On Feb 28, 8:28 pm, LudovicoVan <ju...@diegidio.name> wrote:
> > > > On Feb 28, 1:53 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > > > <snip>
>
> > > > > Therfore, for all n, the first n digits of the diagonal are arbitrary.
>
> > > > OK, I can see that: what diagonal we get depends on the specific list
> > > > given. But, where is that supposed to lead? (What kind of "permutation
> > > > attack" are you envisioning?)
>
> > > The purported antidiag proof claims to only need to see the diagonal
> > > to create it's extra element.
>
> > Yep.
>
> > > The proof claims to work on every permutation of the one set or list.
>
> > Yep, as long as the proof works, it surely works on any list and on
> > any permutation thereof: namely, the anti-diagonal will be missing
> > from the list and from any of its permutations.
>
> You misinterpreted. The point is:
>
> all (anti)diagonals of all the permutations must be provably missing
> from the 1 original list.
But that is quite what I and others have been saying: as long as the
proof works, all anti-diagonals corresponding to all the list
permutations *will* be provably missing from the original list as well
as from any of its permutations.
> > > Then the proof should claim to work with a 1to1 function changing the
> > > order of the rows, as that gives an identical diagonal to a list
> > > permutation.
>
> > No: if you permute the list, the diagonal will be different, and
> > different will be the corresponding anti-diagonal.
>
> Think again.
>
> Constructivist: If I give you the diagonal of one of the permutations
> of my list can you tell me a missing real from my list?
>
> Cardinalitist: Yes!
>
> Constructivist: Then equivalently, I can use a remapping function on
> the rows of my list and give you the diagonal of that list?
>
> Cardinalitist: Yes! If it's a bijection to N.
>
> Constructivist: Then for my one list, I can give you any of a set of
> diagonals?
To be precise: for the one set, whichever the list (the diagonal is a
function of the given list).
> Cardinalitist: Yes!
>
> Constructivist: Then the 1st digit is arbitrary by using a suitable
> list and mapping some row n with row 1 and vice versa so as to only
> change the 1st digit of the diagonal of the original list.
Well, I think you are equivocating on what is "arbitrary" there: the
list is arbitrary, not the diagonal and corresponding anti-diagonal,
which are instead functions of the given list. Plus another nitpick:
if you swap rows 1 and n, that will change the diagonal at two
positions, the 1st and the nth.
And I still cannot see how all that could lead to a disproof of the
diagonal argument.
-LV
Graham Cooper
I thought you asked WHERE IS COOPERS ERROR in the OP?
It clearly describes how to modify only 1 digit.
I think I know where you are not getting it!
There are 2 versions of the Cantor dialog.
------------------------------
VERSION 1
CONSTRUTIVIST: Here is my LIST of all reals.
4343434...
343443...
34543...
3434...
343...
34...
3...
...
CANTORIAN: Here is a missing real. 5545...
CONSTRUCTIVIST: You win!
----------------
VERSION 2
CONSTRUCTIVIST: So I hear you can construct a missing real for any
list just by examining the diagonal?
CANTORIAN: Yep!
CONSTRUCTIVIST: Here is my DIAGONAL. 4563....
CANTORIAN: It's missing 5455.....
CANTORIAN: You win!
-----------------
We're playing Version 2!
Graham Cooper
>
> CANTORIAN: You win!
>
> -----------------
>
> We're playing Version 2
I uhh accidently mixed up the final name tag... funny that!
Should read
CONSTRUCTIVIST: You win! for now..
James Burns
Yes. The same reasoning is there: Missed_by(myList)
is different from every real number on myList.
All we've done with the change is make it harder to
keep track of which myList(n) have been eliminated
as possible locations for Missed_by(myList). But,
by telling us that ONETOONE is a bijection, you
guarantee that every n that was eliminated before
will be eliminated in the new version, that is,
all of them.
Now, here comes a chance for you to show that your claim
that you answer questions was not just blowing smoke.
Suppose that the version with ONETOONE didn't work.
So what? Why does it matter whether the version with
ONETOONE works, as long as the version without it works?
A good proof proves a result, but a bad proof does not
disprove a result.
Graham Cooper
Same_reason_I_underlined_before_that_you_glossed_over.
Don't you think that the fact I can provide you with oo
pseudoDiagonals from the one list is a bit different to giving you 1?
<LIST(1,1), LIST(2,2), LIST(3,3), ...>
Maybe your headinverting counterargument - ahhh who cares it works on
one
could be a tad naive?
James Burns
When I look upthread, there is only one place I see
where you addressed me using underlines that way.
If you think I glossed over your reasons there, too bad.
I addressed them as best I could, given what I could
discern or guess about what you were trying to say.
If that's not good enough for you, you might try
answering questions about /what you mean./
Is the following what you consider this "glossing over"?
(If it is, then you do a much more thorough job of
"glossing over", as your answers are MUCH shorter,
telegraphic really.)
<quote>
!> I will be explicit as I possibly can, try not to
!> feign ignorance.
!>
!> A_number_is_an_explicitly_defined_sequence_of_digits.
!
!
!Ah. So, you're telling me that what the rest of us
!laughingly call "the set of real numbers" actually
!contains no numbers at all?
!
!> There_are_unlimited_orderings_of_the_LIST_so_every_
!> diagonal_digit_is_intractable.
!
!
!"Intractable" meaning what?
!
!It is true that, with a different ordering, a LIST
!will most likely have different anti-diagonals.
!This is not a problem. The claim was never "Here are
!/all/ the numbers not in the LIST." It has always been
!"Here is /at least one/ number not in the LIST."
!
!If you re-order the LIST (call it LIST2), you will be able
!to generate different numbers not necessarily
!anti-diagonals of LIST by the diagonal argument applied
!to LIST2. And, the numbers "previously" anti-diagonals
!of LIST are "no longer" anti-diagonals of LIST2.
!BUT, there is no either-or situation here. The argument
!that the anti-diagonals of LIST are not on LIST has not
!stopped existing while we look at LIST2.
!
!Please explain what is wrong with all of the anti-diagonals
!of LIST plus all of the anti-diagonals of LIST2 not being
!on LIST and also not being on LIST2 -- if that is your
!complaint. (If it isn't, please, at least tell me it isn't,
!so that I can try a different guess at what you mean.)
!
</quote>
> Don't you think that the fact I can provide you with oo
> pseudoDiagonals from the one list is a bit different
> to giving you 1? <LIST(1,1), LIST(2,2), LIST(3,3), ...>
Since you ask whether it's /different/, I'll say yes.
If you had asked whether it /mattered to the proof/
that the size of the reals is larger than the size
of the reals, I would have said NO.
> Maybe your headinverting counterargument -
> ahhh who cares it works on one
> could be a tad naive?
All right, why?! why would "Who cares, it works on one"
be naive? If A is one more than B, then it's more than B.
And if A is an infinite number more than D, then it is
still more than B. Since what we wanted to prove that
A is more than B, why does it matter /how much/ it is
more than B?
Maybe you think you've already answered this question,
but I don't see it.
Mike Terry
Sure - more precisely, there will be a permutation of the original list
which gives a diagonal matching the first n digits given.
>
> That would be super for everyone to see I've cut off the head and only
> the tail remains.
With this particular hydra, the heads are rather unappetizing - all the meat
is in the tails!
Mike.
>
Graham Cooper
The issue is not changing MORE to INFINITELY MORE.
The issue is going from a successful finite trial to
infinite trials and assuming the same conclusion holds.
If I can guess when a dice will roll 6, and I'm right
using batches of 1 roll, 2 rolls, and 3 rolls, then
it supports the claim.
If I take oo rolls and always get a 6, it supports nothing.
Before you dismiss the argument, why not state whether
it holds at all first. That infinite trials can bias
the experiment to provide a null result.
Graham Cooper
> > is true.
>
> > [For certain infinite lists: for all n, the first n digits of the
> > diagonal are arbitrary.]
>
> Sure - more precisely, there will be a permutation of the original list
> which gives a diagonal matching the first n digits given.
>
>
>
> > That would be super for everyone to see I've cut off the head and only
> > the tail remains.
>
> With this particular hydra, the heads are rather unappetizing - all the meat
> is in the tails!
>
> Mike.
>
Mike has done a Daryl and backpeddled.
Why is everyone SAYING the leading digits are arbitrary
then they later clarify GIVEN THAT YOU CAN SUBSTITUE A DIFFERENT LIST.
-------
Mike what happened to your INFINITE COUNTER EXAMPLE argument?
Are you saying it was superfluous since using a permutation is a
different list and hence you cannot blindly swap digits at all now?
Pick an argument and stick to it.
Here is Mike agreeing in principle that you have a degree of freedom
in supplying the pseudoDiagonal.
> > The reason your row swapping argument doesn't work for the
> > infinite case is that you cannot define the meaning of an infinite
> > sequence
> > of swaps - or at least if you assign it some meaning, then it will not
> > necessarily be a "permutation" (1-1 mapping) of the original list.
Now Mike has realised his error and moved to higher ground.
He is specifying a particular Permutation can have a different finite
prefix,
and changing the theorem away from being substitutable or arbitary.
> > [For certain infinite lists: for all n, the first n digits of the
> > diagonal are arbitrary.]
>
> Sure - more precisely, there will be a permutation of the original list
> which gives a diagonal matching the first n digits given.
>
Now every sci.mathematician will cling to the phrase FOR A PERMUTATION
because the leading digits of the GIVEN diagaonal being arbitrary in
the
true sense of the word arbitrary conflicts with the anti-diag being a
valid definition.
James Burns
But *we* don't do that. We start with "infinite trials".
We are just doing straighforward manipulation of
quantifiers. We show "Missed_by(myList)" misses *all* of
the reals in "myList". *Our* proof doesn't involve
a successful finite that gets extended to infinite
trials.
*You* can do that, if you want to. Maybe it will give
you a valid proof, maybe it won't. Valid or invalid,
*your* proof does not change the validity of *our*
proof.
>
> If I can guess when a dice will roll 6, and I'm right
> using batches of 1 roll, 2 rolls, and 3 rolls, then
> it supports the claim.
>
> If I take oo rolls and always get a 6, it supports nothing.
>
> Before you dismiss the argument, why not state whether
> it holds at all first. That infinite trials can bias
> the experiment to provide a null result.
I can't tell whether you're right or wrong, because I
can't tell what you're trying to do. Are you trying
to generate an anti-diagonal that is not in the list?
Are you trying to generate an anti-diagonal that *is*
in the list? Are you trying to show you have a list
containing *all* reals, in spite of the anti-diagonal
argument? Are you off doing something else somewhere
else that I'm not even watching?
Perhaps you can tell what my answer would be, given
that: it doesn't matter where the numbers in the list
come from, dice or ouija board or alien brain-suckers,
*our* argument applies to all numbers, so it applies
to those numbers -- for *all* values of 'those'.
It occurs to me that a slightly different spelling of
the Standard Proof I've been cut-and-pasting would
emphasize how simple, how obvious our use of
quantifiers is.
---
Executive summary: A countable digit sequence can be
chosen so that it dodges each sequence of countably many
fixed countable digit sequences at one or more digits.
To prove: Size(Naturals) < Size(Reals).
(By the Cantorian definition of 'Size', this is
the same as "There are no functions from the
naturals to the reals that hit every real." or
Not exist List in N^R, List(N) = R.)
<proof>
Define the function Missed_by from the set of lists
of reals to real numbers, Missed_by: N^R -> R, by
For all List in N^R, for all n in N,
Missed_by(List)[n] =def= Different_from(List(n)[n]).
Therefore,
for all List in N^R, for all n in N,
Missed_by(List) not= List(n).
For all List in N^R, Missed_by(List) not in List(N).
For all List in N^R, List(N) not= R.
Not exist List in N^R, List(N) = R.
Size(Naturals) < Size(Reals).
</proof>
Graham Cooper
>
> > If I can guess when a dice will roll 6, and I'm right
> > using batches of 1 roll, 2 rolls, and 3 rolls, then
> > it supports the claim.
>
> > If I take oo rolls and always get a 6, it supports nothing.
>
> > Before you dismiss the argument, why not state whether
> > it holds at all first. That infinite trials can bias
> > the experiment to provide a null result.
>
> I can't tell whether you're right or wrong, because I
> can't tell what you're trying to do. Are you trying
> to generate an anti-diagonal that is not in the list?
NO! ANSWER THE QUESTION! IT'S ABOUT 6 SIDED DICE!
L = 1
START
READ LINE L
INC L
MORE LINES? GO BACK TO START
MEA CULPA!
James Burns
ANSWER: "Ducks have feathers."
When you explain your question, I will explain
my answer, and not before then.
Ross A. Finlayson
Hello,
There's an enumeration strategy for real numbers, in 2^n iterations
each of the n-sequences (sequences with n-many elements), are
enumerated. (Or in b^n or b^p in base b, here base 2.) If you
sample a value with resolution 1/n it might be clear to expect that
that's enough to reconstruct half the signal. (Half enough to
reconstruct the signal, where it is plainly Nyquist or 1/2n
resolution.)
Yet when we take enough samples to approximate as we can real numbers,
that's still 1/n of them not 1/2^w of them. (Standard, countable
additivity. Rationals satisfy some definitions of continuous.) These
most certainly approximate our classical geometric expectations.
Then part of the idea is that the matrix is square. It doesn't take
2^n of them in the domain to enumerate n in the range. (Here n = w, w
>= n, n E N.) When functions share these images they're metrified,
square (or correlated).
They still satisfy definitions of the real numbers, these elements of
the range of the equivalency function. They start from zero and go
only upward and then infinitely but never to more than one, the
easiest approximation is one as their maximum and eventual value.
(It's easiest in the sense of, say, conservation principles, in
numbers, technically.) (They go constantly and monotonically from
zero to one.)
Can't say infinity ever gave me headaches, no quite the opposite in
fact. It's important really not to care that much. Heh yet gently,
being quite facile with various modern treatments, for example, the
standard, modern, treatment, the equivalency function bore up so well
in the consideration of these modern, accepted, standard, and
fundamental results, generally I've seen the entire thing as plain,
compelling, and straightforward, as one might well assume mathematics
to be, with the incredible wonder of how it all works out:
mathematics. It has to work with all the standard (classical)
results, the natural/unit equivalency function, else it would be
mathematically inconsistent, and not compelling. Thank you it's quite
compelling. The proofs of these statements are reasonably compelling.
Then, if I were so bold, I would bet that the numbers as I describe
them, in then to the polydimensional, are strong enough to support
various non-classical analyses, of for example effects as we know them
(in conservation). Not really being a gambling man, I'm quite sure of
it.
Give your mind, not your brain, to science.
Regards,
Ross Finlayson
LudovicoVan
wrote:
> On Feb 28, 3:00 am, LudovicoVan <ju...@diegidio.name> wrote:
> > On Feb 28, 6:07 am, "Ross A. Finlayson" <ross.finlay...@gmail.com>
> > wrote:
<snip>
> > > This is all quite standard, thank you. Of course its results follow
> > > in the nonstandard along the lines of verisimilitude of results
> > > established standardly and nonstandardly (in for example Internal Set
> > > Theory, co-consistent with ZFC), a non-real function with standard
> > > context.
>
> > This stuff is giving me head aches! I wish I knew more about ordinal
> > arithmetic...
>
> Hello,
>
> There's an enumeration strategy for real numbers, in 2^n iterations
> each of the n-sequences (sequences with n-many elements), are
> enumerated. (Or in b^n or b^p in base b, here base 2.) If you
> sample a value with resolution 1/n it might be clear to expect that
> that's enough to reconstruct half the signal. (Half enough to
> reconstruct the signal, where it is plainly Nyquist or 1/2n
> resolution.)
>
> Yet when we take enough samples to approximate as we can real numbers,
> that's still 1/n of them not 1/2^w of them. (Standard, countable
> additivity. Rationals satisfy some definitions of continuous.) These
> most certainly approximate our classical geometric expectations.
>
> Then part of the idea is that the matrix is square. It doesn't take
> 2^n of them in the domain to enumerate n in the range. (Here n = w,
> w>= n, n E N.) When functions share these images they're metrified,
> square (or correlated).
>
> They still satisfy definitions of the real numbers, these elements of
> the range of the equivalency function. They start from zero and go
> only upward and then infinitely but never to more than one, the
> easiest approximation is one as their maximum and eventual value.
> (It's easiest in the sense of, say, conservation principles, in
> numbers, technically.) (They go constantly and monotonically from
> zero to one.)
>
> Can't say infinity ever gave me headaches [...]
To be frank it was not really about Infinity: it's James Joyce who is
still way above my head.
-LV
David R Tribble
I had forgotten how much I miss RF's postings.
I've never read James Joyce, but I understand that this is
kind of like the stream-of-consciousness prose he was
famous for.
LudovicoVan
> Ross A. Finlayson wrote:
<snip>
> > Then, if I were so bold, I would bet that the numbers as I describe
> > them, in then to the polydimensional, are strong enough to support
> > various non-classical analyses, of for example effects as we know them
> > (in conservation). Not really being a gambling man, I'm quite sure of it.
>
> I had forgotten how much I miss RF's postings.
>
> I've never read James Joyce, but I understand that this is
> kind of like the stream-of-consciousness prose he was
> famous for.
That he was famous for it is a pretty sure sign that that's not what
he did: James Joyce is the Jimi Hendrix of English language.
-LV
Jesse F. Hughes
> I had forgotten how much I miss RF's postings.
There's no accounting for taste.
--
"If you have a really big idea, you can get a measure of how big it is
by how much people resist the obvious. From what I've seen, I have a
REALLY, REALLY, *REALLY*, BIG DISCOVERY!!!"
--James Harris: If I'm not important, how come people ignore me?
Aatu Koskensilta
> To be frank it was not really about Infinity: it's James Joyce who is
> still way above my head.
You will find, strange as this might sound, that Joyce's _Ulysses_ is
very readable save for the few chapters that are intended to depict a
profound state of inebriation and read appropriately. The same goes for
the supposedly unreadable -- but in reality eminently readable, a true
joy in fact, veritable delight -- _In the Parlour of Alastalo_ by Volter
Kilpi of Finland.
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon man nicht sprechen kann, darüber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
Ilmari Karonen
> LudovicoVan <ju...@diegidio.name> writes:
>
>> To be frank it was not really about Infinity: it's James Joyce who is
>> still way above my head.
>
> You will find, strange as this might sound, that Joyce's _Ulysses_ is
> very readable save for the few chapters that are intended to depict a
> profound state of inebriation and read appropriately. The same goes for
> the supposedly unreadable -- but in reality eminently readable, a true
> joy in fact, veritable delight -- _In the Parlour of Alastalo_ by Volter
> Kilpi of Finland.
For some really unreadable prose, try _Feersum Endjinn_ by the SF
author Iain M. Banks. About one fourth of the book reads like an SMS
typed by a hyperactive dyslexic teenager with a thick accent.
Quote courtesy of Wikipedia:
Woak up. Got dresd. Had brekfast. Spoke wif Ergates thi ant who sed
itz juss been wurk wurk wurk 4 u lately master Bascule, Y dont u ½ a
holiday? & I agreed & that woz how we decided we otter go 2 c Mr
Zoliparia in thi I-ball ov thi gargoyle Rosbrith.
The upside to slogging through all the txtsp34k is that it really is
quite an excellent piece of science fiction. Although, as is typical
of Banks, you'll need to read even the Bascule chapters carefully or
you'll miss details.
--
Ilmari Karonen
To reply by e-mail, please replace ".invalid" with ".net" in address.
Aatu Koskensilta
> For some really unreadable prose, try _Feersum Endjinn_ by the SF
> author Iain M. Banks.
It's not all that unreadable, except in the literal sense that it is
difficult to read, unless, and until one hits on the trick of reading it
out loud in one's mind, upon adopting of which procedure it suddenly all
becomes comprehensible. (Save for some Scottishisms no one except Banks
can be expected to understand, and even he only when he's downed several
mugs of [insert a name of some famous Scottish whiskey].)
> The upside to slogging through all the txtsp34k is that it really is
> quite an excellent piece of science fiction. Although, as is typical
> of Banks, you'll need to read even the Bascule chapters carefully or
> you'll miss details.
It's a nice piece of fiction, yes. Of Banks' books my all-time
favourite is, have no doubt about it, the whimsically dark and amusingly
depressing _Against a Dark Background_. His latest few books are, I'm
afraid and ashamed to say, just utter crap, nothing but lazy,
self-indulgent, puerile, pretentious fuck-you's in the spirit of
I-have-more-money-than-you-so-I-no-longer-need-to-put-any-
effort-in-my-writing-and-can-feed-you-tired-cliches-one-would-find-
very-very-out-of-place-in-my-earlier-work-which-was-unusually-for-
scifi-actually-pretty-decent-literature-but-my-editor-is-too-afraid-
to-edit-out-since-I'm-such-a-bigshot-author. Better than Peter
F. Hamilton, though -- but then, that's not much of an achievement --
who only writes about sex-obsesses brain-dead idiots who become even
more brain-dead at first sight of any cleavage, and consort with
disgusting space-elves.
LudovicoVan
I must admit the last time I have had to read any James Joyce was more
than 10 years ago during my studies in linguistics, when I could
barely speak any English at all: it was mostly a pain of looking up
almost every word in the dictionary. And thank you both for the
further interesting references: my English should be now much
stronger, so it's cool there is something new to have pain with... :)
-LV
Alan Smaill
> Ilmari Karonen <use...@vyznev.invalid> writes:
>
>> For some really unreadable prose, try _Feersum Endjinn_ by the SF
>> author Iain M. Banks.
>
> It's not all that unreadable, except in the literal sense that it is
> difficult to read, unless, and until one hits on the trick of reading it
> out loud in one's mind, upon adopting of which procedure it suddenly all
> becomes comprehensible. (Save for some Scottishisms no one except Banks
> can be expected to understand, and even he only when he's downed several
> mugs of [insert a name of some famous Scottish whiskey].)
yon's "whisky", ye cannae gie us yon "whiskey" *****
--
Alan Smaill