Objection to Cantor's First Proof

[LudovicoVan]

[More a bunch of questions than any solid objection.]

<http://en.wikipedia.org/wiki/Cantor's_first_uncountability_proof>

<< Cantor now breaks the proof into two cases: Either the number of
intervals generated is finite or infinite. >>

Given the assumption of completeness, I cannot see how the number of
intervals can be finite, and not even how the limit could not be an improper
interval (i.e. isn't it always a_oo = b_oo?).

<< If the number of intervals is infinite, let a_oo = lim_{n->oo} a_n.
At this point, Cantor could finish his proof by noting that a_oo is not
contained in the given sequence since for every n, a_oo belongs to the
interior of [a_n, b_n] but x_n does not. >>

Let [ a_n, b_n ] be the sequence of real intervals:

[ a_0, b_0 ] := [ a, b ]
[ a_n, b_n ] c [ a_{n-1}, b_{n-1} ], n > 0

with:

a_n := x_{n'[2n-1]}
b_n := x_{n'[2n]}

where x_{n'[2n-1]} and x_{n'[2n]} are, non respectively, the (2n-1)-th and
(2n)-th entries picked from sequence x per the rules of the game.

Then, we have the property:

(1) E m : A n : n>m -> ~ ( x_n e [ a_n, b_n ] )

i.e. the property that "for every n, x_n does not belong to the interior of
[a_n, b_n]."

From (1) the conclusion allegedly follows.

OTOH, let's consider the limit interval:

[ a_oo, b_oo ] =
= lim_{n->oo} [ a_n, b_n ] =
= lim_{n->oo} [ x_{n'[2n-1]}, x_{n'[2n]} ]

Then, we also have the property:

(2.1) a_oo = lim_{n->oo} a_n = lim_{n->oo} x_{n'[2n-1]}
(2.2) b_oo = lim_{n->oo} b_n = lim_{n->oo} x_{n'[2n]}

The main objection is that the thesis does *not* follow from (1): I think
the conclusion amounts to the same kind of invalid reasoning found in the
standard solution to the balls and vase problem, i.e. "incorrect counting".
Moreover, from property (2) and (I suppose) completeness, I think we are in
fact showing that a_oo (or, b_oo) get picked up from sequence x. --
Roughly speaking, the proof seems to amount to a "trick with indexes".
Otherwise, could anyone formalize the last step of the proof, i.e. the
conclusion, to see which derivation is actually at play?

Thanks,

-LV

[LudovicoVan]

"LudovicoVan" <ju...@diegidio.name> wrote in message
news:junlnt$ugh$1...@speranza.aioe.org...

> Roughly speaking, the proof seems to amount to a "trick with indexes".

"A n, ~ ( x_n e [ a_n, b_n ])" does not contradict the fact that "A n, a_n
(or, b_n) is an element of sequence x".

-LV

[LudovicoVan]

"LudovicoVan" <ju...@diegidio.name> wrote in message
news:junlnt$ugh$1...@speranza.aioe.org...

> Then, we have the property:
>
> (1) E m : A n : n>m -> ~ ( x_n e [ a_n, b_n ] )
>
> i.e. the property that "for every n, x_n does not belong to the interior
> of [a_n, b_n]."

Should be:

(1) E m : A n : n>m -> ~ ( x_n e ] a_n, b_n [ )

-LV

[LudovicoVan]

"LudovicoVan" <ju...@diegidio.name> wrote in message

news:junmgv$vrl$1...@speranza.aioe.org...

With the correction "A n, ~ ( x_n e ] a_n, b_n [)" I'll have to rethink
this...

-LV

[Virgil]

In article <junnm3$28t$1...@speranza.aioe.org>,

*********************************************
A PROOF OF THE UNCOUNTABILITY OF THE REALS
(A variation on Cantor's FIRST proof)

ASSUMPTIONS:

(1) the intersection of a strictly nested sequence of closed real
intervals (the endpoints of each interval being interior points of the
previous interval) is not empty.

(2) A strictly increasing sequence of naturals does not have a natural
as its limit,

(3a) A strictly increasing but bounded sequence of reals has a real
number as a limit, its least upper bound, different from every member of
the sequence.

(3b) A strictly decreasing but bounded sequence of reals has a real
number as a limit, its greatest lower bound, different from every member
of the sequence.

Proof:

If the reals are countable then we may assume each real can be and has
been paired with a natural so that different reals are paired with
different naturals with none of either left out.

Assuming this has been done, take the two reals corresponding to the
lowest naturals as endpoints of a real interval.

It is clear that all the interior points of this real interval must be
paired with naturals larger that paired with its endpoints.

Now take the two reals with the lowest naturals INTERIOR to the previous
interval to be the endpoints of a subinterval of that interval.

It is clear that the interior points of this real interval must be
paired with naturals larger than the naturals paired with its endpoints.

By repeating this process one generates a decreasing, but never empty,
sequence of closed real intervals each of which contains only points
with higher attached naturals than its endpoints have.

The intersection of such a nested sequence of closed intervals is not
empty, but the natural associated with any of its members is necessarily
larger than all of the infinitely many natural numbers associated with
those infinitely many endpoints.

But there is no natural number larger than infinitely many different
natural numbers.

This is a contradiction which can only have been caused by our original
assumption that the reals were countable, so it proves they are not
countable.

QED!

NOTE: While I have never seen this particular proof in the literature of
countability,
it is so obvious that I doubt that it is original with me.
--

[David C. Ullrich]

On Wed, 25 Jul 2012 03:30:20 +0100, "LudovicoVan"
<ju...@diegidio.name> wrote:

>[More a bunch of questions than any solid objection.]
>
><http://en.wikipedia.org/wiki/Cantor's_first_uncountability_proof>

You might reproduce this proof here, or at least state what's being
proved.

> << Cantor now breaks the proof into two cases: Either the number of
>intervals generated is finite or infinite. >>
>
>Given the assumption of completeness, I cannot see how the number of
>intervals can be finite,

What's being proved is this: Given any sequence x_1,... and any
interval [a,b], there exists an element of [a,b] which is not one
of the x_j.

The number of intervals certainly can be finite. At each stage of
the proof we choose the first two x_j's which are elements of
[a_n,b_n] and use them for the endpoints of the next interval.
But there's no reason that [a_n,b_n] has to contain any x_j
at all!

See, it's important to keep straight what's being proved.
The proof does _not_ begin by assuming that x_1,...
is an enumeration of the reals! If it _did_ begin with
that assumption then yes, the number of intervals
would be infinite.

>and not even how the limit could not be an improper
>interval (i.e. isn't it always a_oo = b_oo?).

Certainly not.

You don't give any explanation for what you think is wrong
other than saying "it does not follow" and that it seems like
a trick that reminds you of something else.

_Since_ you're expressing confusing here instead of just
asserting that mathematicians are all wrong, I'll write out
the key points in somewhat more detail than in that
wikipedia article.

Don't tell me what things remind you of and how things
seem. DON'T jump to the end and talk about the
conclusion. Tell me the FIRST step below that you
don't follow:

Asssuming there are infinitely many intervals.
We let a_1 and b_1 be the two first x_j that lie
in the interior of [a,b]. Meaning that a < a_1 < b_1 < b.

To simplify things, let's assume that a_1 = x_25 and b_1 = x_23.

Now this follows:

(*) If x_j lies in the interior of [a_1,b_1] then j > 25.

Why? Because if x_j lies in the interior of [a_1,b_1]
and j <= 25 then the choice of a_1, b_1 would have been
different! We cant't have x_11 in [a_1,b_1], becuause
x_23 and x_25 were the first two x_j lying in [a,b].

Now we choose a_2 and b_2, etc. Maybe a_2 = x_300
and b_2 = x_4999. Now for the same reason as before,
it follows that

(*) If x_j lies in the interior of [a_2,b_2] then j > 4999.

Etc.

Now take that limit. The number a_oo _does_ lie in
[a_1.b_1]. So _if_ a_00 = x_j then j > 25.
And a_00 lies in the interior of [a_2,b_2].
So if a_00 = x_j then j > 4999.

Etc. a_oo cannot be x_j for any j, because if
a_oo = x_j then j > 25 by (*) and also j > 4999
by (**), and also j > 10000 by (***), the first
step I left out. The number j would have to be
larger than infinitely many natral numbers, and
there is no such natural number j.

>
>Thanks,
>
>-LV
>

[Shmuel Metz]

In <junlnt$ugh$1...@speranza.aioe.org>, on 07/25/2012

at 03:30 AM, "LudovicoVan" <ju...@diegidio.name> said:

>Given the assumption of completeness,

There is no such assumption; it's part of the definition of R.

>I cannot see how the number of intervals can be finite,

It can't, but that has to be proven, which Cantor did. It also can't
be infinite, which he also proved.

>I think

No.

--
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

[LudovicoVan]

"David C. Ullrich" <ull...@math.okstate.edu> wrote in message
news:cbuv08t9odnda4gva...@4ax.com...

> On Wed, 25 Jul 2012 03:30:20 +0100, "LudovicoVan"
> <ju...@diegidio.name> wrote:

<snipped>

> > <http://en.wikipedia.org/wiki/Cantor's_first_uncountability_proof>

> > << Cantor now breaks the proof into two cases: Either the
> > number of intervals generated is finite or infinite. >>

> > Given the assumption of completeness, I cannot see how the number
> > of intervals can be finite,
>
> What's being proved is this: Given any sequence x_1,...

Given any sequence (x_n) *of real numbers*. (I would not see what else,
anyway so it is stated by Wikipedia, too.)

> and any interval [a,b],

Again, any interval [ a, b ] of real numbers.

> there exists an element of [a,b] which is not one of the x_j.
>
> The number of intervals certainly can be finite. At each stage of
> the proof we choose the first two x_j's which are elements of
> [a_n,b_n] and use them for the endpoints of the next interval.
> But there's no reason that [a_n,b_n] has to contain any x_j
> at all!

I thought that must be the case because we are dealing with real numbers: in
fact the completeness property is assumed in the proof, for the limits to
exist.

> The proof does _not_ begin by assuming that x_1,...
> is an enumeration of the reals!

Unless I am missing something, it definitely does.

> If it _did_ begin with
> that assumption then yes, the number of intervals
> would be infinite.

OK.

> > and not even how the limit could not be an improper
> > interval (i.e. isn't it always a_oo = b_oo?).
>
> Certainly not.

Hence, unless I am missing something, certainly so.

> > << If the number of intervals is infinite, let a_oo = lim_{n->oo}
> > a_n.
> > At this point, Cantor could finish his proof by noting that a_oo is not
> > contained in the given sequence since for every n, a_oo belongs to the
> > interior of [a_n, b_n] but x_n does not. >>

> > Then, we have the property:
> >
> > (1) E m : A n : n>m -> ~ ( x_n e [ a_n, b_n ] )
> >
> > i.e. the property that "for every n, x_n does not belong to the interior
> > of
> > [a_n, b_n]."

Later corrected to:

(1) E m : A n : n>m -> ~ ( x_n e ] a_n, b_n [ )

> > The main objection is that the thesis does *not* follow from (1): I
> > think
> > the conclusion amounts to the same kind of invalid reasoning found in
> > the
> > standard solution to the balls and vase problem, i.e. "incorrect
> > counting".
> > Moreover, from property (2) and (I suppose) completeness, I think we are
> > in
> > fact showing that a_oo (or, b_oo) get picked up from sequence x. --
> > Roughly speaking, the proof seems to amount to a "trick with indexes".
> > Otherwise, could anyone formalize the last step of the proof, i.e. the
> > conclusion, to see which derivation is actually at play?
>
> You don't give any explanation for what you think is wrong

Actually, you do not formalize the conclusion, while I do have posted a more
precise objection, which was this:

"A n, ~ ( x_n e [ a_n, b_n ])"

does *not* contradict:

"A n, a_n (or, b_n) is an element of sequence x".

That I think exposes the alleged "trick" (a paralogism quite similar to the
"every
ball is eventually removed, hence the vase must end up empty" of the
Ross-Littlewood paradox).

But that objection needs be corrected (as mentioned in another post, too),
because the actual property from which the conclusion allegedly follows is:

(1) E m : A n : ( n > m ) -> ~ ( x_n e ] a_n, b_n [ )

So, I will have to try and work out a correct formalization of the objection
(which, as above, is going to be of the same kind of the "for every ball
removed many more get in, so that the vase is never empty").

> Asssuming there are infinitely many intervals.

> [...]

>
> (*) If x_j lies in the interior of [a_2,b_2] then j > 4999.
>
> Etc.
>
> Now take that limit. The number a_oo _does_ lie in

> [a_1.b_1]. [...]

Yes: as I get it, by completeness, the limit exists and, by construction, it
is internal to all intervals at finite steps (i.e. A n : a_oo e ] a_n, b_n
[), this being a sequence of strictly nested intervals.

> So _if_ a_00 = x_j then j > 25.

> Etc. a_oo cannot be x_j for any j, because

No, I think a_oo cannot be equal to any x_n for n in N because, by
definition, a_oo := lim_{n->oo} a_n, then by the properties of completeness
and of the construction just mentioned above.

> if
> a_oo = x_j then j > 25 by (*) and also j > 4999
> by (**), and also j > 10000 by (***), the first
> step I left out. The number j would have to be
> larger than infinitely many natral numbers, and
> there is no such natural number j.

I'd contend your "induction" is as invalid as the paralogism indicated
above. Namely, you are saying that we "run out of indexes", or something to
that effect. Anyway, that is not a formalization of the conclusion, it is
only a reworded informal presentation of it: I'd need it formally.

-LV

[LudovicoVan]

"Virgil" <vir...@ligriv.com> wrote in message
news:virgil-95F48D....@bignews.usenetmonster.com...

> A PROOF OF THE UNCOUNTABILITY OF THE REALS
> (A variation on Cantor's FIRST proof)

> The intersection of such a nested sequence of closed intervals is not
> empty, but the natural associated with any of its members is necessarily
> larger than all of the infinitely many natural numbers associated with
> those infinitely many endpoints.

Wrong: by completeness (which you do assume), it must be a_oo == b_oo, that
is the limit interval is a singleton (an improper interval). Informally, we
do not "run out of indexes": that is just invalid reasoning.

-LV

[Gus Gassmann]

On Jul 26, 5:37 am, "LudovicoVan" <ju...@diegidio.name> wrote:
> "David C. Ullrich" <ullr...@math.okstate.edu> wrote in messagenews:cbuv08t9odnda4gva...@4ax.com...> On Wed, 25 Jul 2012 03:30:20 +0100, "LudovicoVan"

> > <ju...@diegidio.name> wrote:
>
> <snipped>
>
> > > <http://en.wikipedia.org/wiki/Cantor's_first_uncountability_proof>
> > >     << Cantor now breaks the proof into two cases:  Either the
> > > number of intervals generated is finite or infinite. >>
> > > Given the assumption of completeness, I cannot see how the number
> > > of intervals can be finite,
>
> > What's being proved is this: Given any sequence x_1,...
>
> Given any sequence (x_n) *of real numbers*.  (I would not see what else,
> anyway so it is stated by Wikipedia, too.)
>
> > and any interval [a,b],
>
> Again, any interval [ a, b ] of real numbers.
>
> > there exists an element of [a,b] which is not one of the x_j.
>
> > The number of intervals certainly can be finite. At each stage of
> > the proof we choose the first two x_j's which are elements of
> > [a_n,b_n] and use them for the endpoints of the next interval.
> > But there's no reason that [a_n,b_n] has to contain any x_j
> > at all!
>
> I thought that must be the case because we are dealing with real numbers: in
> fact the completeness property is assumed in the proof, for the limits to
> exist.
>
> > The proof does _not_ begin by assuming that x_1,...
> > is an enumeration of the reals!
>
> Unless I am missing something, it definitely does.

The theorem starts: "Given *any* sequence of real numbers". There is
absolutely no requirement that the sequence enumerate all real numbers
(or try to do so). For instance, pi, 3pi, 2pi, 5pi, 4pi, etc. is a
sequence of real numbers. The theorem has to work for that sequence.

The theorem continues: "Given *any* interval [a, b] of real numbers".
For instance [a, b] = [0, 10] is such an interval. The theorem has to
work for this interval.

The theorem concludes. Given these things, there exists a number in
[a, b] that is *not* a member of the sequence.

So go through the proof: Find the first two numbers in the sequence
that belong to the interior of [a, b]. There don't have to be *any* of
those, but in this case there are: pi and 3pi. Designate the smaller
of these a1 (= pi) and the large b1 (= 3pi). Now find the first two
numbers that lie in the interior of the interval (pi, 3pi). Turns out,
there is only one of those, namely 2pi. So in this case there are only
finitely many intervals, but you can still find a point in (pi, 3pi)
that is different from 2pi. (Right?)

This is of course only an illustration. The proof has to work for
*all* sequences and *all* intervals, and sometimes you will get a
finite number of intervals in the proof, sometimes you will get an
infinite one. That is why Cantor considers two cases.

[LudovicoVan]

"Gus Gassmann" <horand....@gmail.com> wrote in message
news:4990bf35-406e-4835...@w24g2000vby.googlegroups.com...

> The theorem starts: "Given *any* sequence of real numbers". There is
> absolutely no requirement that the sequence enumerate all real numbers
> (or try to do so).

You just miss the point: David said the sequence needn't be a sequence of
real numbers. Nobody has said the sequence must be complete.

> This is of course only an illustration.

Indeed, and it adds nothing to what has already been said: same argument,
same objection. Rather, given the completeness of the reals, isn't the
limit interval necessarily improper?

Indeed, have you bothered to read what was written at all?

-LV

[LudovicoVan]

"Gus Gassmann" <horand....@gmail.com> wrote in message
news:4990bf35-406e-4835...@w24g2000vby.googlegroups.com...

> This is of course only an illustration. The proof has to work for
> *all* sequences and *all* intervals, and sometimes you will get a
> finite number of intervals in the proof, sometimes you will get an
> infinite one. That is why Cantor considers two cases.

OK, maybe I get your point. I will have to think through this though, and
do not have the time right now...

Thanks,

-LV

[Gus Gassmann]

On Jul 26, 11:12 am, "LudovicoVan" <ju...@diegidio.name> wrote:
> "Gus Gassmann" <horand.gassm...@gmail.com> wrote in message

Asshole! I will restore the context of what you so conveniently
snipped:

On Jul 26, 5:37 am, "LudovicoVan" <ju...@diegidio.name> wrote:

> "David C. Ullrich" <ullr...@math.okstate.edu> wrote in messagenews:cbuv08t9odnda4gva...@4ax.com...> "

> > The proof does _not_ begin by assuming that x_1,...
> > is an enumeration of the reals!

> Unless I am missing something, it definitely does.

It is obvious that you *are* missing something, and this is what I
commented on. The wikipedia proof, which you yourself linked to, does
not say anything about an enumeration of reals. Maybe you should read
what you quote, and then read what people respond. If that is within
your grasp, of course.

[LudovicoVan]

"Gus Gassmann" <horand....@gmail.com> wrote in message

news:56c28ea4-8e38-432b...@5g2000vbf.googlegroups.com...

> On Jul 26, 11:12 am, "LudovicoVan" <ju...@diegidio.name> wrote:

>> Indeed, have you bothered to read what was written at all?
>
> Asshole!

Why get excited?

> I will restore the context of what you so conveniently
> snipped:
>
> On Jul 26, 5:37 am, "LudovicoVan" <ju...@diegidio.name> wrote:
>
>> "David C. Ullrich" <ullr...@math.okstate.edu> wrote in
>> messagenews:cbuv08t9odnda4gva...@4ax.com...> "
>
>> > The proof does _not_ begin by assuming that x_1,...
>> > is an enumeration of the reals!
>
>> Unless I am missing something, it definitely does.
>
> It is obvious that you *are* missing something, and this is what I
> commented on. The wikipedia proof, which you yourself linked to, does
> not say anything about an enumeration of reals. Maybe you should read
> what you quote, and then read what people respond. If that is within
> your grasp, of course.

So, as I gather it, "enumeration of reals" to you is a "complete list of
reals". Are you sure? Besides, I thought David was saying something else.
But, as anticipated, I'll have to come back to this discussion later: I am
in a rush now...

-LV

[David C. Ullrich]

On Thu, 26 Jul 2012 15:12:31 +0100, "LudovicoVan"
<ju...@diegidio.name> wrote:

>"Gus Gassmann" <horand....@gmail.com> wrote in message
>news:4990bf35-406e-4835...@w24g2000vby.googlegroups.com...
>
>> The theorem starts: "Given *any* sequence of real numbers". There is
>> absolutely no requirement that the sequence enumerate all real numbers
>> (or try to do so).
>
>You just miss the point: David said the sequence needn't be a sequence of
>real numbers.

No, I didn't say that. I said this: "The proof does _not_ begin by

assuming that x_1,...
is an enumeration of the reals! "

I did not say this: "The proof does _not_ begin by assuming that
x_1,...
is an enumeration of reals! "


Those two sentences mean very different things.

>Nobody has said the sequence must be complete.

Assumming that "complete" means that it contains every
real: No, nobody said that. That's exactly why there
could be only finitely many intervals in the given proof.

[Virgil]

In article <juqvng$euj$1...@speranza.aioe.org>,

"LudovicoVan" <ju...@diegidio.name> wrote:

> "Virgil" <vir...@ligriv.com> wrote in message
> news:virgil-95F48D....@bignews.usenetmonster.com...
>
> > A PROOF OF THE UNCOUNTABILITY OF THE REALS
> > (A variation on Cantor's FIRST proof)
>
> > The intersection of such a nested sequence of closed intervals is not
> > empty, but the natural associated with any of its members is necessarily
> > larger than all of the infinitely many natural numbers associated with
> > those infinitely many endpoints.
>
> Wrong: by completeness (which you do assume), it must be a_oo == b_oo,

Even if that were true, it is clear that neither a_oo nor b_oo can be a
members of the original sequence from which the a_n and b_n were
derived. Thus EVERY sequence of reals STILL omits at least one real, and
Cared(N) < Card(R) as claimed.

Here is my proof again, so anyone can check that LV is wrong:

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

A PROOF OF THE UNCOUNTABILITY OF THE REALS
(A variation on Cantor's FIRST proof)

The intersection of such a nested sequence of closed intervals is not
empty, but the natural associated with any of its members is necessarily
larger than all of the infinitely many natural numbers associated with
those infinitely many endpoints.

[Shmuel Metz]

In <juqvko$epq$1...@speranza.aioe.org>, on 07/26/2012

at 09:37 AM, "LudovicoVan" <ju...@diegidio.name> said:

>I'd contend your "induction" is as invalid as the paralogism
>indicated above. Namely, you are saying that we "run out of
>indexes", or something to that effect.

Liar. He is saying no such thing.

[LudovicoVan]

"David C. Ullrich" <ull...@math.okstate.edu> wrote in message

news:7gn218h33oo3m993t...@4ax.com...

> On Thu, 26 Jul 2012 15:12:31 +0100, "LudovicoVan"
> <ju...@diegidio.name> wrote:
>>"Gus Gassmann" <horand....@gmail.com> wrote in message
>>news:4990bf35-406e-4835...@w24g2000vby.googlegroups.com...
>>
>>> The theorem starts: "Given *any* sequence of real numbers". There is
>>> absolutely no requirement that the sequence enumerate all real numbers
>>> (or try to do so).
>>
>>You just miss the point: David said the sequence needn't be a sequence of
>>real numbers.
>
> No, I didn't say that. I said this: "The proof does _not_ begin by
> assuming that x_1,...
> is an enumeration of the reals! "
>
> I did not say this: "The proof does _not_ begin by assuming that
> x_1,...
> is an enumeration of reals! "
>
> Those two sentences mean very different things.

Do they?? Anyway, my mistake for not knowing that "enumeration" is *not*
equivalent to "sequence". My apologies for the confusion.

-LV

[LudovicoVan]

"Virgil" <vir...@ligriv.com> wrote in message

news:virgil-2AB883....@bignews.usenetmonster.com...

> In article <juqvng$euj$1...@speranza.aioe.org>,
> "LudovicoVan" <ju...@diegidio.name> wrote:
>> "Virgil" <vir...@ligriv.com> wrote in message
>> news:virgil-95F48D....@bignews.usenetmonster.com...
>>
>> > A PROOF OF THE UNCOUNTABILITY OF THE REALS
>> > (A variation on Cantor's FIRST proof)
>>
>> > The intersection of such a nested sequence of closed intervals is not
>> > empty, but the natural associated with any of its members is
>> > necessarily
>> > larger than all of the infinitely many natural numbers associated with
>> > those infinitely many endpoints.
>>
>> Wrong: by completeness (which you do assume), it must be a_oo == b_oo,
>
> Even if that were true,

That must be the case: by your assumption, per absurdo, that we have a
systematic enumeration of all reals (maybe, together with completeness), we
are guaranteed (I think) that we will keep finding end-points for further
nested intervals, ad libitum; on the other hand, again by completeness, the
limit exists, must be a real, and, since the two end-points bound each
other, the limit interval must be improper.

> it is clear that neither a_oo nor b_oo can be a
> members of the original sequence from which the a_n and b_n were
> derived.

I'd agree. One way to see this could be: A n : a_n =/= b_n.

> Thus EVERY sequence of reals STILL omits at least one real, and
> Cared(N) < Card(R) as claimed.

No, that is you who still cannot count (with limits): the omega-th
end-point, a_oo, would be an omega-th entry of the sequence, so nothing is
missed. Informally (as I won't check the formal details now):
card(N*)==card(R*).

> Here is my proof again, so anyone can check that LV is wrong:

LV is ever wrong, but your proof is wrong too.

-LV

[LudovicoVan]

"Gus Gassmann" <horand....@gmail.com> wrote in message

news:56c28ea4-8e38-432b...@5g2000vbf.googlegroups.com...

>> "David C. Ullrich" <ullr...@math.okstate.edu> wrote in
>> messagenews:cbuv08t9odnda4gva...@4ax.com...> "
>
>> > The proof does _not_ begin by assuming that x_1,...
>> > is an enumeration of the reals!
>
>> Unless I am missing something, it definitely does.
>
> It is obvious that you *are* missing something, and this is what I
> commented on. The wikipedia proof, which you yourself linked to, does
> not say anything about an enumeration of reals. Maybe you should read
> what you quote, and then read what people respond. If that is within
> your grasp, of course.

It is within my grasp to note that David's words around the "it's important
to keep straight what's being proved" and your subsequent reply have
unnecessarily introduced the term "enumeration" in this discussion to begin
with. Still thanks, I have got that straight now, plus I have had enough
corrections to already warrant a rewrite of my counter-argument.

-LV

[Shmuel Metz]

In <jurj8k$ugd$1...@speranza.aioe.org>, on 07/26/2012

at 03:12 PM, "LudovicoVan" <ju...@diegidio.name> said:

>You just miss the point: David said the sequence needn't be a
>sequence of real numbers.

Liar.

[Shmuel Metz]

In <7gn218h33oo3m993t...@4ax.com>, on 07/26/2012

at 10:13 AM, David C. Ullrich <ull...@math.okstate.edu> said:

>Assumming that "complete" means that it contains every
>real:

I read "Given the assumption of completeness," as referring to the
theorem that R is complete, but that isn't an assumption either.

[gus gassmann]

Well, alright. Progress has been made. I will grant you that David C.
Ullrich was the first to use the term "enumeration" in this thread,
although he did preface that by a hard-to-miss "_not_". But let's wait
for your rewrite.

[Virgil]

In article <jusne1$k06$2...@speranza.aioe.org>,

There is nothing in my proof that would prevent the successive intervals
from being of the form I_n = [ -1 - 1/n, 1 + 1/n ], thus having limit
[-1, 1], and thus all points of that limit interval un-ennumerated by
the original sequence.

What My proof demonstrates is that, given any sequence of reals, there
is a real which is not a member of that sequence.
--

[Frederick Williams]

LudovicoVan wrote:
>
> "David C. Ullrich" <ull...@math.okstate.edu> wrote in message
> news:7gn218h33oo3m993t...@4ax.com...
> > On Thu, 26 Jul 2012 15:12:31 +0100, "LudovicoVan"
> > <ju...@diegidio.name> wrote:
> >>"Gus Gassmann" <horand....@gmail.com> wrote in message
> >>news:4990bf35-406e-4835...@w24g2000vby.googlegroups.com...
> >>
> >>> The theorem starts: "Given *any* sequence of real numbers". There is
> >>> absolutely no requirement that the sequence enumerate all real numbers
> >>> (or try to do so).
> >>
> >>You just miss the point: David said the sequence needn't be a sequence of
> >>real numbers.
> >
> > No, I didn't say that. I said this: "The proof does _not_ begin by
> > assuming that x_1,...
> > is an enumeration of the reals! "
> >
> > I did not say this: "The proof does _not_ begin by assuming that
> > x_1,...
> > is an enumeration of reals! "
> >
> > Those two sentences mean very different things.
>
> Do they??

Obviously. Look, this: 1, 2, 3, ... is an enumeration of reals. It is
not an enumeration of THE reals. Why not? Because the real 0.5 isn't
included.

> Anyway, my mistake for not knowing that "enumeration" is *not*
> equivalent to "sequence". My apologies for the confusion.

--
The animated figures stand
Adorning every public street
And seem to breathe in stone, or
Move their marble feet.

[Jesse F. Hughes]

Frederick Williams <freddyw...@btinternet.com> writes:

> LudovicoVan wrote:
>>
>> "David C. Ullrich" <ull...@math.okstate.edu> wrote in message
>> news:7gn218h33oo3m993t...@4ax.com...
>> > On Thu, 26 Jul 2012 15:12:31 +0100, "LudovicoVan"
>> > <ju...@diegidio.name> wrote:
>> >>"Gus Gassmann" <horand....@gmail.com> wrote in message
>> >>news:4990bf35-406e-4835...@w24g2000vby.googlegroups.com...
>> >>
>> >>> The theorem starts: "Given *any* sequence of real numbers". There is
>> >>> absolutely no requirement that the sequence enumerate all real numbers
>> >>> (or try to do so).
>> >>
>> >>You just miss the point: David said the sequence needn't be a sequence of
>> >>real numbers.
>> >
>> > No, I didn't say that. I said this: "The proof does _not_ begin by
>> > assuming that x_1,...
>> > is an enumeration of the reals! "
>> >
>> > I did not say this: "The proof does _not_ begin by assuming that
>> > x_1,...
>> > is an enumeration of reals! "
>> >
>> > Those two sentences mean very different things.
>>
>> Do they??
>
> Obviously. Look, this: 1, 2, 3, ... is an enumeration of reals. It is
> not an enumeration of THE reals. Why not? Because the real 0.5 isn't
> included.

Read David's post more carefully. In the second quote, he meant to
replace "enumeration" with "sequence", but he didn't do so.


--
"I deal with reality. It's a brutal reality. But it's the only one
we've got. And people like me, do what it takes. I'm part of a long
line of discoverers. So I do what it takes."
-- James S. Harris channels George W. Bush

[gus gassmann]

On 27/07/2012 9:37 AM, Frederick Williams wrote:

> Obviously. Look, this: 1, 2, 3, ... is an enumeration of reals. It is
> not an enumeration of THE reals. Why not? Because the real 0.5 isn't
> included.

I do have some trouble with that. If not technically wrong, at the very
least it is highly confusing. Here is the definition given by Wikipedia:
"An enumeration of a collection of items is a complete, ordered listing
of all of the items in that collection." In other words, an enumeration
does not exist without mention of a set that is being enumerated. "An
enumeration of reals" suggests very strongly to me that the set that is
intended to be enumerated is the set of real numbers (regardless of
whether that is possible or not), and not some unspecified subset of the
reals. Am I alone in thinking that?

[Frederick Williams]

"Jesse F. Hughes" wrote:

>
> Read David's post more carefully. In the second quote, he meant to
> replace "enumeration" with "sequence", but he didn't do so.

Has he been taken out and shot?

[Frederick Williams]

Probably not. I asked Jesse Hughes if David Ullrich had been taken out
and shot, clearly I'm the one who should be executed. (But not yet. I
mean there's no hurry.)

[dilettante]

"Jesse F. Hughes" <je...@phiwumbda.org> wrote in message
news:87obn1f...@phiwumbda.org...

David will perhaps resolve the issue for us, but I don't think he meant to
replace "enumeration" with "sequence". I think the distinction he was making
was exactly that indicated by Frederick: "enumeration of reals" vs.
"enumeration of the reals", the second phrase implying inclusiveness of the
whole set and the first not.

[dilettante]

"Frederick Williams" <freddyw...@btinternet.com> wrote in message
news:5012A892...@btinternet.com...

I think you plead guilty prematurely. To me, "enumeration of reals"
naturally means "enumeration of some set whose members are real numbers",
while "enumeration of the reals" clearly means "enumeration of the whole set
of real numbers". I suspect that this is the distinction David meant to make
as well, but only he can clear that up for us. In any case, were I your
attorney, I would advise you to refrain from further confessions, and to
take some more mescaline.

[David C. Ullrich]

Yes. Another hint that thats the distinction I was getting at is given
by my post, where I specified that I meant A and not B, where A
and B differ only by the includion of the word "the".

[Aatu Koskensilta]

David C. Ullrich <ull...@math.okstate.edu> writes:

> Yes. Another hint that thats the distinction I was getting at is given
> by my post, where I specified that I meant A and not B, where A
> and B differ only by the includion of the word "the".

This sort of thing can very easily throw off a non-native speaker. For
instance, my only excuse for once saying "definition of an integer" when
I meant "definition of integer" is that there are no articles in
Finnish: for "definition of an integer" we say "kokonaisluvun
m��ritelm�" and "for definition of integer" we say "kokonaisluvun
m��ritelm�", the meaning usually clear from context. It was an
embarrassing and shameful episode in my life, an unforgivable youthful
discretion, and we will never again discuss it.

--
Aatu Koskensilta (aatu.kos...@uta.fi)

"Wovon man nicht sprechen kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus

[David C. Ullrich]

On Fri, 27 Jul 2012 22:18:35 +0300, Aatu Koskensilta
<aatu.kos...@uta.fi> wrote:

>David C. Ullrich <ull...@math.okstate.edu> writes:
>
>> Yes. Another hint that thats the distinction I was getting at is given
>> by my post, where I specified that I meant A and not B, where A
>> and B differ only by the includion of the word "the".
>
> This sort of thing can very easily throw off a non-native speaker. For
>instance, my only excuse for once saying "definition of an integer" when
>I meant "definition of integer" is that there are no articles in
>Finnish: for "definition of an integer" we say "kokonaisluvun
>m��ritelm�" and "for definition of integer" we say "kokonaisluvun
>m��ritelm�", the meaning usually clear from context. It was an
>embarrassing and shameful episode in my life, an unforgivable youthful
>discretion, and we will never again discuss it.

I won't mention it, then. But the prpblem is not just with
non-native speakers. One of my all-time least favorite comps
questions was "Define a Banach space"...

[David Bernier]

(Just kidding)

Have you heard of Contractish, Licensish or TOSish?

I mean by those non-words the form of very precise English often
used in contracts, licenses or Terms of Service agreements/offers ...

David Bernier

[LudovicoVan]

"Aatu Koskensilta" <aatu.kos...@uta.fi> wrote in message
news:87394d2...@uta.fi...

> David C. Ullrich <ull...@math.okstate.edu> writes:
>
>> Yes. Another hint that thats the distinction I was getting at is given
>> by my post, where I specified that I meant A and not B, where A
>> and B differ only by the includion of the word "the".
>
> This sort of thing can very easily throw off a non-native speaker.

Above all when it's totally preposterous. In fact, that bit was clear: 60%+
plus of English comes from Latin. I mostly miss idioms and the phonetics is
utterly different.

-LV

[LudovicoVan]

"Virgil" <vir...@ligriv.com> wrote in message

news:virgil-4F3973....@bignews.usenetmonster.com...

> There is nothing in my proof that would prevent the successive intervals
> from being of the form I_n = [ -1 - 1/n, 1 + 1/n ], thus having limit
> [-1, 1], and thus all points of that limit interval un-ennumerated by
> the original sequence.

Except that you have assumed, per absurdo, that the reals are countable. In
fact, to show that it must be a_oo==b_oo, that and denseness is enough.

Proof (attempted): Assume per absurdo that there is a last interval [a_m,
b_m], with:

a_m = x_{n'[2m-1]}
b_m = x_{n'[2m]}

That implies:

A k : ( k > max(n'[2m-1], n'[2m]) ) -> ~ ( x_k e [a_m, b_m] )

But, by denseness, there are infinitely many reals in [a_m, b_m] and, since
by assumption we are enumerating all reals, all the (infinitely many) reals
in [a_m, b_m] must belong to the sequence (x_n). Now, if there is a last
interval, that would mean that the infinitely many reals in [a_m, b_m] must
have already been enumerated, namely that there must be infinitely many
naturals before min(n'[2m-1], n'[2m]), which is an absurdity. Hence, there
is no last interval.

> What My proof demonstrates is

Your proof is just wrong.

-LV

[Mike Terry]

"LudovicoVan" <ju...@diegidio.name> wrote in message
news:juvcl0$l0a$1...@speranza.aioe.org...

This is NOT an assumption. (Crikey, how many posts have there been already
just clarifying this very point?)

[LudovicoVan]

"Mike Terry" <news.dead.p...@darjeeling.plus.com> wrote in message
news:BoqdndgxE-k3po7N...@brightview.co.uk...

> "LudovicoVan" <ju...@diegidio.name> wrote in message
> news:juvcl0$l0a$1...@speranza.aioe.org...
>> "Virgil" <vir...@ligriv.com> wrote in message
>> news:virgil-4F3973....@bignews.usenetmonster.com...

>> But, by denseness, there are infinitely many reals in [a_m, b_m] and,
> since
>> by assumption we are enumerating all reals,
>
> This is NOT an assumption. (Crikey, how many posts have there been
> already
> just clarifying this very point?)

Then read the bloody thread before adding even more confusion!

-LV

[Jesse F. Hughes]

David C. Ullrich <ull...@math.okstate.edu> writes:

Ah. I looked at those two sentences a dozen times, and never noticed
that the second sentence had "the" in it.

My mistake.

--
Jesse F. Hughes
"Our enemies are innovative and resourceful, and so are we. They never
stop thinking about new ways to harm our country and our people, and
neither do we."-- George W. Bush

[Virgil]

In article <juvcl0$l0a$1...@speranza.aioe.org>,

Is proof by contradiction no longer a valid form of argument?
I.e., assuming something in order to DISprove it.
I had not heard that it is now illegal.

In any case, that assumption is unneccessary, as the argument works
equally well if one only assumes that the set of sequence values is
dense in R, which assumption is met by any sequence of all rationals.
--

[Virgil]

In article <juve6l$o3h$1...@speranza.aioe.org>,

The assumption that the sequence ennumerate all reals is unnecessary, it
is enough if one assumes the sequence is dense in R.
--

[Frederick Williams]

"David C. Ullrich" wrote:
>
> One of my all-time least favorite comps
> questions was "Define a Banach space"...

Comps?

[Mike Terry]

"LudovicoVan" <ju...@diegidio.name> wrote in message

news:juve6l$o3h$1...@speranza.aioe.org...

Ahem, you're right - I hadn't read the thread. I appologise!

I agree with you that for Virgil's proof we must have a_oo = b_oo. However,
Virgil's proof still works in this case, as:
1) a_oo is in the interior of each [a_n, b_n]
2) a_oo = x_k for some k.
3) By construction of [a_n, b_n], this would imply k > n, for all n.
4) Contradiction. (k is a natural number, which of course excludes omega)

Virgil notes that he has never seen his proof in the literature, but it
would seem to be a minor variation of Cantor's first proof as discussed in
the OP for the thread...

Mike.

[LudovicoVan]

"Mike Terry" <news.dead.p...@darjeeling.plus.com> wrote in message

news:QMmdnUjplsBrdI7N...@brightview.co.uk...

> "LudovicoVan" <ju...@diegidio.name> wrote in message
> news:juve6l$o3h$1...@speranza.aioe.org...
>> "Mike Terry" <news.dead.p...@darjeeling.plus.com> wrote in
> message
>> news:BoqdndgxE-k3po7N...@brightview.co.uk...
>> > "LudovicoVan" <ju...@diegidio.name> wrote in message
>> > news:juvcl0$l0a$1...@speranza.aioe.org...
>> >> "Virgil" <vir...@ligriv.com> wrote in message
>> >> news:virgil-4F3973....@bignews.usenetmonster.com...
>>
>> >> But, by denseness, there are infinitely many reals in [a_m, b_m] and,
>> > since
>> >> by assumption we are enumerating all reals,
>> >
>> > This is NOT an assumption. (Crikey, how many posts have there been
>> > already
>> > just clarifying this very point?)
>>
>> Then read the bloody thread before adding even more confusion!
>
> Ahem, you're right - I hadn't read the thread. I appologise!

Accepted. Now piss off, you and Virgil: *that* proof of his is wrong,
period, AND THAT YOU KEEP REPEATING IT AD NAUSEAM won't change it.

EOD here too. (But, please feel to keep discussing the "the", its
importance for international politics, and other such amenities.)

-LV

[David C. Ullrich]

On Sat, 28 Jul 2012 12:00:19 +0100, Frederick Williams
<freddyw...@btinternet.com> wrote:

>"David C. Ullrich" wrote:
>>
>> One of my all-time least favorite comps
>> questions was "Define a Banach space"...
>
>Comps?

Comprehensive exams.

[Virgil]

In article <QMmdnUjplsBrdI7N...@brightview.co.uk>,

The proof still works, mutatis mutandis, if one only assumes that the
values of the sequence are dense in the reals, which, since the
rationals are dense and countable, is known to be possible.

It is a minor revision of Cantor's original proof in either of my two
forms, but enough simpler as to make it a bit more accessible.
--

[Virgil]

In article <jv0qs4$j1a$1...@speranza.aioe.org>,

"LudovicoVan" <ju...@diegidio.name> wrote:

> "Mike Terry" <news.dead.p...@darjeeling.plus.com> wrote in message
> news:QMmdnUjplsBrdI7N...@brightview.co.uk...
> > "LudovicoVan" <ju...@diegidio.name> wrote in message
> > news:juve6l$o3h$1...@speranza.aioe.org...
> >> "Mike Terry" <news.dead.p...@darjeeling.plus.com> wrote in
> > message
> >> news:BoqdndgxE-k3po7N...@brightview.co.uk...
> >> > "LudovicoVan" <ju...@diegidio.name> wrote in message
> >> > news:juvcl0$l0a$1...@speranza.aioe.org...
> >> >> "Virgil" <vir...@ligriv.com> wrote in message
> >> >> news:virgil-4F3973....@bignews.usenetmonster.com...
> >>
> >> >> But, by denseness, there are infinitely many reals in [a_m, b_m] and,
> >> > since
> >> >> by assumption we are enumerating all reals,
> >> >
> >> > This is NOT an assumption. (Crikey, how many posts have there been
> >> > already
> >> > just clarifying this very point?)
> >>
> >> Then read the bloody thread before adding even more confusion!
> >
> > Ahem, you're right - I hadn't read the thread. I appologise!
>
> Accepted. Now piss off, you and Virgil: *that* proof of his is wrong,
> period, AND THAT YOU KEEP REPEATING IT AD NAUSEAM won't change it.

What's wrong with this version, which does NOT assume any surjection
from N to R?

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

A PROOF OF THE UNCOUNTABILITY OF THE REALS
(A variation on Cantor's FIRST proof)

ASSUMPTIONS:

(1) the intersection of a strictly nested sequence of closed real
intervals (the endpoints of each interval being interior points of the
previous interval) is not empty.

(2) A strictly increasing sequence of naturals does not have a natural
as its limit,

(3a) A strictly increasing but bounded sequence of reals has a real
number as a limit, its least upper bound, different from every member of
the sequence.

(3b) A strictly decreasing but bounded sequence of reals has a real
number as a limit, its greatest lower bound, different from every member
of the sequence.

Proof:

It is clear, using ennumerations of the rationals as an example, that
there are seqeunces whose sets of values are dense subsets of the reals.

It is equally clear that if ANY sequence is to cover all reals its set
of values must be dense in the reals.

Thus if every such sequence (whose values are dense in the reals) fails
to include all reals then there is no surjecton from N to R and R is
uncountable.

SO assume any such sequence whose values are dense in R.

The reals corresponding to first two naturals are endpoints of a real
interval.

The next two points of the sequence inside THAT interval form a
sub-interval, and so, because of density, ad infinitum.

The intersection of all those nested intervals cannot be empty, nor be a
point of the sequence.

Thus no mapping from N to R can be onto R

QED!
--

[Frederick Williams]

I know nothing of these things beyond what I read here:
http://en.wikipedia.org/wiki/Comprehensive_examination. For example,
this:

"Comprehensive examinations are typically based on a reading list agreed
upon by the student and his or her committee, which is staffed by the
primary supervisor and several advisors, normally professors at the
university, but not necessarily in the same faculty. This reading list
may comprise dozens or hundreds of books and other works.

There is no standard definition for what such exams entail, with some
universities having almost no exam, whilst at other universities the
process is quite rigorous. The exams thus take a number of forms,
including an informal meeting of just a few hours, a critical review of
one's academic portfolio, the submission of an academic paper which may
take several hours or months to write, or a series of proctored exams
taking anywhere from a few to as many as thirty-six hours."

So, in your case was it hundreds of books and other works and thirty-six
hours of exams?

[David C. Ullrich]

On Sat, 28 Jul 2012 22:15:30 +0100, Frederick Williams

Was talking about exams written by colleagues since I've been
a professor. The ones I took were so long ago they're just a blur...

A few books, in both cases.

[LudovicoVan]

"LudovicoVan" <ju...@diegidio.name> wrote in message

news:junlnt$ugh$1...@speranza.aioe.org...
> [More a bunch of questions than any solid objection.]
>
> <http://en.wikipedia.org/wiki/Cantor's_first_uncountability_proof>
>
> << Cantor now breaks the proof into two cases: Either the number of
> intervals generated is finite or infinite. >>
>
> << If the number of intervals is infinite, let a_oo = lim_{n->oo} a_n.
> At this point, Cantor could finish his proof by noting that a_oo is not
> contained in the given sequence since for every n, a_oo belongs to the
> interior of [a_n, b_n] but x_n does not. >>

We can try and close this one too (in fact, no claim that there are no
mistakes, and there are some open issues left, too, so corrections and
feedback welcome).

The basic idea is this: the proof must work for any sequence of real
numbers, but we will provide a counter-example by assuming that we are
given, in particular, an *enumeration* of the reals. We will first proof
that the limit interval in this case must be improper, then we will argument
how this does *not* entail that there is still a missing real.

Definitions:

Let (x_n) be a sequence of real numbers.

Let [a_n, b_n] be a sequence of real intervals s.t.:

[a_0, b_0] := [a, b]
[a_n, b_n] c [a_{n-1}, b_{n-1}], for n > 0

with:

a_n := x_{n'[2n-1]}
b_n := x_{n'[2n]}

where x_{n'[2n-1]} and x_{n'[2n]} are, non respectively, the (2n-1)-th and
(2n)-th entries picked from (x_n) per the rules of the game.

Let the limit interval be:

[ a_oo, b_oo ] :=
= lim_{n->oo} [a_n, b_n] =
= [ lim_{n->oo} a_n, lim_{n->oo} b_n ]

Theorem (1): Given (x_n) an *enumeration* of real numbers (plus denseness),

the limit interval must be improper.

Proof: Assume per absurdo that there is a last interval [a_m, b_m], with:

a_m = x_{n'[2m-1]}
b_m = x_{n'[2m]}

By construction, it must be a_m =/= b_m, so the interval is proper. By
denseness, there are infinitely many reals in ]a_m, b_m[. Since, per
hypothesis, we are enumerating all reals, these (infinitely many) reals
in ]a_m, b_m[ must precede a_m and b_m in (x_n), otherwise [a_m, b_m] could
not be the last interval, i.e. they should form a subsequence (x_k) of (x_n)
s.t. A k : k < min(n'[2m-1], n'[2m]). But, that entails that there should
be infinitely many naturals that precede min(n'[2m-1], n'[2m]), which is an
absurdity. Hence, there can be no last interval.
QED.

Now, we know that, by completeness, a_oo (or, b_oo) must be a real number,
so we are demanded to show how could a_oo ever belong to the given
enumeration, but we argument that this demand is rather an absurdity in
itself. ((In fact, I have some doubts that completeness even justifies that
a_oo is a real number, but this could just make the present argument
unnecessary and the overall objection would stay as correct.))

That demand is, more specifically, an instance of the "incorrect counting"
that results from an improper mix of standard and extended settings: an
omega-th end-point, a_oo, would necessarily be drawn from an omega-th entry
of the sequence! Formally, we have the following property:

A m : a_m e (x_n) & b_m e (x_n)

That works not only for n and m in N, but also for n and m in N*.
Informally, that property, which is a plain consequence of the construction
rules, states that *every* end-point is drawn from the given sequence, i.e.
nothing can *ever* be "missing", not even by extension.

To sum it up, by choosing an enumeration of the real numbers as our
sequence, we (seem to) have provided a counter-example to Cantor's First
Proof.

-LV

[Virgil]

In article <jv1rd1$slh$3...@speranza.aioe.org>,

"LudovicoVan" <ju...@diegidio.name> wrote:

> "LudovicoVan" <ju...@diegidio.name> wrote in message
> news:junlnt$ugh$1...@speranza.aioe.org...
> > [More a bunch of questions than any solid objection.]
> >
> > <http://en.wikipedia.org/wiki/Cantor's_first_uncountability_proof>
> >
> > << Cantor now breaks the proof into two cases: Either the number of
> > intervals generated is finite or infinite. >>
> >
> > << If the number of intervals is infinite, let a_oo = lim_{n->oo} a_n.
> > At this point, Cantor could finish his proof by noting that a_oo is not
> > contained in the given sequence since for every n, a_oo belongs to the
> > interior of [a_n, b_n] but x_n does not. >>
>
> We can try and close this one too (in fact, no claim that there are no
> mistakes, and there are some open issues left, too, so corrections and
> feedback welcome).
>
> The basic idea is this: the proof must work for any sequence of real
> numbers, but we will provide a counter-example by assuming that we are
> given, in particular, an *enumeration* of the reals. We will first proof
> that the limit interval in this case must be improper, then we will argument
> how this does *not* entail that there is still a missing real.

I have no idea what you men by an "improper" interval in this context.

http://www-old.me.gatech.edu/~ywang/research/interval/WhyGeneralizedInter
val.html
defines an "improper" interval as one like [2,1] in which the lower
limit is greater than th upper limit.

>
> Definitions:
>
> Let (x_n) be a sequence of real numbers.
>
> Let [a_n, b_n] be a sequence of real intervals s.t.:
>
> [a_0, b_0] := [a, b]
> [a_n, b_n] c [a_{n-1}, b_{n-1}], for n > 0
>
> with:
>
> a_n := x_{n'[2n-1]}
> b_n := x_{n'[2n]}
>
> where x_{n'[2n-1]} and x_{n'[2n]} are, non respectively, the (2n-1)-th and
> (2n)-th entries picked from (x_n) per the rules of the game.

What rules of what game?

>
> Let the limit interval be:

Until the rules of the game are made explicit, there can be no evidence
that any limit interval actually can exist.

So until those "rules of the game" are made explicit, the rest is
nonsense.
--

[Mike Terry]

"LudovicoVan" <ju...@diegidio.name> wrote in message

news:jv1rd1$slh$3...@speranza.aioe.org...

You have shown that if the sequence (x_n) which enumerates the reals, there
will be infinitely many constructed intervals [a_m, b_m]. (This is
correct.)

Also, with some more work you could show

lim(n-->oo) a_n = lim(n-->oo) b_n

Letting a = lim(n-->oo) a_n,
b = lim(n-->oo) b_n
we are asked to examine the interval [a,b] = {a} = {b} (i.e. a singleton
set)

Note that a is not "the omegath entry in the sequence (x_n)". Neither is it
"the omegath entry in the sequence (a_n)" or "the omegath entry in the
sequence (b_n)". It is simply a real number, which is the limit as above.

The sequences used in Cantor's proof are mappings from N to R, not from N*
to R, and NONE OF THEM HAVE ENTRIES CORRESPONDING TO AN ARGUMENT OF OMEGA.

I.e. they are regular "sequences" as generally understood by most
mathematicians...

>
> Now, we know that, by completeness, a_oo (or, b_oo) must be a real number,
> so we are demanded to show how could a_oo ever belong to the given
> enumeration, but we argument that this demand is rather an absurdity in
> itself. ((In fact, I have some doubts that completeness even justifies
that
> a_oo is a real number, but this could just make the present argument
> unnecessary and the overall objection would stay as correct.))
>
> That demand is, more specifically, an instance of the "incorrect counting"
> that results from an improper mix of standard and extended settings: an
> omega-th end-point, a_oo, would necessarily be drawn from an omega-th
entry
> of the sequence!

There is no "improper mixing of standard and extended settings", because in
Cantor's proof there are no extended settings involved, i.e. there is no mix
at all. You are the only person introducing "extended settings". As I
noted above, a_oo [=a in my notation] is not an omegath end-point of
anything, despite the confusing notation, and there is no requirement to
introduce an extra omegath entry in any sequence.

Of course, if you really *want* to, you could extend sequence (x_n) or (a_n)
or (b_n) by adding an omegath entry x_oo = a etc.. This does not help you,
because Cantor's claim is that a was missing from the *original* sequence,
not from your new extended sequence, so the Cantor proof still holds.

Regards,
Mike.

[LudovicoVan]

"Mike Terry" <news.dead.p...@darjeeling.plus.com> wrote in message

news:WOCdnf-bJP1FEYnN...@brightview.co.uk...

> The sequences used in Cantor's proof are mappings
> from N to R, not from N* to R,

The problem is reasoning as if they were mappings from N to R*.

> and NONE OF THEM HAVE ENTRIES CORRESPONDING
> TO AN ARGUMENT OF OMEGA.

Indeed.

-LV

[Ross A. Finlayson]

On Jul 28, 6:07 pm, "Mike Terry"

EF, a function (and its scalar multiples), in the context of Cantor's
nested intervals result, sees different results than any other
function, particularly not the conclusion that it's forever
inexhausted.

The number (or count) of "intervals" is one and it's of EF(0) and
EF(1).

Quite simple, that is.

Regards,

Ross Finlayson

[LudovicoVan]

"Ross A. Finlayson" <ross.fi...@gmail.com> wrote in message
news:e430c4ba-26d9-42bc...@rq10g2000pbb.googlegroups.com...

> EF, a function (and its scalar multiples), in the context of Cantor's
> nested intervals result, sees different results than any other
> function, particularly not the conclusion that it's forever
> inexhausted.
>
> The number (or count) of "intervals" is one and it's of EF(0) and
> EF(1).
>
> Quite simple, that is.

What is EF? And what is CDF?

-LV

[Ross A. Finlayson]

On Jul 28, 6:50 pm, "LudovicoVan" <ju...@diegidio.name> wrote:
> "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote in messagenews:e430c4ba-26d9-42bc...@rq10g2000pbb.googlegroups.com...

>
> > EF, a function (and its scalar multiples), in the context of Cantor's
> > nested intervals result, sees different results than any other
> > function, particularly not the conclusion that it's forever
> > inexhausted.
>
> > The number (or count) of "intervals" is one and it's of EF(0) and
> > EF(1).
>
> > Quite simple, that is.
>
> What is EF?  And what is CDF?
>
> -LV
>
>

EF is the natural/unit equivalency function. A CDF you can find
defined in a textbook on probability.

Basically EF is a function.

EF(n) = n/d, d->oo, n, d E N

It has various properties.

EF(0) = 0
EF(n+1) > EF(n)
lim n->oo EF(n) = 1

Its domain is the (set or collection of) natural integers, its range
(or co-image) the unit interval of reals.

And, in each of the known number-theoretic or topological theorems
otherwise uncountability of the reals (eg the antidiagonal or nested
interval arguments), it sees a different result, and conclusion.

As it's also a CDF, being monotonically increasing and defined for the
domain over the range [0,1], that as well simply establishes a
corresponding distribution, of said domain. Here, as the difference
between elements (given a constant difference in elements of the
domain) is constant: a uniform distribution.

Regards,

Ross Finlayson

[Virgil]

In article
<ecb2554b-7475-4aa2...@a7g2000pbm.googlegroups.com>,

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

> On Jul 28, 6:50 pm, "LudovicoVan" <ju...@diegidio.name> wrote:
> > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote in
> > messagenews:e430c4ba-26d9-42bc...@rq10g2000pbb.googlegroups.c
> > om...
> >
> > > EF, a function (and its scalar multiples), in the context of Cantor's
> > > nested intervals result, sees different results than any other
> > > function, particularly not the conclusion that it's forever
> > > inexhausted.
> >
> > > The number (or count) of "intervals" is one and it's of EF(0) and
> > > EF(1).
> >
> > > Quite simple, that is.
> >
> > What is EF?  And what is CDF?
> >
> > -LV
> >
> >
>
>
> EF is the natural/unit equivalency function. A CDF you can find
> defined in a textbook on probability.
>
> Basically EF is a function.
>
> EF(n) = n/d, d->oo, n, d E N
>
> It has various properties.
>
> EF(0) = 0
> EF(n+1) > EF(n)
> lim n->oo EF(n) = 1
>
> Its domain is the (set or collection of) natural integers, its range
> (or co-image) the unit interval of reals.

Where do you prove that any such function from N to [0,1] as you alleged
EF exists?

I have seen proofs that convince me that your EF does not exist,
at least not as you describe it above.
--

[Virgil]

In article <jv235n$bgn$1...@speranza.aioe.org>,

"LudovicoVan" <ju...@diegidio.name> wrote:

> "Mike Terry" <news.dead.p...@darjeeling.plus.com> wrote in message
> news:WOCdnf-bJP1FEYnN...@brightview.co.uk...
>
> > The sequences used in Cantor's proof are mappings
> > from N to R, not from N* to R,
>
> The problem is reasoning as if they were mappings from N to R*.

Then cease doing so!

>
> > and NONE OF THEM HAVE ENTRIES CORRESPONDING
> > TO AN ARGUMENT OF OMEGA.
>
> Indeed.
>
> -LV
>

--

[David C. Ullrich]

How in the world is this the negation of "the limit interval is
proper"?

>By construction, it must be a_m =/= b_m, so the interval is proper. By
>denseness, there are infinitely many reals in ]a_m, b_m[. Since, per
>hypothesis, we are enumerating all reals, these (infinitely many) reals
>in ]a_m, b_m[ must precede a_m and b_m in (x_n), otherwise [a_m, b_m] could
>not be the last interval, i.e. they should form a subsequence (x_k) of (x_n)
>s.t. A k : k < min(n'[2m-1], n'[2m]). But, that entails that there should
>be infinitely many naturals that precede min(n'[2m-1], n'[2m]), which is an
>absurdity. Hence, there can be no last interval.
>QED.
>
>Now, we know that, by completeness, a_oo (or, b_oo) must be a real number,
>so we are demanded to show how could a_oo ever belong to the given
>enumeration, but we argument that this demand is rather an absurdity in
>itself. ((In fact, I have some doubts that completeness even justifies that
>a_oo is a real number, but this could just make the present argument
>unnecessary and the overall objection would stay as correct.))
>
>That demand is, more specifically, an instance of the "incorrect counting"
>that results from an improper mix of standard and extended settings: an
>omega-th end-point, a_oo, would necessarily be drawn from an omega-th entry
>of the sequence! Formally, we have the following property:
>
> A m : a_m e (x_n) & b_m e (x_n)
>
>That works not only for n and m in N, but also for n and m in N*.

This is nonsense. The notation "a_oo" is just a convenient label for
the limit of a_n. There's no reason whatever to think that a_oo
is an element of the sequence (a_n).

It's actually bad notation. Take the proof, and instead of calling
the limit interval [a_oo,b_oo] call it [c,d] instead.

>Informally, that property, which is a plain consequence of the construction
>rules, states that *every* end-point is drawn from the given sequence, i.e.
>nothing can *ever* be "missing", not even by extension.
>
>To sum it up, by choosing an enumeration of the real numbers as our
>sequence, we (seem to) have provided a counter-example to Cantor's First
>Proof.

It only seems that way to you.

>-LV
>

[David C. Ullrich]

On Sun, 29 Jul 2012 02:20:56 +0100, "LudovicoVan"
<ju...@diegidio.name> wrote:

>"Mike Terry" <news.dead.p...@darjeeling.plus.com> wrote in message
>news:WOCdnf-bJP1FEYnN...@brightview.co.uk...
>
>> The sequences used in Cantor's proof are mappings
>> from N to R, not from N* to R,
>
>The problem is reasoning as if they were mappings from N to R*.

There is no such reasoning in the actual proof. The theorem is about
sequences. By definition, a sequence of reals is a mapping from
N to R. There is no N* and no R* anywhere in the proof.

>
>> and NONE OF THEM HAVE ENTRIES CORRESPONDING
>> TO AN ARGUMENT OF OMEGA.
>
>Indeed.

"Indeed"? You didn't quite notice that this shows that you're
speaking nonsense?

>-LV
>

[Ross A. Finlayson]

On Jul 28, 7:58 pm, Virgil <vir...@ligriv.com> wrote:
> In article
> <ecb2554b-7475-4aa2-9197-0828b7043...@a7g2000pbm.googlegroups.com>,

You've seen proofs in a theory that you'll agree is, at best,
incomplete.

Yes, and you also accept that there's no universe in your set theory,
where, its existence, or anything in it, would demonstrate otherwise.
(Universe would be its own powerset, and a model of physical objects
is of mathematical objects.)

And, yes, I've described why Dedekind/Cauchy/Eudoxus is insufficient
to represent all the real numbers, of the continuum of real numbers.

Then, this EF is standardly modeled by real functions, as are other
non-real functions useful to real analysis, and application, and has
its concomitant properties, standardly and non.

Yep, pretty much everybody has seen a proof that EF doesn't exist.
It's a nice big sign on the pasture fence saying "there's no other
side". Where delta-epsilonics is correct and as we know the basis
for much and most of modern real analysis, still we know that there is
more to the infinitesimal analysis than that careful corraling.

Now, how many of the natural integers would you say are even multiples
of two? It's half.

Oh, and by the way, applications of solely transfinite cardinals are
yet to be found (asymptotics suffice).

No, Hancher, we left the discussion whether EF is a function: that it
is. Your memory is short, poor, or deficient: that's generous. And
that's recent.

Regards,

Ross Finlayson

[Virgil]

In article
<450fa4c3-c27a-488a...@c4g2000pba.googlegroups.com>,

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

At best, your 'theory' is nonsense.
--

[Shmuel Metz]

In <juujak$9tr$1...@dont-email.me>, on 07/27/2012
at 12:32 PM, "dilettante" <n...@nonono.no> said:

>I think you plead guilty prematurely. To me, "enumeration of reals"
>naturally means "enumeration of some set whose members are real
>numbers", while "enumeration of the reals" clearly means
>"enumeration of the whole set of real numbers".

That's my reading.

--
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

[Shmuel Metz]

In <virgil-7EDF48....@bignews.usenetmonster.com>, on
07/27/2012

at 10:21 PM, Virgil <vir...@ligriv.com> said:

>Is proof by contradiction no longer a valid form of argument?

To an intuitionist?

[Shmuel Metz]

In <juvajj$h2s$1...@speranza.aioe.org>, on 07/28/2012
at 01:09 AM, "LudovicoVan" <ju...@diegidio.name> said:

>Above all when it's totally preposterous. In fact, that bit was
>clear: 60%+ plus of English comes from Latin.

The attempts of English grammarians to cram English into Latin grammar
do not alter the fact that English is a Germanic language.

[Shmuel Metz]

In <jv1rd1$slh$3...@speranza.aioe.org>, on 07/29/2012
at 12:08 AM, "LudovicoVan" <ju...@diegidio.name> said:

>Definitions:

You're missing the definition of "the rules of the game".

>Now, we know that, by completeness, a_oo (or, b_oo) must be a real
>number, so we are demanded to show how could a_oo ever belong to the
>given enumeration, but we argument that this demand is rather an
>absurdity in itself.

What are you smoking? There are no "absurd" or "demand" in
Mathematics, only proofs, and it has been proven the intersection of a
proper nested sequence of closed intervals does not contain any of the
end points. The fact that it has been proven trumps the fact that you
claim to find it absurd.

>That demand is, more specifically, an instance of the "incorrect
>counting" that results from an improper mix of standard and
>extended settings: an omega-th end-point, a_oo,

You haven't defined "an omega-th end-point"; the fact that you used
"oo" as part of a name has nothing to do with Omega but is purely
arbitrary.

> A m : a_m e (x_n) & b_m e (x_n)
>That works not only for n and m in N, but also for n and m in N*.

No.

>To sum it up,

You understand Logic as poorly as you understand Mathematics.

[Marshall]

On Sunday, July 29, 2012 7:27:17 PM UTC-7, Seymour J. Shmuel Metz wrote:
>
> You understand Logic as poorly as you understand Mathematics.

But he makes up for it with his hair-trigger temper!


Marshall

[LudovicoVan]

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

news:ecb2554b-7475-4aa2...@a7g2000pbm.googlegroups.com...

> On Jul 28, 6:50 pm, "LudovicoVan" <ju...@diegidio.name> wrote:

<snip>

>> What is EF? And what is CDF?
>

> EF is the natural/unit equivalency function. A CDF you can find
> defined in a textbook on probability.
>

> Basically EF is a function. [...]

>
> EF(n) = n/d, d->oo, n, d E N

Interesting, thanks.

-LV

[LudovicoVan]

"David C. Ullrich" <ull...@math.okstate.edu> wrote in message
news:b0ma1855kfm7l2prt...@4ax.com...

> On Sun, 29 Jul 2012 02:20:56 +0100, "LudovicoVan"
> <ju...@diegidio.name> wrote:
>
>>"Mike Terry" <news.dead.p...@darjeeling.plus.com> wrote in
>>message
>>news:WOCdnf-bJP1FEYnN...@brightview.co.uk...
>>
>>> The sequences used in Cantor's proof are mappings
>>> from N to R, not from N* to R,
>>
>>The problem is reasoning as if they were mappings from N to R*.
>
> There is no such reasoning in the actual proof. The theorem is about
> sequences. By definition, a sequence of reals is a mapping from
> N to R. There is no N* and no R* anywhere in the proof.

I know very well what the theorem is about and do agree with you: a_oo :=
lim_{n->oo} a_n. Now, please, explain to Virgil and co. that that entails
there is no such thing as "a number a_oo missing from (x_n)".

-LV

[LudovicoVan]

"David C. Ullrich" <ull...@math.okstate.edu> wrote in message

news:rmla1854mjs59n523...@4ax.com...

> On Sun, 29 Jul 2012 00:08:06 +0100, "LudovicoVan"
> <ju...@diegidio.name> wrote:

<snipped>

>> Formally, we have the following property:
>>
>> A m : a_m e (x_n) & b_m e (x_n)
>>
>>That works not only for n and m in N, but also for n and m in N*.
>
> This is nonsense. The notation "a_oo" is just a convenient label for

Drop the extended part if you are uncomfortable with it. The rest still
holds.

> There's no reason whatever to think that a_oo
> is an element of the sequence (a_n).

Now that is sloppy: if one goes on and considers such thing as a number a_oo
missing from sequence (x_n), then one does have to use the extended setting.
In the present problem that is even obvious because all end-points are drawn
from (x_n) to begin with (as expressed by the above property for both
settings). Then, still nothing is missing, because the sequence (x_n)
should be extended to begin with.

-LV

[Aatu Koskensilta]

"LudovicoVan" <ju...@diegidio.name> writes:

> "Aatu Koskensilta" <aatu.kos...@uta.fi> wrote in message
> news:87394d2...@uta.fi...
>> David C. Ullrich <ull...@math.okstate.edu> writes:
>>
>>> Yes. Another hint that thats the distinction I was getting at is given
>>> by my post, where I specified that I meant A and not B, where A
>>> and B differ only by the includion of the word "the".
>>
>> This sort of thing can very easily throw off a non-native speaker.

>
> Above all when it's totally preposterous.

I'm afraid I don't follow. What do you think preposterous?

--
Aatu Koskensilta (aatu.kos...@uta.fi)

"Wovon man nicht sprechen kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus

[Shmuel Metz]

In <jv6121$7g6$1...@speranza.aioe.org>, on 07/30/2012

at 02:09 PM, "LudovicoVan" <ju...@diegidio.name> said:

>Now that is sloppy: if one goes on and considers such thing as a
>number a_oo missing from sequence (x_n), then one does have to
>use the extended setting.

Utter balderdash.

[David C. Ullrich]

You're not making any sense.


>-LV
>

[Virgil]

In article <jv6121$7g6$1...@speranza.aioe.org>,

If that were so then |N as an ordered set (sequence) would have to
contain oo.
--

[Virgil]

In article <jv60mr$6i1$2...@speranza.aioe.org>,

In the real number system it is quite possible to have a strictly
increasing infinite sequence which is bounded above by every member of a
strictly decreasing infinite sequence so that there is at least one real
between them which is not a member of either sequence.

And for any sequence that is dense in the reals there MUST be such a
pair of strictly increasing and decreasing subsequences with at least
one real between them which is not in the original sequence.

At least if one is using the standard set of naturals and the standard
set of reals.

I have no idea what set of naturals or set of reals LV is basing his
arguments on.
--

[Virgil]

In article <jv60ml$6i1$1...@speranza.aioe.org>,

Which means that EF(n) < epsilon for every epsilon > 0.

But Ross would have you believe that

SUM_(n = 1..oo) EF(n) = 1.
>
> Interesting, thanks.
>
> -LV
>
--

[Ross A. Finlayson]

On Jul 30, 12:47 pm, Virgil <vir...@ligriv.com> wrote:
> In article <jv60ml$6i...@speranza.aioe.org>,
>
>  "LudovicoVan" <ju...@diegidio.name> wrote:
> > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote in message

> >news:ecb2554b-7475-4aa2...@a7g2000pbm.googlegroups.com...
> > > On Jul 28, 6:50 pm, "LudovicoVan" <ju...@diegidio.name> wrote:
> > <snip>
>
> > >> What is EF?  And what is CDF?
>
> > > EF is the natural/unit equivalency function.  A CDF you can find
> > > defined in a textbook on probability.
>
> > > Basically EF is a function. [...]
>
> > > EF(n) = n/d, d->oo, n, d E N
>
> Which means that EF(n) < epsilon for every epsilon > 0.
>
> But Ross would have you believe that
>
>    SUM_(n = 1..oo) EF(n) = 1.
>
> > Interesting, thanks.
>
> > -LV
>
> --

No, INT_(n = 1 ... oo) EF(n) = 1. Now, this might be surprising
because it looks like a staircase for finite values of d that sees the
integral go to 1/2, but, it's 1.

Also, lim_d->oo SUM_(n=1 ... d) 1/d = 1. That's the p.d.f. of a
uniform distribution over the naturals, 1/oo. (Yes I know that's more
implicit in notation.)

Then, for the first point, basically via an analog of the intermediate
value theorem as EF(0) = 0 and lim_n->oo EF(n) = 1, Exists n s.t.
EF(n) > epsilon for epsilon in [0,1).

Then, as the notation becomes more implicit, SUM_(n = 1..oo) EF(n) =
oo/2. This is basically from that, where REF + EF = 1, that for half
of the domain of oo, the sum is oo/2.

Have that EF_d (n) is n/d for d E N. Then, REF_1 + EF_1 looks like
the unit square, (REF + EF)_2 is two half squares, etcetera. Yet, in
utter extremis, where it's a unit line segment in y at each natural n
E N, the integral is 2. Similarly, as described above, the points in
a line are 1-sided and contribute only half as much, evaluating the
integral over the line, as the 2-sided points each on a line.

Then, in the polydimensional, it gets quite the bit richer in
possibilities.

And, this may well see near to direct application in, for example,
mathematical physics.

These reals are R^bar^umlaut as I describe them.

Regards,

Ross Finlayson

[Virgil]

In article
<5a5948dd-f1d0-4018...@wt8g2000pbb.googlegroups.com>,

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

> On Jul 30, 12:47�pm, Virgil <vir...@ligriv.com> wrote:

> > But Ross would have you believe that
> >
> > � �SUM_(n = 1..oo) EF(n) = 1.
> >
> > > Interesting, thanks.
> >
> > > -LV
> >
> > --
>
>
> No, INT_(n = 1 ... oo) EF(n) = 1.

Then give me any of the values of EF(n) for any of the n.

For any uniform distribution over all members of N
one would have to have EF(1) = EF(2) =...= EF(n) + ...
so that INT_(n = 1 ... m) EF(n) = m*EF(1) for all naturals m.

Thus INT_(n = 1 ... oo) EF(n) = oo*EF(1), if it exists at all.

So tell us, please, the real value of EF(1).
Or EF(n) for n any member of N.
--

[Ross A. Finlayson]

On Jul 30, 7:41 pm, Virgil <vir...@ligriv.com> wrote:
> In article

> <5a5948dd-f1d0-4018-aae6-0c7395705...@wt8g2000pbb.googlegroups.com>,

>  "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:
>
> > On Jul 30, 12:47 pm, Virgil <vir...@ligriv.com> wrote:
> > > But Ross would have you believe that
>
> > > SUM_(n = 1..oo) EF(n) = 1.
>
> > > > Interesting, thanks.
>
> > > > -LV
>
> > > --
>
> > No, INT_(n = 1 ... oo) EF(n) = 1.
>
> Then give me any of the values of EF(n) for any of the n.
>
> For any uniform distribution over all members of N
> one would have to have EF(1) = EF(2) =...= EF(n) + ...
> so that INT_(n = 1 ... m) EF(n) = m*EF(1) for all naturals m.
>
> Thus INT_(n = 1 ... oo) EF(n) = oo*EF(1), if it exists at  all.
>
> So tell us, please, the real value of EF(1).
> Or EF(n) for n any member of N.
> --
>
>

Actually, they define real values between zero and one, those
elements.

EF(1) = EF(1)

Empty to full, the unit line segment: defined by EF the natural/unit
equivalency function.

Reals: complete ordered field and partially ordered ring with rather
restricted transfer principle.

Then, in the description of sums and products of iota-values, where
iota is conveniently a letter and a word with plain meaning as here
basically a semi-atomic bit, there's a reasonable difference between
products and iterated sums in iota-values. Here, iota the
infinitesimal, or EF(1), is simply the next in the normal ordering of
the reals, well-ordered, from the origin.

Regards,

Ross Finlayson

[Virgil]

In article
<67bf4363-74ca-4f80...@l6g2000pbf.googlegroups.com>,

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

Then EF(1) = EF(2) = EF(3) = ...= EF(n) = ..., for all n in N
thus EF(n) < epsilon for every positive real epsilon
thus EF(n) = 0 for and n in N.

At least with the real reals.
--

[LudovicoVan]

"Virgil" <vir...@ligriv.com> wrote in message
news:virgil-71471F....@bignews.usenetmonster.com...

> In article <jv6121$7g6$1...@speranza.aioe.org>,
> "LudovicoVan" <ju...@diegidio.name> wrote:
>> "David C. Ullrich" <ull...@math.okstate.edu> wrote in message
>> news:rmla1854mjs59n523...@4ax.com...
>> > On Sun, 29 Jul 2012 00:08:06 +0100, "LudovicoVan"
>> > <ju...@diegidio.name> wrote:
>> <snipped>
>>

>> > There's no reason whatever to think that a_oo
>> > is an element of the sequence (a_n).
>>
>> Now that is sloppy: if one goes on and considers such thing as a number
>> a_oo
>> missing from sequence (x_n), then one does have to use the extended
>> setting.
>> In the present problem that is even obvious because all end-points are
>> drawn
>> from (x_n) to begin with (as expressed by the above property for both
>> settings). Then, still nothing is missing, because the sequence (x_n)
>> should be extended to begin with.
>
> If that were so then |N as an ordered set (sequence) would have to
> contain oo.

Still your basic mistake: to contain oo, that would be N* := N u { oo },
then you have an a_oo (in R*).

I'll state the main point again: my counter-argument works either with an
enumeration from N to R (supposedly, the original Cantorian setting), or
with an enumeration from N* to R*. The reason why Cantor's First Proof
fails is because of invalid reasoning with putative sequences from N to R*
(so that an a_oo looks like it is missing from the enumeration).

-LV

[LudovicoVan]

"LudovicoVan" <ju...@diegidio.name> wrote in message

news:jv89ao$95r$1...@speranza.aioe.org...

> "Virgil" <vir...@ligriv.com> wrote in message
> news:virgil-71471F....@bignews.usenetmonster.com...
>> In article <jv6121$7g6$1...@speranza.aioe.org>,
>> "LudovicoVan" <ju...@diegidio.name> wrote:
>>> "David C. Ullrich" <ull...@math.okstate.edu> wrote in message
>>> news:rmla1854mjs59n523...@4ax.com...
>>> > On Sun, 29 Jul 2012 00:08:06 +0100, "LudovicoVan"
>>> > <ju...@diegidio.name> wrote:
>>> <snipped>
>>>
>>> > There's no reason whatever to think that a_oo
>>> > is an element of the sequence (a_n).
>>>
>>> Now that is sloppy: if one goes on and considers such thing as a number
>>> a_oo
>>> missing from sequence (x_n), then one does have to use the extended
>>> setting.
>>> In the present problem that is even obvious because all end-points are
>>> drawn
>>> from (x_n) to begin with (as expressed by the above property for both
>>> settings). Then, still nothing is missing, because the sequence (x_n)
>>> should be extended to begin with.
>>
>> If that were so then |N as an ordered set (sequence) would have to
>> contain oo.
>
> Still your basic mistake: to contain oo, that would be N* := N u { oo },
> then you have an a_oo (in R*).

That may be confusing. For some reason, I have in mind R* := [-oo, +oo],
although that's maybe irrelevant here.

-LV

[LudovicoVan]

"LudovicoVan" <ju...@diegidio.name> wrote in message

news:jv89ao$95r$1...@speranza.aioe.org...

That's wrongly put: the point just boils down to the fact that, to consider
a_oo as a real number that belongs or does not belong to (x_n), we need
consider sequences over N*.

-LV

[Virgil]

In article <jv89ao$95r$1...@speranza.aioe.org>,

"LudovicoVan" <ju...@diegidio.name> wrote:

> I'll state the main point again: my counter-argument works either with an
> enumeration from N to R (supposedly, the original Cantorian setting), or
> with an enumeration from N* to R*.

Perhaps so in WMytheology, or some similar corruption of standard
mathematics, but not at all in standard mathematics.
--

[Virgil]

In article <jv8mt2$ok0$1...@speranza.aioe.org>,

"LudovicoVan" <ju...@diegidio.name> wrote:

> > I'll state the main point again: my counter-argument works either with an
> > enumeration from N to R (supposedly, the original Cantorian setting), or
> > with an enumeration from N* to R*. The reason why Cantor's First Proof
> > fails is because of invalid reasoning with putative sequences from N to R*
> > (so that an a_oo looks like it is missing from the enumeration).
>
> That's wrongly put: the point just boils down to the fact that, to consider
> a_oo as a real number that belongs or does not belong to (x_n), we need
> consider sequences over N*.

You may need to but those who are basing their analysis on standard
mathematics do not.

It is well known in standard mathematics that a nested sequence of real
intervals, with each being a proper subinterval of all its predecessors,
has a non-empty intersection, which may be an interval of positive
length, in which case your alleged a_oo is a real an interval of
positive length.
--

[Virgil]

In article <jv8a46$bsq$1...@speranza.aioe.org>,

"LudovicoVan" <ju...@diegidio.name> wrote:

> > Still your basic mistake: to contain oo, that would be N* := N u { oo },
> > then you have an a_oo (in R*).
>
> That may be confusing. For some reason, I have in mind R* := [-oo, +oo],
> although that's maybe irrelevant here.
>
> -LV

It is, but that has never stopped yo before!
--

[William Hughes]

On Jul 31, 10:34 am, "LudovicoVan" <ju...@diegidio.name> wrote:
> "LudovicoVan" <ju...@diegidio.name> wrote in message
>
> news:jv89ao$95r$1...@speranza.aioe.org...
>
>
>
>
>
>
>
>
>
> > "Virgil" <vir...@ligriv.com> wrote in message
> >news:virgil-71471F....@bignews.usenetmonster.com...

> >> In article <jv6121$7g...@speranza.aioe.org>,
> >> "LudovicoVan" <ju...@diegidio.name> wrote:
> >>> "David C. Ullrich" <ullr...@math.okstate.edu> wrote in message

Your problem is that you are mixing up the proof by contradiction and
the
proof that there in no surjection from N to R.

Forget about surjections and a_oo

Let D be the set of functions from N to R (D is certainly
non empty)

Let f be an element of D (note there is no assumption that
f is a surjection)

Theorem: There is an element b of R such that for all n in N b <>
f(n)

Construct b by the method of Cantor's first proof.

We have now shown that if f is an element of D then f is not a
surjection.

The final proof by contradiction is left as an exercise to the reader.

[Ross A. Finlayson]

On Jul 31, 2:42 am, "LudovicoVan" <ju...@diegidio.name> wrote:
> "Virgil" <vir...@ligriv.com> wrote in message
>
> news:virgil-71471F....@bignews.usenetmonster.com...
>
>
>
>
>

> > In article <jv6121$7g...@speranza.aioe.org>,
> > "LudovicoVan" <ju...@diegidio.name> wrote:
> >> "David C. Ullrich" <ullr...@math.okstate.edu> wrote in message

You can have natural integers with the one-point compactification of
the point at infinity (say, N+, or the point oo) without having the
hyperintegers (N*, or *N).

Some would have that as a consequence of quantification that the
natural integers do have a point at infinity. That's simply non-
Archimedean and is well known and a regular object of discourse, in
modern times as in for hundreds of years.

Analysis in the meromorphic regularly has points at infinity.

Robinson's hyperreals don't have much beyond what the standard reals
have in analytical character, though their existence is consistent
with them, in regards to, for example, the co-consistency of ZFC and
IST. In fact they add nothing to real analysis. But, they're there
(in "standard" and "modern" mathematics).

Axiomless system of natural deduction: dually-self-infraconsistent.

Regards,

Ross Finlayson

[Ross A. Finlayson]

On Jul 31, 12:48 pm, Virgil <vir...@ligriv.com> wrote:
> In article <jv8mt2$ok...@speranza.aioe.org>,

Huh, well how exactly the opposite that is to what you wrote before.

http://www.tiki-lounge.com/~raf/finlayson_injectrationals.pdf


Congratulations, you contradict yourself, you flaming troll.

Regards,

Ross Finlayson

[Virgil]

In article
<cef3510f-6eca-4e42...@id7g2000pbc.googlegroups.com>,

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

Actually not at all different from what wrote before.
But what I wrote seems to be quite different from what you claim to have
read.
--

[Virgil]

In article
<a0dbb9de-c422-40e6...@iw9g2000pbc.googlegroups.com>,

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

> You can have natural integers with the one-point compactification of
> the point at infinity (say, N+, or the point oo) without having the
> hyperintegers (N*, or *N).

When talking about sequences, I prefer to have the naturals without any
such extraneous entities as you eem bent on appending.

>
> Some would have that as a consequence of quantification that the
> natural integers do have a point at infinity.

No one who knows what they are talking about would say that.

But Ross is not of their number.
--

[LudovicoVan]

"Ross A. Finlayson" <ross.fi...@gmail.com> wrote in message
news:a0dbb9de-c422-40e6...@iw9g2000pbc.googlegroups.com...

> On Jul 31, 2:42 am, "LudovicoVan" <ju...@diegidio.name> wrote:
>> "Virgil" <vir...@ligriv.com> wrote in message
>> news:virgil-71471F....@bignews.usenetmonster.com...

> > > If that were so then |N as an ordered set (sequence) would have to
> > > contain oo.
> >
> > Still your basic mistake: to contain oo, that would be N* := N u { oo },
> > then you have an a_oo (in R*).
> >

> > I'll state the main point again: [I should have left it as it was,
> > this comment was totally incorrect.]

>
> You can have natural integers with the one-point compactification of
> the point at infinity (say, N+, or the point oo) without having the
> hyperintegers (N*, or *N).

That is what I was talking about: N* := N u { oo }, not the hyperintegers.

-LV

[Jim Burns]

On 8/1/2012 4:44 AM, LudovicoVan wrote:
> "Ross A. Finlayson" <ross.fi...@gmail.com> wrote in message
> news:a0dbb9de-c422-40e6...@iw9g2000pbc.googlegroups.com...

>> You can have natural integers with the one-point compactification of
>> the point at infinity (say, N+, or the point oo) without having the
>> hyperintegers (N*, or *N).
>
> That is what I was talking about: N* := N u { oo }, not the hyperintegers.

The impression I draw from your posts is that you accept that
there _are no_ functions from N _onto_ R, but that you wish to show
that there _are_ functions from (N u { oo }) _onto_ R.
Is this impression correct?

Assuming that is what you want:

Consider the bijection g: N -> N*, where
g(n) = { oo, n=0
{ n-1, otherwise

A function f: N* -> R is onto _if and only if_
the function f': N -> R defined by f'(n) = f(g(n)) is onto.

That is to say, they are the same problem, with slightly
different decorations.

[Shmuel Metz]

In <jv8mt2$ok0$1...@speranza.aioe.org>, on 07/31/2012

at 02:34 PM, "LudovicoVan" <ju...@diegidio.name> said:

>That's wrongly put: the point just boils down to the fact that,
>to consider a_oo as a real number that belongs or does not
>belong to (x_n), we need consider sequences over N*.

It's not a fact, it's a false assumption.

[LudovicoVan]

"Jim Burns" <burn...@osu.edu> wrote in message
news:501918E4...@osu.edu...
<snipped>

> The impression I draw from your posts is that you accept that
> there _are no_ functions from N _onto_ R

No, I don't. Let's make a quick recap: I have *proven* (no correction, no
mistake) that if we are given an *enumeration* of the reals, the limit
interval is degenerate (it is a singleton: I had wrongly called this
interval "improper"). In a standard setting, with sequences as functions
N->R, I claim that that is enough to prove that Cantor's First Theorem does
not hold in general. It is to the subsequent objection that a_oo (the limit
end-point) itself would not be in the given enumeration that we reply: no,
that needs an extended setting (the sequence would be extended, i.e. a
function N*->R), then nothing is missing anyway.

> That is to say, they are the same problem, with slightly
> different decorations.

It is not as simple as that: without at least compactification some problems
are simply mis-modeled (and the standard definitions of limit inf and sup
are just broken). Also, speaking in general, with transfinite ordinals the
structures we get are much finer.

-LV

[William Hughes]

On Aug 1, 10:16 am, "LudovicoVan" <ju...@diegidio.name> wrote:

<snip>

> ...I have *proven* (no correction, no

> mistake) that if we are given an *enumeration* of the reals, the limit
> interval is degenerate

True, but if we are given an *enumeration*
of the reals then you are the pope.
If you start out with a false premise you
can prove anything.

[LudovicoVan]

"William Hughes" wrote in message
news:fff4a0df-e110-4310...@h5g2000vbl.googlegroups.com...

You simply get a zero in logic: if Cantor's arguments fail, *you* haven't
proved anything at all. Then, for an explicit enumeration, just wait a
little bit...

-LV

[William Hughes]

On Aug 1, 12:49 pm, "LudovicoVan" <ju...@diegidio.name> wrote:
> "William Hughes"  wrote in message
>
> news:fff4a0df-e110-4310...@h5g2000vbl.googlegroups.com...
>
> > On Aug 1, 10:16 am, "LudovicoVan" <ju...@diegidio.name> wrote:
> > <snip>
>
> > > ...I have *proven* (no correction, no
> > > mistake) that if we are given an *enumeration* of the reals, the limit
> > > interval is degenerate
>
> > True, but if we are given an *enumeration*
> > of the reals then you are the pope.
> > If you start out with a false premise you
> > can prove anything.
>
> You simply get a zero in logic: if Cantor's arguments fail, *you* haven't
> proved anything at all.

But Cantor's arguments do not fail. Cantor shows that given a
function
from N to R, there is an element b of R not in the range of the
function.
You introduce the incorrect assumption
"there exists an enumeration" early. If you do this you can get many
contradictions, including the final interval is both non-degenerate
and
degenerate at the same time. Do not introduce the assumption until
after you have shown that any function from N to R defines an element
b
not in the range.

[Jim Burns]

On 8/1/2012 9:16 AM, LudovicoVan wrote:
> "Jim Burns" <burn...@osu.edu> wrote in message
> news:501918E4...@osu.edu...

>> The impression I draw from your posts is that you accept that
>> there _are no_ functions from N _onto_ R
>
> No, I don't. Let's make a quick recap: I have *proven* (no correction,
> no mistake) that if we are given an *enumeration* of the reals, the
> limit interval is degenerate (it is a singleton: I had wrongly called
> this interval "improper"). In a standard setting, with sequences as
> functions N->R, I claim that that is enough to prove that Cantor's First
> Theorem does not hold in general.

Since an enumeration of the reals would be a counter-example to
Cantor's First Theorem, it is to be expected that that would be
enough to prove it does not hold. If I assume that I have an
example where 1+1=3, that would be enough to prove that it is
not a theorem of PA that 1+1=2. But I don't.

Do you have an example of a complete enumeration of the reals?

> It is to the subsequent objection
> that a_oo (the limit end-point) itself would not be in the given
> enumeration that we reply: no, that needs an extended setting (the
> sequence would be extended, i.e. a function N*->R), then nothing is
> missing anyway.

In this new enumeration, a_oo would not be missing. That is not the
same as showing nothing is missing.

Do you agree that there exist bijections between N and N* ?

Do you agree that, if g: N -> N* is a bijection, and
f: N* -> R, f': N-> R, with f'(n) = f(g(n)), then
either f and f' are both onto or neither are?

The original argument still applies to the new enumeration,
so there is some real not in the enumeration, not a_oo but
some other real.

Also, remember that the original enumeration was assumed to be
complete. Despite this, a_oo was proven to not be in that enumeration.
Would you not agree that this makes the assumption of completeness
problematic?

>> That is to say, they are the same problem, with slightly
>> different decorations.
>
> It is not as simple as that:

All I say here is that every enumeration f: N* -> R corresponds to
an enumeration f': N -> R and that either f and f' are both onto
or neither onto. It really is very nearly that simple. I doubt it
it takes more than a couple lines to prove it.

> without at least compactification some
> problems are simply mis-modeled (and the standard definitions of
> limit inf and sup are just broken). Also, speaking in general, with
> transfinite ordinals the structures we get are much finer.

I don't see how compactification, lim sup, lim inf and transfinite
ordinals affect this point. All I see that is needed is the definition
of function, the definition of bijection, the definition of surjection,
and just a bit of logic.