Cantor diagonalization and base 2

[George Cornelius]

Cantor's proof of the nondenumberability of the reals is
generally given for a base 10 number representation. This
is perfectly fine since the number system base in use has
little significance, but, still, for purity's sake, there
would seem to be an advantage in removing dependence on a
rather arbitrarily chosen number system base.

Since the proof as typically given relies on the
existence of at least three different digit values in
generating the counterexample from the digits on the
diagonal, it would appear that 3 would be the smallest
base for which the proof would apply without significant
modification. That the construction technique always
produces elements of a set closely related to the Cantor
set would be an interesting byproduct of that choice,
and one could consider this the solution to the problem
of what would be the most natural, and least arbitrary,
of the bases in which to couch the proof.

That having been dispensed with, what would be the simplest
way (short of bunching up the bits into groups of n, making
it essentially base-2^n) of adapting the proof to base 2
(remembering that the usual proof must avoid generating
a sequence ending in all b's, to infinity, where b is
the number system base minus 1, generally achieved by
generating a number which contains no b's at all)?

I'll propose one obvious adaptation:

o as usual, assume all reals in [0,1) are written in
some sequence down the page and represented as binary
fractions, with 'terminating' sequences chosen so as
to end in an infinite string of all 0's and not all 1's.

o pull from this list all of the numbers 2^-(2n),
n>=1 . Start building a new list by placing each
2^-(2n) value at position 2n, i.e., such that the
even positions all contain a number with only
a single 1, that 1 residing on the diagonal.

o fill the odd positions of the new list with all
of the remaining reals, in the same order as before,
but with, of course, the 2^-(2n) subsequence eliminated.

o now apply the usual diagonalization process,
replacing zeros by ones and ones by zeros.

o Note that every even position is a zero, so the
number so generated satisfies the usual requirements,
including not having a terminal string of all ones,
and the proof is complete.

Still slightly awkward, though, with not much to
recommend it over base 3 or base 10.

So how can we improve it? Can you propose a simple
and natural base 2 version of Cantor's proof?

[William Elliot]

> Can you propose a simple and natural base 2 version of Cantor's proof?

Yes.

[calvin]

On Oct 4, 4:00 am, William Elliot <ma...@panix.com> wrote:
> > Can you propose a simple and natural base 2 version of Cantor's proof?
>
> Yes.

The way I first heard the base 2 diagonal proof long
ago was simply to assume there is a countable list
containing all possible binary numbers of the form .xxx ...
where x is 0 or 1, and it doesn't matter how many
times the same number might be repeated in one way
or another. All that matters is that we assume that
all binary numbers exprressed by a binary point followed
by 0's and 1's are contained in the countable list.

Then flip the diagonal, and obviously that number is
not contained in the list. What's wrong with that?
Why isn't it obvious?

[calvin]

An answer I once saw was that the diagonal might be
repeating, and therefore the flipped diagonal would
be repeating, and that could be equal to a number that
was in the list.

The problem with that argument is that the probability
of the diagonal being repeating is absolute zero, not
small but zero, and things with zero probability don't
happen.

[Arturo Magidin]

Because the flipped number may be a second presentation of a number already on the list.

For example, suppose your the first number on the list is .1000000.... and that after that, the nth number on the list has a 0 in the nth position. Then the "flipped number" will be .0111111....

which seems fine, until you realize that the number whose binary expansion is .011111... is none other than the number whose binary expansion is .100000....

This is exactly the same phenomenon you get in decimal notation, which makes .2 be the same as .199999...... and why one usually avoids replacing numbers with 9s in the decimal notation version of Cantor's proof.

That is: the problem is that it is no longer true that just because the constructed number differs from a given number on the list in a single digit, then they must be distinct numbers; it's no longer true because you have no way to guarantee that the constructed number does not have two distinct representations.

--
Arturo Magidin

[Arturo Magidin]

I cannot make heads nor tails of your final paragraph. You are *given* a list. There is no probability. I can certainly produce a list in which the straighforward version of Cantor's diagonal argument fails to produce a number not on the list, so I do not see how you can claim that such a thing does not happen. Namely, take the list a_n where a_1=.10000000 and a_n, with n>1, is the number whose first n-1 digits are 1 and all the rest are 0; that is,

a_1 = .100000000...
a_2 = .100000000...
a_3 = .110000000...
a_4 = .111000000....
etc.
(If you demand that the list have no repeats, replace a_2 with .10100000....)

Then the number produced by "flipping the diagonal" is .0111111..... which is just another binary presentation of a_1, hence is already on the list.

(Yes, the list is *obviously* not exhaustive, but that's not the point; the point is that the argument fails to produce a number not on the list)

So... how can the "probability of the diagonal being repeating" be "absolute zero" (whatever that means) if "things with zero probability don't happen", and yet here we see that actually happening?

--
Arturo Magidin

[Arturo Magidin]

On Thursday, October 4, 2012 2:37:44 AM UTC-5, George Cornelius wrote:


>
> That having been dispensed with, what would be the simplest
>
> way (short of bunching up the bits into groups of n, making
>
> it essentially base-2^n) of adapting the proof to base 2
>
> (remembering that the usual proof must avoid generating
>
> a sequence ending in all b's, to infinity, where b is
>
> the number system base minus 1, generally achieved by
>
> generating a number which contains no b's at all)?

Usually you also want to avoid numbers that have tails of 0s... unless you specify whether you want your list to display numbers with a particular kind of representation when two are available. For example, in base 3, you need to specify that your numbers will be displayed using finite presentations (tails of 0s) when two are available, so that your constructed number can use 0s and 1s, and you are not worried that your constructed number may occur on the list with a tail of 2s. (So if you don't want to make such specifications, then you need to worry about both base 2 and base 3). This is not a requirement you need to make in any base other than 2 or 3, since you can always construct your number using, say, 1's and 2's and thus ensure that it has a unique presentation.

The simplest base-2 adaptation *is* to use two-digit blocks.

>
>
>
> I'll propose one obvious adaptation:
>
>
>
> o as usual, assume all reals in [0,1) are written in
>
> some sequence down the page and represented as binary
>
> fractions, with 'terminating' sequences chosen so as
>
> to end in an infinite string of all 0's and not all 1's.
>
>
>
> o pull from this list all of the numbers 2^-(2n),
>
> n>=1 . Start building a new list by placing each
>
> 2^-(2n) value at position 2n, i.e., such that the
>
> even positions all contain a number with only
>
> a single 1, that 1 residing on the diagonal.
>
>
>
> o fill the odd positions of the new list with all
>
> of the remaining reals, in the same order as before,
>
> but with, of course, the 2^-(2n) subsequence eliminated.

You don't need to eliminate anything. Keep the numbers if they already appeared, it doesn't matter.

[George Cornelius]

Arturo Magidin wrote:
> On Thursday, October 4, 2012 2:37:44 AM UTC-5, George Cornelius wrote:
>
>
>>
>> That having been dispensed with, what would be the simplest
>> way (short of bunching up the bits into groups of n, making

>> it essentially base-2^n) of adapting the proof to base 2 [...]

> Usually you also want to avoid numbers that have tails of 0s... unless you
> specify whether you want your list to display numbers with a particular

> kind of representation when two are available. [...]

And you'll note that, in the base-2 case, that is exactly what
I did.

> The simplest base-2 adaptation *is* to use two-digit blocks.

In which case it would be base 4 and I'm still left with
the conclusion that the preferred system is base 3. But
base 4 is good as well, since, as you stated, you can then
exclude both zeros and threes from the generated number.

>> o fill the odd positions of the new list with all
>> of the remaining reals, in the same order as before,
>> but with, of course, the 2^-(2n) subsequence eliminated.
>
> You don't need to eliminate anything. Keep the numbers if they already
> appeared, it doesn't matter.

Yes, you are right, of course. I knew that adding to the list
was OK, but was thinking that I would then have to bring in
additional firepower, such as the theorem that a denumberable
plus a denumerable is a denumerable. But all that is really
needed is to state that it is clear that the generated number
is not present in either of the the two subsequences, and
in particular not present in the original list.

[Ross A. Finlayson]

On Oct 4, 9:03 am, George Cornelius <cornel...@eisner.decus.org>
wrote:

nondenumerability = uncountability


nonzero =/= zero


Is there any difference between denumerability and countability? The
Engllish language only has so many words that work on numbers,
countable is simply countable instead of denumerable, say. It is the
style.


EF(n) = n/d
n-> d, d->oo, n e d+


constructs this function of naturals


EF(0) = .0000...
EF(1) = .0000...
...
EF(d-) = .1111...
EF(d) = .1111...


The range goes from zero to one, where EF(1) > EF(0) and EF(d-) <
EF(d).


The antidiagonal from EF(0) is .111... < EF(n). The antidiagonal is
then on the list, as it is EF(d), as constructed, always on the end of
the list, there that being some value >= 1.0.


Then, Sum_n=1^d 1/d = 1. This is so for the halves, thirds, and so on
of the unit line segment. Simply constructed these multiply or scale
over what becomes the ray then line, of these, the complete ordered
field is built of rational approximations, then. Building rational
approximations can be of those.


Here, d+ means d+1, d- = d-1, R+ means the positive real numbers, and
R means all of them. The arrows mean goes to and oo is shorthand for
infinity. The ellipse has the same value going on forever from
there.


Then, 1 and 1.0 have the same value, when there is only one 1.0.


Regards,

Ross Finlayson

[Arturo Magidin]

On Thursday, October 4, 2012 11:03:09 AM UTC-5, George Cornelius wrote:
> Arturo Magidin wrote:
>
> > On Thursday, October 4, 2012 2:37:44 AM UTC-5, George Cornelius wrote:
>
> >
>
> >
>
> >>
>
> >> That having been dispensed with, what would be the simplest
>
> >> way (short of bunching up the bits into groups of n, making
>
> >> it essentially base-2^n) of adapting the proof to base 2 [...]
>
>
>
>
>
> > Usually you also want to avoid numbers that have tails of 0s... unless you
>
> > specify whether you want your list to display numbers with a particular
>
> > kind of representation when two are available. [...]
>
>
>
> And you'll note that, in the base-2 case, that is exactly what
>
> I did.

Indeed. But I thought that you wanted a unified approach that worked on *all* bases; for base 4 and up you don't have to worry about dual representation at all provided you make your definition of the constructed number correctly. For base 3 you *do* need to worry about dual presentations, so you must also specify properties for the given list. That's all I was trying to point out: there actually are two bases that are troublesome if you are not careful, namely base 2 *and* base 3.

--
Arturo Magidin

[calvin]

Why isn't that the point? You are supposed to produce an
exhaustive list so that you can show (reductio ad absurdum)
that a new number you come up with is not on the list. If
your list is not assumed exhaustive then you've prepared nothing
subject to a possible contradiction. All you have done here
is create an admittedly incomplete list and then shown that
a flipped diagonal is not on that particular list. So what?

I'm saying that you must first assume that an exhaustive list
exists, and it would be crazy to assume that the diagonal of
such a list was repeating. However, I concede that what I am
trying to say cannot be put into mathematical terms.

[George Cornelius]

Arturo Magidin wrote:
> You don't need to eliminate anything. Keep the numbers if they already
> appeared, it doesn't matter.

This insight, then, leads to a simplified version of
the originally proposed base 2 proof:

o Start with a list as originally specified. The
individual digits can be designated in the usual way
as x_j_k for the digit in the jth row and the kth
column.

o Digits of generated number y are then:

y_(2k-1) = 1 - x_k_(2k-1)

and

y_2k = 0

for all k>0.

No need for insertions into the original list, and
this y meets all of the the requirements for the
reductio ad absurdum proof. We've just changed
from using a diagonal to using what we might call
a pseudodiagonal.

[Frederick Williams]

calvin wrote:
>
> On Oct 4, 10:58 am, Arturo Magidin <magi...@member.ams.org> wrote:

> [...] I can certainly produce a list in which the straighforward version of Cantor's diagonal argument fails to produce a number not on the list, so I do not see how you can claim that such a thing does not happen. Namely, take the list a_n where a_1=.10000000 and a_n, with n>1, is the number whose first n-1 digits are 1 and all the rest are 0; that is,

> >
> > a_1 = .100000000...
> > a_2 = .100000000...
> > a_3 = .110000000...
> > a_4 = .111000000....
> > etc.
> > (If you demand that the list have no repeats, replace a_2 with .10100000....)
> >
> > Then the number produced by "flipping the diagonal" is .0111111..... which is just another binary presentation of a_1, hence is already on the list.
> >
> > (Yes, the list is *obviously* not exhaustive, but that's not the point;
>
> Why isn't that the point? You are supposed to produce an
> exhaustive list so that you can show (reductio ad absurdum)
> that a new number you come up with is not on the list. If
> your list is not assumed exhaustive then you've prepared nothing
> subject to a possible contradiction. All you have done here
> is create an admittedly incomplete list and then shown that
> a flipped diagonal is not on that particular list. So what?

Surely what Arturo Magidin is doing is expanding on a particular step in
the proof. The numeral .0111111..... is not in the list, but that is
the same number as .100000000... which is on the list. So just flipping
the digits won't do: you have to take into account that one number may
have two representations. Isn't Arturo Magidin simply stressing that
point?

--
Where are the songs of Summer?--With the sun,
Oping the dusky eyelids of the south,
Till shade and silence waken up as one,
And morning sings with a warm odorous mouth.

[George Cornelius]

Ross A. Finlayson wrote:
> nondenumerability = uncountability

> nonzero =/= zero

Am I guilty of imprecision? No doubt. After all,
{1,2,3} could be said to be nondenumerable.

I would ordinarily have used the term uncountability
for exactly that reason, but had seen the phrase
'nondenumerablity of the reals' in an earlier
(and rather questionable) post, so used that to
pose my problem.

But we members of the various priesthoods do so
love the Latin sounding phrases - nondenumerabiles
sunt - so the usage is hard to resist.

Arguably the usual formulation of the Cantor proof
requires modification when applied to a finite list, so
the proof itself is ordinarily a proof of
nondenumerability.

Or perhaps I used the wrong term outright in the post
you were replying to, where I referred to theorems
I decided I would not actually have needed after all?

[Jesse F. Hughes]

William Elliot <ma...@panix.com> writes:

>> Can you propose a simple and natural base 2 version of Cantor's proof?
>
> Yes.

Do you work at being an asshole, or does it come naturally?

Seriously, what the fuck is wrong with you? Seriously.

For the OP, I imagine that others have already answered in more detail,
but the basic idea is to diagonalize on pairs of digits (bits, properly
speaking). Thus, given a listing like so:



We group the binary sequences like so:

0.0000101010101....
^^
0.1010001111111....
^^
0.1111101111111....
^^

And then we simply construct a number such that the n'th pair of bits
differs from the n'th pair of bits in the constructed number -- *and* we
don't fall into a dual representation trap. This is not difficult, but
I omit the details unless no one else answers your question
satisfactorily.

To William: Do you see? Is that so hard? If it is, why not just remain
silent? Being an asshole is a willful act. You don't have to do it,
you son of a bitch.

With due apologies to your mother, who has surely suffered enough.

--
I was driving down the interstate through Winslow, Arizona,
I had Seven Vices on my mind --
Sloth and Avarice, Fornication, Television,
Whiskey, Beer and Wine. -- Austin Lounge Lizards

[Virgil]

In article
<f2685b36-abb3-4044...@j14g2000yqb.googlegroups.com>,

calvin <cri...@windstream.net> wrote:

> > (Yes, the list is *obviously* not exhaustive, but that's not the point;
>
> Why isn't that the point? You are supposed to produce an
> exhaustive list so that you can show (reductio ad absurdum)
> that a new number you come up with is not on the list. If
> your list is not assumed exhaustive then you've prepared nothing
> subject to a possible contradiction. All you have done here
> is create an admittedly incomplete list and then shown that
> a flipped diagonal is not on that particular list. So what?
>
> I'm saying that you must first assume that an exhaustive list
> exists, and it would be crazy to assume that the diagonal of
> such a list was repeating. However, I concede that what I am
> trying to say cannot be put into mathematical terms.

The point is that the proof of incompleteness works for ANY list,
whether or not it is claimed to be complete, thus showing that ANY list
is incomplete. Thus there cannot be any such thing as a complete list,
and thus no point in assuming that a list is complete.

What works for ALL lists must automatically work for lists assumed to be
complete lists. So that assumption of completeness is irrelevant.
--

[Jesse F. Hughes]

George Cornelius <corn...@eisner.decus.org> writes:

> Ross A. Finlayson wrote:
>> nondenumerability = uncountability
>
>> nonzero =/= zero
>
> Am I guilty of imprecision? No doubt. After all,
> {1,2,3} could be said to be nondenumerable.

Oh, don't you fret. Ross is a crank, plain and simple.

Please don't concern yourself too much with his bleatings.

I mean this in all generosity and I encourage you to take the time to
read his recent posts to determine for yourself whether you think that
he is the kind of poster you should respond to.

For myself, I've decided that years ago.

--
Jesse F. Hughes
"That's what's brutal about mathematics! When you're wrong, you can
have spent years, and lots of effort, and come out at the end with
nothing." -- James S. Harris on the path of self-discovery (?)

[Arturo Magidin]

First: you are incorrect in claiming that you must proceed by contradiction. In its simplest form, what we are trying to establish is:

Given a list of real numbers, there exists a real number not on the list.

If you examine your "argument by contradiction" carefully, you will find that in fact it is a "fake" argument by contradiction: the hypothesis that your list is complete is completely and utterly unnecessary until the very last line of the argument... and then it is only used to reach a contradiction by comparing it with the *direct* conclusion you have found, namely, that there is a number not on the list. If you remove the first line ("Assume the list is complete") and the last line ("which contradicts our hypothesis that the list is complete") you get a perfectly valid, perfectly fine, argument of the statement above.

Second, there reason it is not the point is this: one is trying to present a *general* argument that holds *for all given lists* (or if you prefer, for all given lists that are assumed to be complete). However, if there are situations in which the general argument does not hold, then what you have is an incomplete argument. It doesn't matter that for any specific instance I give you of a list in which the general argument fails you can device an ad hoc argument, *for that list only*, that shows the list to be incomplete, because we still have the problem that you do not have a **general** argument that works for **all** lists.

Your assertion was that there *is* no list for which the general argument can fail to produce a number not on the list; but that assertion is plainly false, since I can produce lists for which the argument given *does* fail. I know that *other*, specially tailored arguments will work for *those* lists, but it's immaterial because it still invalidates the implicit claim that the general argument *always* works.

That's why it's beside the point, even if your contention that one must assume that the list is complete were valid. It is *false* that it is impossible for the given list to produce a number with dual representation.

--
Arturo Magidin

[Ross A. Finlayson]

On Oct 4, 11:53 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

What I dispute that. If I was a crank, I would be telling you
something that you could prove in the real world or empirically,
obviously false.


Then in terms of some absurdist mathematical crank who works in any
mathematics that is readily applied that is more clear, crankiness.
Here I won't say that transfinite cardinals are used to prove untrue
things about the objects of real analysis instead that there is not
the applied in that, the pure mathematics as realm of the mathematical
crank.


Here then in calling me a mathematical crank, that is where in
discussion about the antidiagonal argument I point to structural
features of particular forms of ordering of the real numbers that is
most directly to show there IS a function between the naturals and
reals. Then, that it works out that it does (satisfy numerical
properties as expected and also as is), is not unexpected. This is
where, if a crank is arguing that there are true things that are not
true things of the standard, they would have to have that to not be a
crank.


No basically, in the interests of the applied, I am interested in EF
and its structure directly and most directly in real analysis and
infinitesimal analysis.


Then you could see that when the crank encounters disagreement, or
where the sincere proponent of alternative encounters disagreement,
then generally in logic we have structure to validate the logic. Here
the crank adheres to fallacious reasoning or as well flip-flops, while
the sincere proponent is for reasons of validity in concern.


Then whether or not you see a crank I don't, which would also be
another sign of a crank, contrasted with a crackpot or troll, the
crackpot is incompetent, the crank is wrong, and the troll isn't
right. The sincere proponent as rational would acknowledge his own
incompetence, if simply as a possibility, thus requiring adequate
satisfaction of the argument, to maintain it.


Crank, crackpot, incompetent troll: to each his own.


Not being a troll they don't know me from one anyways.


Regards,


Ross Finlayson

[Arturo Magidin]

On Thursday, October 4, 2012 12:37:49 PM UTC-5, George Cornelius wrote:
> Arturo Magidin wrote:
>
> > You don't need to eliminate anything. Keep the numbers if they already
>
> > appeared, it doesn't matter.
>
>
>
> This insight, then, leads to a simplified version of
>
> the originally proposed base 2 proof:
>
>
>
> o Start with a list as originally specified. The
>
> individual digits can be designated in the usual way
>
> as x_j_k for the digit in the jth row and the kth
>
> column.
>
>
>
> o Digits of generated number y are then:
>
>
>
> y_(2k-1) = 1 - x_k_(2k-1)

You want absolute values here: y_{2k-1} = |1-x_{k,2k-1}|.

>
>
> and
>
>
>
> y_2k = 0
>
>
>
> for all k>0.
>
>
>
> No need for insertions into the original list, and
>
> this y meets all of the the requirements for the
>
> reductio ad absurdum proof. We've just changed
>
> from using a diagonal to using what we might call
>
> a pseudodiagonal.

Which, if you think about it, amounts simply to working with pairs of digits instead of single digits, thus moving back to the "base 4" trick for dealing with base 2. What is happening is that you are looking at the digits 2 at a time, and you are changing digits pairs of the form 1x to 00, and digit pairs of the form 0x to 10.

--
Arturo Magidin

[George Cornelius]

Arturo Magidin wrote:
> On Thursday, October 4, 2012 12:37:49 PM UTC-5, George Cornelius wrote:

>> y_(2k-1) = 1 - x_k_(2k-1)
>
> You want absolute values here: y_{2k-1} = |1-x_{k,2k-1}|.

But for binary digit x, 1-x, often called its
complement, is a binary digit as well.

>> No need for insertions into the original list, and
>> this y meets all of the the requirements for the
>> reductio ad absurdum proof. We've just changed
>> from using a diagonal to using what we might call
>> a pseudodiagonal.
>
> Which, if you think about it, amounts simply to working with pairs of
> digits instead of single digits, thus moving back to the "base 4" trick
> for dealing with base 2. What is happening is that you are looking at the
> digits 2 at a time, and you are changing digits pairs of the form 1x to
> 00, and digit pairs of the form 0x to 10.

Of course. I started out by thinking I would just restate
the original mapping algebraically, and had pretty much
forgotten that in order to simplify the proof I changed it
to be the equivalent of the insertion of a row of 1's
at each even position in the list.

Ultimately we have constructed two methods of adapting the
proof to base two by inserting rows of all 1's (a formal step
only; we have already ruled out a row of all ones as an
allowable canonical representation of any of our reals):

o insert a row of 1's at each even position

or

o insert a row of 1's at position 2^k for each k>=0

The diagonalization is the same in each case: you generate
a sequence of digits by taking the digitwise complement of
the diagonal, yielding the canonical representation of a
real number having the desired properties.

The 2^k insertion solution then becomes the preferred binary
based one, since the even rows technique has no advantage
over Cantor's proof applied directly to base 4.

In the end you choose any of the bases up to base 4 as
your preferred one, with the base 2 proof being slightly
more complex than the others.

[NOTE] For the 2^k case as well you could skip all discussion
of insertions by instead noting that the equivalent is, for each
k>0, to apply complementation to digit j = k + 1 + L(k) of
row k, L(k) being the greatest integer less than or equal to
log_2 k, and making this the jth digit of the result; while
setting all remaining digit positions of the generated result
to zero.

[Ross A. Finlayson]

On Oct 4, 11:04 am, George Cornelius <cornel...@eisner.decus.org>
wrote:

Well it's quite alright for example where two cardinals being
equipollent, equivalent, and the same size are being the same thing,
die Unendlichkeit oder die Unbegrentzheit, here the same as another,
different.


Yes quite, go on. My Latin suffers if only for that I really only
speak English, and phrases and patois of various European and Asiatic
languages, non-denumerabilis est, probably more French than Latin.
Then, my Italian is rather Spanish but my Spanish is Catalonian (or,
Mexican). That said, caveat lector being the only Latin phrase
recently here, here basically in the immobile celestial spheres they
are mobile. From the Latin, for the foundations, here that
complements atomism, Greek in Latin.


In the finite cases with the completion of the lists or the product
spaces of the numbers here as expansions for the diagonal method, the
list is much longer than a square where the diagonal is the same
length as an expansion as a member of the list, so in the product
space. I think you allude to that in these modifications of the
finite list, of the elements of the product space.


Heh no then it is simply style, denumerable or countable, here, I
wonder: do you start counting with one, or zero? Then, it could just
be used for denumerable and countable starting with one or zero, the
counting integers.


At least, it is surely if I thought there was anything wrong with that
and you asked I would tell you, no, it was also good to establish that
non-zero, is not zero.


Regards,


Ross Finlayson

[William Elliot]

On Thu, 4 Oct 2012, calvin wrote:
> On Oct 4, 4:00 am, William Elliot <ma...@panix.com> wrote:

> > > Can you propose a simple and natural base 2 version of Cantor's proof?
> > Yes.

First we show that the set S, of binary sequences is uncountable.

Assume S is countabke and s_k the k-th binary sequence s_k:N -> 2^N
Define the diagonal sequence d(k) = s_k(k) = 1 if s_k(k) = 0, = 0 otherwise.

We see that the sequence d isn't in the list.
Thus the set of binary seqences is uncountable.

Let I = [0,1) written in binary decimal form
without ending in forever repleting 1's.
Note each r in I has a unique binary decimal expression.

Show that the set J, of numbers in [0,1) that be written in binary
decimal form ending in forever repleting 1's is countable.

Since S = I \/ J, I cannot be countable
because the union of two countable sets is countable.








and n_kj the

[Virgil]

In article
<184c5531-88fa-4f0f...@q9g2000pbo.googlegroups.com>,

"Ross A. Finlayson" <ross.fi...@gmail.com> wrote:

> On Oct 4, 11:53 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> > George Cornelius <cornel...@eisner.decus.org> writes:
> > > Ross A. Finlayson wrote:
> > >> nondenumerability = uncountability
> >
> > >> nonzero =/= zero
> >
> > > Am I guilty of imprecision?  No doubt.  After all,
> > > {1,2,3} could be said to be nondenumerable.
> >
> > Oh, don't you fret.  Ross is a crank, plain and simple.
> >
> > Please don't concern yourself too much with his bleatings.
> >
> > I mean this in all generosity and I encourage you to take the time to
> > read his recent posts to determine for yourself whether you think that
> > he is the kind of poster you should respond to.
> >
> > For myself, I've decided that years ago.
> >
> > --
> > Jesse F. Hughes
> > "That's what's brutal about mathematics!  When you're wrong, you can
> > have spent years, and lots of effort, and come out at the end with
> > nothing." -- James S. Harris on the path of self-discovery (?)
> >
> >
>
>
>
> What I dispute that. If I was a crank, I would be telling you
> something that you could prove in the real world or empirically,
> obviously false.

Every crank does dispute his crankdom.

And on occasion you have told us things provably false.

Or at least tried to tell us such things.
--

[Virgil]

In article <878vbma...@phiwumbda.org>,

"Jesse F. Hughes" <je...@phiwumbda.org> wrote:

> William Elliot <ma...@panix.com> writes:
>
> >> Can you propose a simple and natural base 2 version of Cantor's proof?
> >
> > Yes.
>
> Do you work at being an asshole, or does it come naturally?
>
> Seriously, what the fuck is wrong with you? Seriously.
>
> For the OP, I imagine that others have already answered in more detail,
> but the basic idea is to diagonalize on pairs of digits (bits, properly
> speaking). Thus, given a listing like so:
>
>
>
> We group the binary sequences like so:
>
> 0.0000101010101....
> ^^
> 0.1010001111111....
> ^^
> 0.1111101111111....
> ^^
>
> And then we simply construct a number such that the n'th pair of bits
> differs from the n'th pair of bits in the constructed number -- *and* we
> don't fall into a dual representation trap. This is not difficult, but
> I omit the details unless no one else answers your question
> satisfactorily.

As Arturo Magidin has already noted, one has to be careful to use only
01 and 10 as replacements in such binary sequences, because allowing
either 00 or 11 would allow numbers with dual representation.

Similarly in bases greater than 3, the "anti-diagoal" should not be
allowed to contain either 0 or base-minus-one as a digit.

[Ross A. Finlayson]

On Oct 4, 8:24 pm, Virgil <vir...@ligriv.com> wrote:
> In article
> <184c5531-88fa-4f0f-b4c1-2507d2bd9...@q9g2000pbo.googlegroups.com>,

I'd be happy that you remind me of them or personally.

Thanks,

Ross Finlayson

[Virgil]

In article
<eb40df3f-c49a-4e17...@c6g2000pba.googlegroups.com>,

"Ross A. Finlayson" <ross.fi...@gmail.com> wrote:

If you don't even remember them, they must have been false!
--

[Ross A. Finlayson]

On Oct 4, 8:24 pm, Virgil <vir...@ligriv.com> wrote:
> In article
> <184c5531-88fa-4f0f-b4c1-2507d2bd9...@q9g2000pbo.googlegroups.com>,

No, I'd be happy to examine any particular example where you claim
something is false, and I claim and uphold that it is true.


For example, ever expanding the antidiagonal rule until it is defined
on the range of EF, in extremis defines the range of EF, again
defining the antidiagonal of EF as 1.0. Once again, the conscientious
mathematician sees that EF is unique among these other functions,
while sharing properties of its range with the numerical continuum of
the real numbers, from the natural continuum of the natural numbers.


For example, this here particularly, about all the ways to span digits
to generate an everywhere-non-diagonal, was addressed as above. And,
it was as simple to dispatch, then.


Or, here for Hughes, the drinking quote example.


I acknowledge my limits but dispute your trolling. I'm a
conscientious mathematician, EF is a function.


Regards,


Ross Finlayson

[Frederick Williams]

"Ross A. Finlayson" wrote:
>
> [...] Once again, the conscientious

> mathematician sees that EF is unique among these other functions,
> while sharing properties of its range with the numerical continuum of
> the real numbers, from the natural continuum of the natural numbers.

How do you define 'continuum'?

[Ross A. Finlayson]

On Oct 5, 6:26 am, Frederick Williams <freddywilli...@btinternet.com>
wrote:

Williams, I'd hope to define the continuum as Spinoza does, that has
seemed to me as rational a definition of any, of the most basic
fundamental underpinning of all structure, at once to the totally
definitive itself as the space, with the natural numbers as a
continuum, then the continuum of real numbers and all products of the
space. I see Spinoza's definition of the natural numbers as the
continuum, as part of the definition.


Then, with the continuum being unbroken and complete, still we can
look to the individua that comprise it, then how those would be, here
in the reals iota-values or points, variously, and to structures built
of the completed pieces. There's a rational composition.


Then as above I describe a simple space-filling curve as a natural
continuum.


I'm a Platonist, it's much already there this continuum as we define
it.


Regards,


Ross Finlayson

[Virgil]

In article
<b6420270-abfa-4586...@pz10g2000pbb.googlegroups.com>,

"Ross A. Finlayson" <ross.fi...@gmail.com> wrote:

Your EFing is no part of standard matheatics.
--

[Virgil]

In article
<134df24c-3fa1-4d11...@q5g2000pbk.googlegroups.com>,

"Ross A. Finlayson" <ross.fi...@gmail.com> wrote:

Like the your alleged "continuum" of natural numbers?

Perhaps it would be easier for us to understand what YOU mean by
"continuum" if you were to give us a few examples of things which you
do NOT regard as continuua.
--

[Shmuel Metz]

In <Pine.NEB.4.64.12...@panix1.panix.com>, on 10/04/2012

at 07:00 PM, William Elliot <ma...@panix.com> said:

>First we show that the set S, of binary sequences is uncountable.

At that point you've done all of the heavy lifting and don't need the
diagonalization proof.

--
Shmuel (Seymour J.) Metz, SysProg and JOAT <http://patriot.net/~shmuel>

Unsolicited bulk E-mail subject to legal action. I reserve the
right to publicly post or ridicule any abusive E-mail. Reply to
domain Patriot dot net user shmuel+news to contact me. Do not
reply to spam...@library.lspace.org

[Ross A. Finlayson]

On Oct 5, 1:11 pm, Virgil <vir...@ligriv.com> wrote:
> In article

> <134df24c-3fa1-4d11-a497-a4d582116...@q5g2000pbk.googlegroups.com>,

The naturals are a continuum for induction. The reals are a continuum
for course. Obviously the naturals aren't the continuum in real
space, the reals are the continuum. (And they don't need any more to
be the continuum and any more would already be in it, elements of
the.)

Here the definition is for the continuum to infinity, items in a line
are sequential or contiguous, continuity has for the infinitely many
items their part in the continuum. So, a finite number of integers is
not the continuum. All the finite integers is a continuum.

Continuity defines many features for example real analysis. Then, I
read of the word individua, from Eco. Then, the rationals are not a
continuum.

Then, where the two finite objects is the universe it is the
continuum.

Mine, rather, "of".

"Of", Virgil, "Of".

That's Spinoza's.

Pronounced "E.F.". It's a function, it doesn't care if I care. EF.

For a continuum, it's the function.

The. (Pronounced "the".)

It, Virgil, it's it.

"Virgil, you think you kept set theory, but I took real analysis, and
I kept set theory."

Regards,

Ross Finlayson

[Virgil]

In article
<e6c3be90-0b0c-4f35...@q5g2000pbk.googlegroups.com>,

"Ross A. Finlayson" <ross.fi...@gmail.com> wrote:

> The naturals are a continuum for induction.

Websterąs Concise Electronic Dictionary

1 sense(s) for łcontinuum˛

1. con·tin·u·um
€ (noun) [con·tin·ua]
­ sequence or progression (as of qualities) showing only minute
differences from one to another

http://www.thefreedictionary.com/continuum
con·tin·u·um  (kn-tny-m)
n. pl. con·tin·u·a (-tny-) or con·tin·u·ums
1. A continuous extent, succession, or whole, no part of which can be
distinguished from neighboring parts except by arbitrary division.
2. Mathematics
a. A set having the same number of points as all the real numbers in an
interval.
b. The set of all real numbers.


And many other definitions. none of which are compatible with the
natural numbers being in any way a continuum.
--

[Virgil]

In article
<e6c3be90-0b0c-4f35...@q5g2000pbk.googlegroups.com>,

"Ross A. Finlayson" <ross.fi...@gmail.com> wrote:

> The naturals are a continuum for induction.

Not for any standard meaning of "continuum".
--

[Frederick Williams]

Yeah, that's why I asked!

> Perhaps it would be easier for us to understand what YOU mean by
> "continuum" if you were to give us a few examples of things which you
> do NOT regard as continuua.
> --
>
>

[Frederick Williams]

"Ross A. Finlayson" wrote:

>
> The naturals are a continuum for induction. The reals are a continuum
> for course. Obviously the naturals aren't the continuum in real
> space, the reals are the continuum. (And they don't need any more to
> be the continuum and any more would already be in it, elements of
> the.)

A continuum is a compact connected T2 space.

What topology are you putting on N to make it a continuum?

[Ross A. Finlayson]

On Oct 6, 7:42 am, Frederick Williams <freddywilli...@btinternet.com>
wrote:

Let's see, it's compact because there's a point at infinity, connected
from defining the natural integers from that, T2 and a space, I'll
agree that you can define a compact, connected, T2 space as a
continuum and that Cantor, Spinoza, and so on define it from
antiquity. A topological continuum, say, has a topology, apparently
that one in terms of the properties of its elements. That's the
topology of the reals (with points at infinity or in a compact
segment) which is founded on elements of an inductive framework. Here
I don't know "T2", this is two properties that begin with T or the
second T property. (Compared to, say, measured Banach spaces.)

http://en.wikipedia.org/wiki/Topology
http://en.wikipedia.org/wiki/Topology_glossary

Then there's a space over the naturals.

Sure it's just that. Structurally, there's a space, here in R dots,
that as a continuum it has properties of compactness and
connectedness, the set of natural integers (compared to just, say,
inductive routine) The naturals are compact (here with a point at
infinity though I say they have that already). They're connected in
this space that defines open subsets over them, as they do not as they
don't.

Williams though here using support from Spinoza or Borges for the
infinite counting numbers themselves as a continuum, still I fully
intend that it satisfies its standard meaning, what is a definition of
the continuum.

What then is "T2"?

Thanks,

Ross Finlayson

[Frederick Williams]

"Ross A. Finlayson" wrote:
>
> On Oct 6, 7:42 am, Frederick Williams <freddywilli...@btinternet.com>
> wrote:

> >
> > A continuum is a compact connected T2 space.
> >
> > What topology are you putting on N to make it a continuum?

No reply then.

> What then is "T2"?

A space is T2, or T_2 if you prefer, if it is separated or Hausdorf.

[Ross A. Finlayson]

On Oct 6, 10:30 am, Frederick Williams <freddywilli...@btinternet.com>

The finite sets are the open sets and their complements are the closed
sets. Or vice versa, either of these is a topology on the naturals.

That's an example note they have one.

Then for being Hausdorff all the points have disjoint neighborhoods.
The neighborhoods are only the points. Then for any finite
neighborhood the naturals are not Hausdorff, because if two points'
immediately share as neighbors the points between them then there
doesn't exist neighborhoods of each that are disjoint and separate
from each other. Defined though for the rationals, which aren't
connected, those defining points in normal order on the unit line
segment, those are Hausdorff, in their space, of their topology.

Why thank you I haven't much looked into algebraic topology for some
time. Now having alluded to my utter incompetence in not knowing or
remembering what T_2 was, still I'm quite content to define this space
on the naturals (finite and infinite) and see from thats' usual
topology the space there, then the naturals have a topology (in finite
and infinite here, distance, counted between them).

So, you ask, there already is one, was I already particularly aware of
that, "the topology on the naturals from defining open and closed as
finite and infinite", more from having implicitly compact naturals
(from Russell, for example) and now so that you at least claim I don't
define a topology for the naturals, there it is.

Regards,

Ross Finlayson

[Arturo Magidin]

On Saturday, October 6, 2012 3:57:28 PM UTC-5, Ross A. Finlayson wrote:
> On Oct 6, 10:30 am, Frederick Williams <freddywilli...@btinternet.com>
>
> wrote:
>
> > "Ross A. Finlayson" wrote:
>
> >
>
> > > On Oct 6, 7:42 am, Frederick Williams <freddywilli...@btinternet.com>
>
> > > wrote:
>
> >
>
> > > > A continuum is a compact connected T2 space.
>
> >
>
> > > > What topology are you putting on N to make it a continuum?
>
> >
>
> > No reply then.
>
> >
>
> > > What then is "T2"?
>
> >
>
> > A space is T2, or T_2 if you prefer, if it is separated or Hausdorf.
>
> >
>
> > --
>
> > Where are the songs of Summer?--With the sun,
>
> > Oping the dusky eyelids of the south,
>
> > Till shade and silence waken up as one,
>
> > And morning sings with a warm odorous mouth.
>
> >
>
> >
>
>
>
>
>
> The finite sets are the open sets and their complements are the closed
>
> sets.

That's not a topology: open sets are not closed under arbitrary unions.

> Or vice versa, either of these is a topology on the naturals.

"Vice-versa" is called the cofinite topology; that *is* a topology, but...

>
>
>
> That's an example note they have one.
>
>
>
> Then for being Hausdorff all the points have disjoint neighborhoods.

Nope. In the cofinite topology, no two points have disjoint neighborhoods (that would require two subsets, X and Y, of the naturals, each with finite complement, to be disjoint; that is impossible).

The cofinite topology is a well-known example of a T_1 topology that is not T_2 (given any two points x and y, there are neighborhoods X and Y of x and y, respectively, such that x is not in Y and y is not in X; but there do not exist disjoint neighborhoods).

>
> The neighborhoods are only the points.

No; "neighborhood of x" is any set that contains an open set U that contains x.

Really... wouldn't it be easier to say "I have no idea what I'm talking about" than waste time and space with blatantly incorrect assertions?

--
Arturo Magidin

[Virgil]

In article
<48063903-4a94-4411...@ql4g2000pbc.googlegroups.com>,

"Ross A. Finlayson" <ross.fi...@gmail.com> wrote:

an example of Ross' incompetence at its finest!
--

[Virgil]

In article <eb60a499-b787-4e0f...@googlegroups.com>,

Arturo has made my point showning precisely how incompetent Ross is.
--

[Ross A. Finlayson]

Here simply those neighborhoods of naturals that are disjoint are the
points themselves, they don't have neighborhoods or it's a different
space than the simple one built from the properties of their
adjacency.

Here I plainly agree those are features of those things, except back
to the first one, the finite topology, open sets are open under open
unions.

Professor Magidin I plainly would thank you to remind me of given the
associated structural definitions, there are these plain features that
of are course well-covered and familiar to many where often grad
students take a couple years of topology, I am happy here to adhere to
the most non-controversial and well-defined, definitions.

And, I'm quite familiar with them. thank you. For example in
algebraic topology there are some forty or more named theorems and I
know not their content. Yet, thank you, I'm quite familiar with them.

So, then was where the neighborhoods of two points in R, are for R
dots, two points infinitely far apart, that is simple. Notice how
this again builds from the simple definition that why yes that the
naturals (finite and infinite) so support the continuum.

Then though it seems I would append 80 pages of definitions with my
reply. Or, I'd be happy to refer you to relevant standard, and agreed-
upon definitions, here of for example that the countable infinity is
already complete/compact and then these other features along generally
why it is a good idea that the other functions are for uncountably
many to be connected, whether or not the continuum is countable, in
terms of working functions for the continuum in the continuous, and
the discrete (where finite combinatorics is perfectly modeled and
complete rationally).

No, I don't see much in that either. Here simply working reversible
definitions and construction of open and closed sets in topology, the
intersection is so that the space as well is open, then into for
example covers, in the finite topology.

Regards,

Ross Finlayson

[Virgil]

In article
<6f4b6ee6-f3ca-4d57...@q5g2000pbk.googlegroups.com>,

"Ross A. Finlayson" <ross.fi...@gmail.com> wrote:

The only interesting topology on the set of naturals has infinite
subsets as a basis of the open sets.

>
> Here I plainly agree those are features of those things, except back
> to the first one, the finite topology, open sets are open under open
> unions.

What distinguishes an open union from some other kind of union if all
the sets involved are open sets anyway?

>
> Professor Magidin I plainly would thank you to remind me of given the
> associated structural definitions, there are these plain features that
> of are course well-covered and familiar to many where often grad
> students take a couple years of topology, I am happy here to adhere to
> the most non-controversial and well-defined, definitions.
>
> And, I'm quite familiar with them. thank you. For example in
> algebraic topology there are some forty or more named theorems and I
> know not their content. Yet, thank you, I'm quite familiar with them.

In standard mathematics, being familiar with a theorem includes knowing
a good deal about its content.

In Ross-Land, apparently, familiarity cedes content.

>
> So, then was where the neighborhoods of two points in R, are for R
> dots, two points infinitely far apart, that is simple. Notice how
> this again builds from the simple definition that why yes that the
> naturals (finite and infinite) so support the continuum.

I do NOT notice any such thing. In particular, I do not "notice" any
non-finite naturals.
--

[Arturo Magidin]

This paragraph either uses standard terminology in completely non-standard ways, or else is nonsense as written. Either way, I have neither the time nor the inclination to try to untangle it or to educate you.

>
>
>
> Here I plainly agree those are features of those things, except back
>
> to the first one, the finite topology, open sets are open under open
>
> unions.

Again, the above is patent nonsense.

>
>
>
> Professor Magidin I plainly would thank you to remind me of given the
>
> associated structural definitions, there are these plain features that
>
> of are course well-covered and familiar to many where often grad
>
> students take a couple years of topology, I am happy here to adhere to
>
> the most non-controversial and well-defined, definitions.

There *are* no "controversial" definitions, and there *are* no "well-defined definitions" here. There are only *basic* definitions. If you don't know them, that's not a slight on yourself, nor is it a problem in and of itself.

It only becomes a problem when, *even though you do not know nor understand the definitions* you nonetheless pretend to speak about the subject as if you did. *That's* the problem. If you don't know what a topology is (and I could plainly drop the "if", since you demonstarted quite clearly that you do not), then the sane, reasonable, smart answer is to say "I don't know the definitions well enough" before proceeding. The dumb, ignorant, stupid way to proceed is to mouth off and then pretend that the reason your statements were incorrect has something to do with "controversy" or "issues" in the subject, rather than your own abject ignorance.

>
> And, I'm quite familiar with them. thank you.

Obviously, you are not. You have demonstrated this through your statements. *Saying* you are quite familiar with them does not endow you with such familiarity; it only shows that you are too proud to admit your ignorance.

> For example in
>
> algebraic topology there are some forty or more named theorems and I
>
> know not their content. Yet, thank you, I'm quite familiar with them.

For example, here; this is complete, utter, absolute nonsense. One cannot simultaneously be "quite familiar with them" *and* "know not their content". Either you are familiar, or you don't know their content, but not both at the same time.

> So, then was where the neighborhoods of two points in R, are for R
>
> dots, two points infinitely far apart, that is simple. Notice how
>
> this again builds from the simple definition that why yes that the
>
> naturals (finite and infinite) so support the continuum.

This is utter nonsense, again.

> Then though it seems I would append 80 pages of definitions with my
>
> reply.

No need to append 80 pages of definitions. Either you use language the way it is used by everyone else,you invent your own language and pretend that anybody who does not understand you is just being difficult; or, in your case, you **pretend** that you are using the language like everybody else, even though your very words betray that this is quite simply not the case.

You do *not* know what a topology is. You do *not* know what a neighborhood is. You do *not* know what "compact" means. You do *not* know what a "continuum" is. And you *know* that this is the case, but prefer to try to bluff your way past that ignorance. In the end, you are just like the kid in the playground saying "I know what it means, but you tell me so I can check that you do."

> Or, I'd be happy to refer you to relevant standard, and agreed-
>
> upon definitions, here of for example that the countable infinity is
>
> already complete/compact and then these other features along generally
>
> why it is a good idea that the other functions are for uncountably
>
> many to be connected, whether or not the continuum is countable, in
>
> terms of working functions for the continuum in the continuous, and
>
> the discrete (where finite combinatorics is perfectly modeled and
>
> complete rationally).

Kindly do not refer me to anything; it is patently clear that you are ignorant about the subject on which you attempted to pontificate. It is also patently clear that you are incapable of admitting such obvious, glaring mistakes as the ones you committed. I have no interest in continuing any talking past you, nor in wasting my time trying to educate the uneducable. Now, you can continue to pretend to know you know what you are talking about, but let me break the sad news to you: you aren't fooling anyone except yourself.

You can avoid replying to this message, or mentioning my name in the future. I have nothing else to say to you.

--
Arturo Magidin

[Ross A. Finlayson]

The open or finite ones in the finite topology, their finite or open
union is finite, open. Obviously an infinite or closed union of
finite sets may be closed, infinite. Why would it not? An infinite
union of finite sets is infinite, in what would be the finite
topology, on the naturals.

I don't much care to reply except to note that is, for example, true,
my statement, for what would be, a or the finite topology.

Then, simply to maintain what I see as true (here as defined), it's
about 80 pages of definitions.

"Really... wouldn't it be easier to say "I have no idea what I'm
talking about" than waste time and space with blatantly incorrect
assertions?"

What will be, will be. Que sera, sera.

Then, what would be the finite topology? What would be the
consequence of a space over an inductive set?

Then for the set itself of naturals, as infinite it is closed though
so it is not a topology, what would be the finite topology on N, with
sets of naturals or ordinals for union and intersection. Infinite
collections of finite sets could be an infinite set that is not the
set of all of them, i.e. unions of finite/open sets that are infinite/
closed, so I agree that is not a topology on N, thank you Arturo, what
would be the finite topology. The definition of topology on a set
that I will share has the empty set and the set being open in the
topology, the topology defines open and closed for sets closed under
union and intersection. Then for example the empty set and universal
set have a disjoint neighborhood and any union of them is open,
besides their disjoint is in the set. Then, for the finite sets being
defined, to have a set of finite sets, in a topology, these are the
finite naturals, for example as ordinals.

Then the structure is on sets of naturals. Then, the union is their
sum, intersection partial sum, in working toward what would be the
finite topology on the naturals.

Then, I guess the difference is, for what features there are that
define a topology here for example for topology's definition of a
continuum, I'm looking at how the naturals have that, you having
decided already whether or not they do that they don't. Then, the co-
finite topology you note exists, my impression as well is to build the
finite topology, here as a simple way to bridge the concern. This can
rather simply be built (for example in classical topology or for that
matter, geometry).

Also there's a consideration on building the topology on N, and
otherwise that finite values are open, not sets but values, their
intersection is always disjoint and their union has, only their union
being open, otherwise it would be a finite sum or the finite topology,
compared to what would be the topology on finite sets of naturals a la
the co-finite topology, a finite point topology.

Then, and I'm sure a casual reader who, bereft of the time, sees your
reply as invalidating as nonsense, might not notice for example that
it is one word that is underdefined, where the simple provision of
that fuller definition leaves what was "nonsense" instead a well-
formed "statement".

As well, I'm not too surprised if the point topology, escaped your
attention.

So, I could understand why, whether or not to present that to another
as dogma they can repeat without understanding, no, I could understand
why you would call that nonsense before claiming to fit it together.

Here, then, it is to what would be a continuum of natural integers,
having drifted into the features of the natural integers finite and
infinite, it is simply constructivist why the natural counting
integers are as good a way as any to sequence any inductive process,
here obviously through all of them. Then, its feature satisfy being
dense, for example around points to start in the establishment of
monotone functions from N to R[0,1], in building those.

Then, I have a constructivist response for that.

For Arturo, I certainly humbly accept your opinion and will simply
treat it as advice, for example having nothing to do with your recent
identity hijacking on sci.math, not attribute your poor attitude to
social causes. And to hell with those bastards or simply one guy
with a troll rampage fetish, Magidin I wrote here because it's easy to
find it later.

Then, the constructivist reponse, back to the definition of continuum
and topology's definition of continuum, here then noting the
structural features of the continuum, the implications of the
definition carry through to topology and here algebraic topology and
modern algebraic topology, and vice-versa.

Then for me to apologize it is for how the definition of topology on
the finite sets, for the discrete construction, work, if here
otherwise the general course is of the topology, there is a lesser
structure in the theory of topology here that the set itself is not
open. Then, the empty set is maybe not so either, just as example of
that two different structural features see that the other (non-
trivial) theorems about the topology or what would be the topology or
here, any other structure that it is, defining the separation of sets
into all those that are open together, and all those that are closed
together.

Then, with no infinite union, that is not an open set, so, it is only
for a finite union that it is an open set. Fine, that is feature,
here that for the closure of all the operations of union and
intersection, the sets are still elements of the set, which is itself
open. Yet, there are all the infinite subsets of that, that are not
open, in "what would be" the finite topology. Then how are those
finite? Here union is over any of them, each finite, if only over all
of them, then it IS all of them. The only infinite set of just the
numbers themselves is all of them!

Union is add, intersection is = 0. Definitions.

Finite point number topology, or, finite number point topology.

Yes, then to establish the idea of the finite topology, here to make
the union of two elements of the topology is their sum, then the
intersection is zero or here the empty set. For any two the union is
always open. For all of them, it is defined open. Recursively all
the unions are open. The only infinite union is all of them. To get
to that you have to add to them through them.
Then, the union only being defined for finitely many, there is not
infinity in the finite topology, then the union is open and the
intersection is always disjoint. But, any, as a product, is still in
the set, compared to that the union of any two numbers, as sets, as
numbers, might be not a number. This is for carrying the theorem
through.

Ah, yes then, quite.

"Here simply those neighborhoods of naturals that are disjoint are the
points themselves, they don't have neighborhoods or it's a different
space than the simple one built from the properties of their
adjacency."

Regards,

Ross Finlayson

[Virgil]

In article
<bb2e8c1c-3edc-4c36...@j2g2000pbg.googlegroups.com>,

"Ross A. Finlayson" <ross.fi...@gmail.com> wrote:

It would, so why won't Ross do it? Unwarranted arrogance!

>
> The open or finite ones in the finite topology, their finite or open
> union is finite, open.

By a "finite topology" on N you mean:
1, one with a finite number of open sets, or
2. one with every finite set being open?

These are quite different , but essentially your only options.

> Obviously an infinite or closed union of
> finite sets may be closed, infinite.

An infinite union and a closed union are not at all the same thing

> Why would it not? An infinite
> union of finite sets is infinite, in what would be the finite
> topology, on the naturals.

Which "the finite topology' is that?

>
> I don't much care to reply except to note that is, for example, true,
> my statement, for what would be, a or the finite topology.

As usual noting only what is not necessarily true.

>
> Then, simply to maintain what I see as true (here as defined), it's
> about 80 pages of definitions.

What you "see as true" seems to have little realation to reality.

>
> "Really... wouldn't it be easier to say "I have no idea what I'm
> talking about" than waste time and space with blatantly incorrect
> assertions?"

It would be, so why don't you try it?

>
> What will be, will be. Que sera, sera.
>
> Then, what would be the finite topology?

I asked you first!

> What would be the
> consequence of a space over an inductive set?

What do you man by a "space over an inductive set" or even a "space over
a set"?
>
The rest of Ross' stream of unconsciousness, being too muddled to be at
all parsible, I have deleted.
--

[Peter Webb]

"George Cornelius" wrote in message
news:506d3cc8$0$63202$815e...@news.qwest.net...

Cantor's proof of the nondenumberability of the reals is
generally given for a base 10 number representation. This
is perfectly fine since the number system base in use has
little significance, but, still, for purity's sake, there
would seem to be an advantage in removing dependence on a
rather arbitrarily chosen number system base.

_______________________________________________________
Indeed, but choosing base 2 doesn't help.

Better to remove any requirement to have any base at all. You can do this by
using the continued fraction representation instead of the decimal or binary
representation, just make the nth term of the continued fraction different.
That eliminates any reference to bases (as continued fractions were designed
to do).

[George Cornelius]

William Elliot wrote:

> On Thu, 4 Oct 2012, calvin wrote:
>> On Oct 4, 4:00 am, William Elliot <ma...@panix.com> wrote:
>
>> > > Can you propose a simple and natural base 2 version of Cantor's
>> > > proof?
>> > Yes.

>
> First we show that the set S, of binary sequences is uncountable.
>

> Assume S is countabke and s_k the k-th binary sequence s_k:N -> 2^N
> Define the diagonal sequence d(k) = s_k(k) = 1 if s_k(k) = 0, = 0
> otherwise.
>
> We see that the sequence d isn't in the list.
> Thus the set of binary seqences is uncountable.
>
> Let I = [0,1) written in binary decimal form
> without ending in forever repleting 1's.
> Note each r in I has a unique binary decimal expression.
>
> Show that the set J, of numbers in [0,1) that be written in binary
> decimal form ending in forever repleting 1's is countable.
>
> Since S = I \/ J, I cannot be countable
> because the union of two countable sets is countable.
>

OK, sure. Use diagonalization where it works best,
to show that the set of all infinite sequences of
elements of the set {0,1} is uncountable.

Then bring out a stock theorem to show by reductio
ad absurdum that we can ignore the countable
collection of excluded duplicate representations
by proving that uncountable - countable =
uncountable.

You could even do the first part as a stock theorem.
After all, we know, by diagonalization in fact,
that | 2 ^ X | > | X | for any set (or cardinal) X.

I mentioned, for a slightly different context,
that I was trying to avoid the theorem that you
used. Doesn't detract from yours being a great
answer, just that it loses some of the incredible
simplicity of the Cantor proof. I am convinced
I could teach the proof to a fifth grader; the
theory of cardinality would have to wait (at
least one year!).

So was Cantor smarter than a fifth grader?
How could we even question it? But his proof,
like many great insights, makes us think that
it is something anyone could have come up with
with just a few moments' thought.

[George Cornelius]

Can't address continued fractions - a gap in my
mathematics background - but it looks plausible.

The reason for examining base 2 is that - assuming
you want to preserve Cantor's approach pretty much
intact - it is a candidate for most natural of
choices; even more so in today's world of the
ubiquity of binary.

Another reason for base 2 arose in this discussion
by accident. Mr. Elliot came up with a two-stage
proof regarding which I commented that his choice to
apply diagonalization to the set of all sequences
of binary values - sequences of zeros and ones -
could also be done as a mere consequence of the
theorem that | 2 ^ X | > | X | for set or cardinal
X. I further mentioned that I believed the standard
proof of the theorem was Cantor diagonalization.

But it is only in base 2 that we have that natural
bijection between power sets and infinite sequences
of digits.

[George Cornelius]

Ross A. Finlayson wrote:
> Yes quite, go on. My Latin suffers if only for that I really only
> speak English, and phrases and patois of various European and Asiatic
> languages, non-denumerabilis est

I said Latin *sounding* phrases were appealing to me. To
each his own.

Today [*] my skill consists of typing something like

'define innumerabilis'

into Google and finding everything I might need among the first
few matches, e.g.,

http://en.wiktionary.org/wiki/innumerabilis ,

with my subsequent thinking in this case being that the plural
version sounded more impressive than the more applicable
singular one.

No matches on denumerabilis, though.

George Cornelius

[*] I recently ran across in my father's papers a certificate
for me placing second in some Latin competition, a paper on which
the teacher had written in first place and then crossed it out.
I'm thinking I was her favorite and she was making it clear
that she knew I could have studied harder for the exam and
done better. 'Course if you have a surname that goes back to
not just Roman times but Etruscan as well your teacher might
think you should be more of a star than you actually were.

[Graham Cooper]

On Oct 4, 5:37 pm, George Cornelius <cornel...@eisner.decus.org>
wrote:

> Cantor's proof of the nondenumberability of the reals is
> generally given for a base 10 number representation.  This
> is perfectly fine since the number system base in use has

Incorrect!

ALL(function):N->R EXIST(real):R ALL(natural):N
function(natural) =/= real

It is a 2OL proof of a_Certain_Representation of Reals
using a_Certain_Set_of_Operations on that representation.

Using TCP-IP for instance, Transmission Control Protocol on the
Internet, infinite streams are bundled into arbitrary size packets.

f(1) = 0. 314 159 265 ...
f(2) = 0. 278 182 845 ...
f(3) = 0. 333 333 333 ...
f(4) = 0. 123445 54321 ...
f(5) = 0. 11 11 11 11 ...
...

The anti-diagonal never produces a unique packet of any size.

*************

Here Jesse H Hughes is clearly mistaken, though he never follows up on
this.

https://groups.google.com/group/sci.math/browse_thread/thread/e1b7c9ac9a1bc17/81cf093a3e5957ee

Is there a List of All Novels somewhere?

NOVEL 1 = In the beginning, the Universe was...
NOVEL 2 = They were the best of times, they...
NOVEL 3 = Friends, Romans, Countrymen...
NOVEL 4 = The Rebel Alliance raced across...
...

Diagonal Novel = In were countrymen raced...

Virgil came up with

Unlisted Novel = Over where Cityfolk slept...

> Despite how many authors submit a novel, despite every poss. word
> sequence for all lengths being submitted, Virgil just goes through the
> stack of papers and writes something completely new!

[JESSE]
As long as only countably many authors submit novels, yes, he does.


WRONG!

Just by forcing the infinite word sequences to be segmented (into
sentences with a full stop), it is obvious that Virgil can never write
1 new sentence that has not already been submitted.

#despite every poss. word sequence for all lengths being submitted

THIS PROVES ANTIDIAGONALISTION CANNOT PRODUCE A NEW SEQUENCE OF
DIGITS.


Herc

[MoeBlee]

On Oct 8, 7:18 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:

> Diagonal Novel

Who knew that all this time the Great American Novel was on the
diagonal! No wonder all the typing monkeys missed it!

MoeBlee

[Graham Cooper]

f(1) = 0. 314 159 265 ...
f(2) = 0. 278 182 845 ...
f(3) = 0. 333 333 333 ...
f(4) = 0. 123445 54321 ...
f(5) = 0. 11 11 11 11 ...
...

The anti-diagonal never produces a unique packet of any size.

*************

> No wonder all the typing monkeys missed it!
>
> MoeBlee

that would explain a lot around here!

Herc

[Michael Stemper]

In article <48063903-4a94-4411...@ql4g2000pbc.googlegroups.com>, "Ross A. Finlayson" <ross.fi...@gmail.com> writes:

>On Oct 6, 7:42=A0am, Frederick Williams <freddywilli...@btinternet.com> wrote:
>> "Ross A. Finlayson" wrote:

>> > The naturals are a continuum for induction. The reals are a continuum
>> > for course. Obviously the naturals aren't the continuum in real
>> > space, the reals are the continuum. (And they don't need any more to
>> > be the continuum and any more would already be in it, elements of
>> > the.)
>>
>> A continuum is a compact connected T2 space.
>>
>> What topology are you putting on N to make it a continuum?

>Let's see, it's compact because there's a point at infinity,

There is? Which natural number would that be?

--
Michael F. Stemper
#include <Standard_Disclaimer>
Life's too important to take seriously.

[Michael Stemper]

In article <bbdecbb8-615e-4d60...@y8g2000yqy.googlegroups.com>, MoeBlee <mode...@gmail.com> writes:

>On Oct 8, 7:18=A0pm, Graham Cooper <grahamcoop...@gmail.com> wrote:

>> Diagonal Novel
>
>Who knew that all this time the Great American Novel was on the
>diagonal! No wonder all the typing monkeys missed it!

Great "American" novel? Wasn't that written by Rowling, and set on Diagon
Alley?

[Curlytop]

Graham Cooper set the following eddies spiralling through the space-time
continuum:

> (snip the bull)

>
> f(1) = 0. 314 159 265 ...
> f(2) = 0. 278 182 845 ...
> f(3) = 0. 333 333 333 ...
> f(4) = 0. 123445 54321 ...
> f(5) = 0. 11 11 11 11 ...
> ...

> (nip more bull)

Every supposed demonstration of Cantor diagonalisation invokes a
*disordered* subset of the supposedly greater-than-infinite set of all
reals, as here. If somebody would only put them into some kind of order,
the absurdity of Cantor's proposition will become immediately apparent to
all.
--
ξ: ) Proud to be curly

Interchange the alphabetic letter groups to reply

[Graham Cooper]

On Oct 10, 5:37 am, Curlytop <pvstownsend.zyx....@ntlworld.com> wrote:
> Graham Cooper set the following eddies spiralling through the space-time
> continuum:
>
> > (snip the bull)
>
> > f(1) = 0. 314 159 265 ...
> > f(2) = 0. 278 182 845 ...
> > f(3) = 0. 333 333 333 ...
> > f(4) = 0. 123445 54321 ...
> > f(5) = 0. 11 11 11 11 ...
> > ...
> > (nip more bull)
>
> Every supposed demonstration of Cantor diagonalisation invokes a
> *disordered* subset of the supposedly greater-than-infinite set of all
> reals, as here. If somebody would only put them into some kind of order,
> the absurdity of Cantor's proposition will become immediately apparent to
> all.
> --

Dth digit of the Nth real is UTM(N,D) (mod 10)


Yes but then you get a logical proof, people like their
*constructable* bigger infinities.

IF you used a FINITE ANTIDIAG() FUNCTION

e.g.
FUNCTION ANTIDIAG( DIAGONAL )
{
RETURN DIAGONAL + 0.111111111....
}

Then it would input a HYPERREAL DIAGONAL
(DIAGONAL doesn't appear on any computable row)

HYPERREAL <DIAG>
+ 0.1111111111... <ANTI>
------------------------
= HYPERREAL <MISSING>

i.e. the computable reals list is not missing any computable real.


ALTERNATIVELY

HYPERREAL <DIAG>
+ HYPERREAL <oo ANTI>
------------------------
= COMPUTABLE <MISSING>




So EITHER WAY
ANTIDIAG() is FINITE --> no missing computable real
ANTIDIAG() is INFINITE --> no computable missing real


Herc

[Michael Stemper]

In article <virgil-7A3612....@bignews.usenetmonster.com>, Virgil <vir...@ligriv.com> writes:
>In article <bb2e8c1c-3edc-4c36...@j2g2000pbg.googlegroups.com>, "Ross A. Finlayson" <ross.fi...@gmail.com> wrote:

>It would, so why won't Ross do it? Unwarranted arrogance!
>>
>> The open or finite ones in the finite topology, their finite or open
>> union is finite, open.
>
>By a "finite topology" on N you mean:
>1, one with a finite number of open sets, or
>2. one with every finite set being open?

If every finite set is open, wouldn't that lead to *all* sets being
open? The union of an arbitrarly collection of open sets is an open
set, isn't it?

[Virgil]

In article <k524j5$jpf$1...@dont-email.me>,

mste...@walkabout.empros.com (Michael Stemper) wrote:

> In article <virgil-7A3612....@bignews.usenetmonster.com>, Virgil
> <vir...@ligriv.com> writes:
> >In article
> ><bb2e8c1c-3edc-4c36...@j2g2000pbg.googlegroups.com>, "Ross A.
> >Finlayson" <ross.fi...@gmail.com> wrote:
>
> >It would, so why won't Ross do it? Unwarranted arrogance!
> >>
> >> The open or finite ones in the finite topology, their finite or open
> >> union is finite, open.
> >

> >By a "finite topology" on N you (Ross) mean:

> >1, one with a finite number of open sets, or
> >2. one with every finite set being open?
>
> If every finite set is open, wouldn't that lead to *all* sets being
> open? The union of an arbitrarly collection of open sets is an open
> set, isn't it?

Yes, but since Ross never expressed his choice, and neither choice would
have supported his wild claims, it hardly matters.
--

[Virgil]

In article <k51uee$1r6$3...@dont-email.me>,

Curlytop <pvstownse...@ntlworld.com> wrote:

> Graham Cooper set the following eddies spiralling through the space-time
> continuum:
>
> > (snip the bull)
> >
> > f(1) = 0. 314 159 265 ...
> > f(2) = 0. 278 182 845 ...
> > f(3) = 0. 333 333 333 ...
> > f(4) = 0. 123445 54321 ...
> > f(5) = 0. 11 11 11 11 ...
> > ...
> > (nip more bull)
>
> Every supposed demonstration of Cantor diagonalisation invokes a
> *disordered* subset of the supposedly greater-than-infinite set of all
> reals, as here. If somebody would only put them into some kind of order,
> the absurdity of Cantor's proposition will become immediately apparent to
> all.

The difficulty being that Cantor has shown that any and every attempt to
put them in serial order necessarily leaves some of them out.

His proof can, in fact, be extended to show that there are at least as
many left out as included.
--

[Virgil]

In article <k51kok$80o$1...@dont-email.me>,

mste...@walkabout.empros.com (Michael Stemper) wrote:

> In article
> <48063903-4a94-4411...@ql4g2000pbc.googlegroups.com>, "Ross A.
> Finlayson" <ross.fi...@gmail.com> writes:
> >On Oct 6, 7:42=A0am, Frederick Williams <freddywilli...@btinternet.com>
> >wrote:
> >> "Ross A. Finlayson" wrote:
>
> >> > The naturals are a continuum for induction. The reals are a continuum
> >> > for course. Obviously the naturals aren't the continuum in real
> >> > space, the reals are the continuum. (And they don't need any more to
> >> > be the continuum and any more would already be in it, elements of
> >> > the.)
> >>
> >> A continuum is a compact connected T2 space.
> >>
> >> What topology are you putting on N to make it a continuum?
>
> >Let's see, it's compact because there's a point at infinity,
>
> There is? Which natural number would that be?

And Finlayson flubs again!
--

[Graham Cooper]

> > Every supposed demonstration of Cantor diagonalisation invokes a
> > *disordered* subset of the supposedly greater-than-infinite set of all
> > reals, as here. If somebody would only put them into some kind of order,
> > the absurdity of Cantor's proposition will become immediately apparent to
> > all.
>
> The difficulty being that Cantor has shown that any and every attempt to
> put them in serial order necessarily leaves some of them out.

Dth digit of the Nth real is UTM(N,D) (mod 10)

>

> His proof can, in fact, be extended to show that there are at least as
> many left out as included.
>

They are COMPUTABLE!

The Hypothetical Anti-Diagonal is RANDOM!

There is no FINITE REPRESENTATION of an INFINITE RANDOM SEQUENCE.

There is NOTHING *REAL* MISSING!

**************

YOU CAN EXTEND your *PROOF* by testing the 2 POSSIBILITIES.

ANTIDIAG() FUNCTION IS FINITE
ANTIDIAG() IS INFINITE



Herc

[George Cornelius]

Jesse F. Hughes wrote:
> George Cornelius <corn...@eisner.decus.org> writes:
>> Ross A. Finlayson wrote:
>>> nondenumerability = uncountability
>>
>>> nonzero =/= zero
>>
>> Am I guilty of imprecision? No doubt. After all,
>> {1,2,3} could be said to be nondenumerable.
>
> Oh, don't you fret. Ross is a crank, plain and simple.

Thanks for the heads up. Wasn't too difficult to come
to that conclusion on my part as well, but I went ahead
and answered concerning the fact that I knew from the
outset that a certain issue in the wording of the problem
was present while not really being any serious error.

Amazing that I plus two of the other posters seem to have
worked out some good solutions to the the original problem -
collaborative mathematics via newsgroup - but there's been
an incredible amount of piling on by people like Ross who
are contributing lots of heat and very little light.

George Cornelius

[Ross A. Finlayson]

On Oct 9, 5:15 pm, Virgil <vir...@ligriv.com> wrote:
> In article <k51kok$80...@dont-email.me>,

>  mstem...@walkabout.empros.com (Michael Stemper) wrote:
>
>
>
>
>
> > In article

> > <48063903-4a94-4411-9aaf-a9a0ec68c...@ql4g2000pbc.googlegroups.com>, "Ross A.

> > Finlayson" <ross.finlay...@gmail.com> writes:
> > >On Oct 6, 7:42=A0am, Frederick Williams <freddywilli...@btinternet.com>
> > >wrote:
> > >> "Ross A. Finlayson" wrote:
>
> > >> > The naturals are a continuum for induction. The reals are a continuum
> > >> > for course. Obviously the naturals aren't the continuum in real
> > >> > space, the reals are the continuum. (And they don't need any more to
> > >> > be the continuum and any more would already be in it, elements of
> > >> > the.)
>
> > >> A continuum is a compact connected T2 space.
>
> > >> What topology are you putting on N to make it a continuum?
>
> > >Let's see, it's compact because there's a point at infinity,
>
> > There is? Which natural number would that be?
>
> And Finlayson flubs again!
> --

Great, leave you alone for a few days and come back to you cawing up a
storm and throwing shit all about. Not surprising. Get an original
thought. Because your fuddy-duddy bowtie and shortsleeves doesn't add
up to anything new.

Quit bullying me, I'm bigger than you.

It seems not to matter so much that you're a cretin lacking
imagination as willfully in denial of true features of the numbers,
that care not one iota your opinion of them.

It doesn't matter what I say those things are, they are as they are.
And, it doesn't matter what you say, they are as they are.


Regards,

Ross Finlayson

[Ross A. Finlayson]

On Oct 9, 9:52 am, mstem...@walkabout.empros.com (Michael Stemper)
wrote:

> In article <48063903-4a94-4411-9aaf-a9a0ec68c...@ql4g2000pbc.googlegroups.com>, "Ross A. Finlayson" <ross.finlay...@gmail.com> writes:
>
> >On Oct 6, 7:42=A0am, Frederick Williams <freddywilli...@btinternet.com> wrote:
> >> "Ross A. Finlayson" wrote:
> >> > The naturals are a continuum for induction. The reals are a continuum
> >> > for course. Obviously the naturals aren't the continuum in real
> >> > space, the reals are the continuum. (And they don't need any more to
> >> > be the continuum and any more would already be in it, elements of
> >> > the.)
>
> >> A continuum is a compact connected T2 space.
>
> >> What topology are you putting on N to make it a continuum?
> >Let's see, it's compact because there's a point at infinity,
>
> There is? Which natural number would that be?
>
> --
> Michael F. Stemper
> #include <Standard_Disclaimer>
> Life's too important to take seriously.
>
>

Yes, EF passes continued fractions, too.

Then about the naturals being complete, and compact, they go to
infinity. Yawn there's a bit more to it than that.

Then about defining a topology, or the nearest thing casually enough,
it is as above. That's just a casual observation compared to what
other features of the numbers are of interest, for the applied, in the
real.

Regards,

Ross Finlayson

[Ross A. Finlayson]

On Oct 10, 5:37 pm, George Cornelius <cornel...@eisner.decus.org>
wrote:
> Jesse F. Hughes wrote:

Yes, plain and simple, these numbers have true features you don't
know. And Goedel proves that to you, as you accept, not a hypocrite.

Higher order thinking skills: get some.

Uses in the applied for transfinite cardinals: find some. Make you
famous.

Get some.

Antidiagonal argument: doesn't work in base two. To get around that
you define the elements on the range of EF. There's quite a
background in the development, good luck with that.

Regards,

Ross Finlayson