Arbitrariness in set theory: "Every" is not "all".

[WM]

Every initial sequence of digits D_n = d1d2...dn of the antidiagonal D differs from the first n entries L_n of the Cantor-list L. From this it is concluded that D differs from all entries of the Cantor-list and therefore D is not in the list. ∀n ∈ ℕ: D_n ∉ L_n → D ∉ L.

Every initial sequence of digits D_n = d1d2...dn of the antidiagonal D differs from every element of the set X of all irrational numbers because it is rational. This is not used to argue that D is not in the set X of irrational numbers. ∀n ∈ ℕ: D_n ∉ X → D_ ∉ X.

Every rational number can be indexed by a natural number. From this it is concluded that all rational numbers can be indexed by natural numbers.

Every natural number leaves the overwhelming majority of rationals without index. From this it is not concluded that all natural numbers leave the overwhelming majority of rationals without index.

Regards, WM

[Dan Christensen]

On Sunday, July 26, 2020 at 4:17:00 AM UTC-4, WM wrote:

>
> Every natural number leaves the overwhelming majority of rationals without index. From this it is not concluded that all natural numbers leave the overwhelming majority of rationals without index.
>

Check our the visual presentation of an indexing of the positive rational numbers at
https://en.wikipedia.org/wiki/Rational_number#/media/File:Diagonal_argument.svg

You would skip those rational numbers not lowest terms (shown in red). So the indexing would be:

1. 1/1
2. 2/1
3. 1/2
4. 1/3
5. 3/1 (skipping 2/2 since not in lowest terms)
6. 4/1

and so on.

I hope this helps.


Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

[Dan Christensen]

On Sunday, July 26, 2020 at 4:17:00 AM UTC-4, WM wrote:

Anyway, you don't need to construct an indexing of all rational numbers in Cantor's diagonal proof. You don't even need to prove one exists. You are proving only that no countable list of decimal expansions in the interval (0, 1) will include every such number.

[Ross A. Finlayson]

... given that there isn't a function that admits no re-ordering,
that the anti-diagonal's always at the end of the list,
that is embodied by the equivalency function,
a particular unique example among mathematical functions,
that's also interesting for its probabilistic and
differential character, and not just boring.

[Mostowski Collapse]

You could use Lewis logic S1, etc.. which
has negation -A and impossibility ~A.

And then define:

forall x A(x) :<=> -exists x A(x)

each x A(x) :<=> ~exists x A(x)

You would then have neither:

/* wont hold */
1) forall x A(x) => each x A(x)

/* wont hold */
2) each x A(x) => forall x A(x)

2) would hold if the modal logic has the
principle "T". 1) would hold if the modal
logic would degenerate to classical logic.

But if the Barcan Formula holds, we would have:

3) forall x -~A(x) => each x A(x)

[Mostowski Collapse]

Corr.: Typo

forall x A(x) :<=> -exists x -A(x)

each x A(x) :<=> ~exists x -A(x)

Now I am too lazy to figure out whats going on.

[Mostowski Collapse]

Actually you could have 4 different quantifiers:

forall x A(x) :<=> -exists x -A(x)

any x A(x) :<=> -exists x ~A(x)

each x A(x) :<=> ~exists x -A(x)

tutti x A(x) :<=> ~exists x ~A(x)

LMAO!

[Peter]

Mostowski Collapse wrote:
> Actually you could have 4 different quantifiers:
>
> forall x A(x) :<=> -exists x -A(x)
>
> any x A(x) :<=> -exists x ~A(x)
>
> each x A(x) :<=> ~exists x -A(x)
>
> tutti x A(x) :<=> ~exists x ~A(x)
>
> LMAO!
>

See http://matwbn.icm.edu.pl/ksiazki/fm/fm44/fm4412.pdf from the person
after whom your collapse is named.

>

[Mostowski Collapse]

The Mostowski paper shows some counting quantifiers.
I am more interested in modal quantifier, so
that "forall" and "each" would be different modes.

Not all counting quantifiers satisfy:

G |- A(t)
--------------
G |- Qx A(x)

But interestingly in modal logic we would have:

G |- A(t)
--------------------
G |- exists x A(x)

And then by necessity:

G |- exists x A(x)
------------------------
[] G |- [] exists x A(x)

Strange.

[Mostowski Collapse]

So assume the "existential" quantifiers were
exists x A(x) and <>exists x A(x), where <>
is the possibility operator.

Then the two "universal" quantifiers could be,
where - is negation:

forall x A(x) :<=> -exists x -A(x)

each x A(x) :<=> -<>exists x -A(x)

But Lewis had impossibility operator
~F, which we would define nowadays
as ~F := -<>F. So that it would be:

each x A(x) <=> ~exists x -A(x)

[Mostowski Collapse]

Now we can do the necessity game again,
this time:

G |- A(x) x notin G
-------------------
G |- forall x A(x)

And then:

G |- forall x A(x)
------------------------
[] G |- [] forall x A(x)

But [] forall x A(x) <=> [] -exists x -A(x)
<=> -<>exists x -A(x). So im summary we have:

G |- A(x) x notin G
-------------------
[] G |- each x A(x)

LoL

Mostowski Collapse schrieb:

[Peter]

Mostowski Collapse wrote:
> Now we can do the necessity game again,
> this time:
>
>      G |- A(x)           x notin G
>     -------------------
>     G |- forall x A(x)
>
> And then:
>
>     G |- forall x A(x)
>   ------------------------
>   [] G |- [] forall x A(x)
>
> But [] forall x A(x) <=> [] -exists x -A(x)
> <=> -<>exists x -A(x). So im summary we have:
>
>          G |- A(x)           x notin G
>     -------------------
>     [] G |- each x A(x)
>
> LoL

Ruth Barcan Marcus/de re and de dicto modalities...
ah it's all coming back to me...

[Mostowski Collapse]

The possible world could be the WM FISONs,
finite initial segements. We would have
that there exists a varying maximum n:

/* satisfiable in each world */
forall m (n >= m)

But not a varying quark maximum n:

/* not satisfiable in each world */
each m (n >= m)

So the each quantifier simulates the infinite
natural numbers N, although our possible world
are only finite.

So you could do natural numbers N with the
each quantifier, although you dont have
natural numbers as a whole.

Proof:
In each possible world, this here is false

forall n exists m (n < m)

respectively: -exists n -exists m (n < m)

respectively: -exists n forall m (n >= m)

but this here is true in each possible world:

forall n <>exists m (n < m)

respectively: -exists n -<>exists m (n < m)

respectively: -exists n each m (n >= m)

Maybe should make possibility <> behave that it
only looks into the WM future, i.e. from a
smaller segment to larger segments.

Peter schrieb:

[Jim Burns]

On 7/26/2020 4:16 AM, WM wrote:

> Every initial sequence of digits D_n = d1d2...dn of the
> antidiagonal D differs from the first n entries L_n of
> the Cantor-list L.
> From this

No, not from that.
This is why it's not enough for you to _agree_ with arguments
you don't actually understand, not if you intend to share
opinions about them.

*Not* an initial finite sequence of digits, a *single digit* d_n
of D is enough to prove that D -- the whole D -- does not equal
the n_th entry L(n).

There's more than one way to do this. For example, let the
_difference_ between the n_th digit d_n of D and the n_th digit
L(n)_n of the n_th entry be 5, in either direction.
Then *the closest* that D and L_n can be is 4/10^n apart.

> From this it is concluded that D differs from all entries
> of the Cantor-list and therefore D is not in the list.
> ∀n ∈ ℕ: D_n ∉ L_n → D ∉ L.

( abs( d_n - L(n)_n ) = 5 ) ->
( abs(D - L(n)) >= 4/10^n )

( forall n e N, abs(D - L(n)) >= 4/10^n ) ->
( not exists n e N, L(n) = D )

> Every initial sequence of digits D_n = d1d2...dn of the
> antidiagonal D differs from every element of the set X
> of all irrational numbers because it is rational.

Every finite initial d1d2...dn segment of d1d2...
is the finite initial segment of both rational and
irrational numbers.

Therefore, we _should not_ conclude from any finite initial
segment that its continuation is either rational or irrational.
Neither concluding D is rational, when it could be irrational,
nor concluding D is irrational, when it could be rational
are _valid_ (in the "true under all conditions" sense).

> This is not used to argue that D is not in the set X of
> irrational numbers. ∀n ∈ ℕ: D_n ∉ X → D_ ∉ X.

The passive voice here obscures who is saying what.
Are _you_ (WM) saying that we _should_ conclude from a finite
initial segment d1d2...dn that d1d2... is _rational_ ?
Are _you_ (WM) saying that we _should_ conclude from a finite
initial segment d1d2...dn that d1d2... is _irrational_ ?

"It is argued" that we are not justified in concluding
either of those (argued by essentially everyone else).

What do you (WM) argue? And how do you justify it?

Do you understand the importance of being able to
_justify_ one's claims?

[WM]

Am Montag, 27. Juli 2020 20:34:09 UTC+2 schrieb Jim Burns:
> On 7/26/2020 4:16 AM, WM wrote:
>
> > Every initial sequence of digits D_n = d1d2...dn of the
> > antidiagonal D differs from the first n entries L_n of
> > the Cantor-list L.
> > From this
>

> *Not* an initial finite sequence of digits, a *single digit* d_n
> of D is enough to prove that D -- the whole D -- does not equal
> the n_th entry L(n).

But that is impossible to prove, because every checked digit belongs to an initial sequence of digits D_n = d1d2...dn.

>
> There's more than one way to do this. For example, let the
> _difference_ between the n_th digit d_n of D and the n_th digit
> L(n)_n of the n_th entry be 5, in either direction.
> Then *the closest* that D and L_n can be is 4/10^n apart.

You should at least try to understand that every n belongs to a finite initial segment. More is not possible to check.

This is why it's not enough for you to _agree_ with arguments you don't actually understand, not if you intend to share opinions about them.
>

> > From this it is concluded that D differs from all entries
> > of the Cantor-list and therefore D is not in the list.

So you agree with all I wrote but only say that you deny it.

> > ∀n ∈ ℕ: D_n ∉ L_n → D ∉ L.

Just this is an invalid claim from every finite initial segment to the complete sequence.

>
> Every finite initial d1d2...dn segment of d1d2...
> is the finite initial segment of both rational and
> irrational numbers.

Irrationalo numbers have no digit representation and therefore have no finite initial segment.

>
> Therefore, we _should not_ conclude from any finite initial
> segment that its continuation is either rational or irrational.
> Neither concluding D is rational, when it could be irrational,
> nor concluding D is irrational, when it could be rational
> are _valid_ (in the "true under all conditions" sense).
>
> > This is not used to argue that D is not in the set X of
> > irrational numbers. ∀n ∈ ℕ: D_n ∉ X → D_ ∉ X.
>
> The passive voice here obscures who is saying what.
> Are _you_ (WM) saying that we _should_ conclude from a finite
> initial segment d1d2...dn that d1d2... is _rational_ ?
> Are _you_ (WM) saying that we _should_ conclude from a finite
> initial segment d1d2...dn that d1d2... is _irrational_ ?

I see that you don't understand this simple arguing again. I say what every sober mind should know: Every digit represents a rational. Infinitely many digits represent infinitely many rational numbers.

Regards, WM

[Mostowski Collapse]

Whats your new connective and logic that could
express "forall" and "each" side by side. Verbal
hand waving will lead you to nowhere.

Thats neighter how philosophy nor math works.
It will only show your Cantor psychosis.

"Psychosis is an abnormal condition of the mind
that results in difficulties determining what
is real and what is not real."
https://en.wikipedia.org/wiki/Psychosis

LoL

[Mostowski Collapse]

That we can have simulatenously:

∀n ∈ ℕ: D_n ∉ X

D ∈ X

Is a form of dedekind completness,
where a cut is also a set of rational
numbers, but the correspond real

number need not be rational.

https://en.wikipedia.org/wiki/Completeness_of_the_real_numbers#Dedekind_completeness

But only a pyschotic crazy crank
can mix dedekind completness with
diagonal argument AND then

claim he has two different quantifiers.
Its rather some brain cells are missing
in Augsburg crank institute.

LMAO!

[Jim Burns]

On 7/27/2020 5:34 PM, WM wrote:
> Am Montag, 27. Juli 2020 20:34:09 UTC+2
> schrieb Jim Burns:
>> On 7/26/2020 4:16 AM, WM wrote:

>>> Every initial sequence of digits D_n = d1d2...dn of the
>>> antidiagonal D differs from the first n entries L_n of
>>> the Cantor-list L.
>>> From this
>>
>> *Not* an initial finite sequence of digits, a *single digit* d_n
>> of D is enough to prove that D -- the whole D -- does not equal
>> the n_th entry L(n).
>
> But that is impossible to prove,

Let the 8_th digit d_8 of D be 2.
Let the 8_th digit L(8)_8 of L(8) be 7.

The closest that D and L(8) can be would be if
-- all the digits _before_ the eighth were equal, and
-- all the digits in the lower number raise the value
as much as possible (are all 9's)
-- all the digits in the higher number lower the value
as much as possible (are all 0's)

For example, we might have
D = 0.12345672999...
L(8) = 0.12345677000...

Or, we might have
D = 3.14159622999...
L(8) = 3.14159267000...

Or, something else.

In every close-as-possible case,
abs(D - L(8)) = 0.00000004

This is why we can say, _no matter what D and L(8) are_
if abs( d_8 - L(8)_8 ) = 5
then abs(D - L(8)) >= 0.00000004

Generalizing over _all_ the digits,
if abs( d_n - L(n)_n ) = 5
then abs(D - L(n)) >= 4/10^n

Would you like of proof of this to set beside all
the other proofs that you've snipped?

> because every checked digit belongs to an initial sequence
> of digits D_n = d1d2...dn.

Note that
if abs( d_n - L(n)_n ) = 5
then abs(D - L(n)) >= 4/10^n
AND
every digit belongs to a finite initial segment of digits.

It's almost as though you don't know what "because" means.

>> There's more than one way to do this. For example, let the
>> _difference_ between the n_th digit d_n of D and the n_th digit
>> L(n)_n of the n_th entry be 5, in either direction.
>> Then *the closest* that D and L_n can be is 4/10^n apart.
>
> You should at least try to understand that every n belongs to
> a finite initial segment. More is not possible to check.
>
> This is why it's not enough for you to _agree_ with arguments
> you don't actually understand, not if you intend to share
> opinions about them.

You're not very good at irony, are you?
It would have worked better for you if you hadn't snipped
me saying exactly that in my previous post.

And also, of course, if I didn't understand Cantor's diagonal
argument, which I do and you don't.

"Every n belongs to a finite initial segment" is irrelevant.

[Mostowski Collapse]

WMs Pyschosis in a nutshell. "Yesterday I
was in the park, I first saw a white swan

(something of the form (A->B)), later I saw
a black swan (something of the form (A->~B).

Conclusion set theory is inconsistent."

LoL

[WM]

Am Dienstag, 28. Juli 2020 00:34:20 UTC+2 schrieb Jim Burns:

> every digit belongs to a finite initial segment of digits.

Therefore it represents a rational number.

>
> It's almost as though you don't know what "because" means.

Every digit represents a rational number and every sequence of digits that you can define represents a rational number, because no digit is simultaneously definable and representing an irrational number. Note Cantor's theorem B: "Among the numbers of the set there is always a smallest one."

> > You should at least try to understand that every n belongs to
> > a finite initial segment. More is not possible to check.
> >
> > This is why it's not enough for you to _agree_ with arguments
> > you don't actually understand, not if you intend to share
> > opinions about them.
>
> You're not very good at irony, are you?

That is not my profession.

>
> "Every n belongs to a finite initial segment" is irrelevant.

Says the believer in absurdity. Better you should claim to be Christ. Then we could check whether you can go over water.

Regards, WM

[Ross A. Finlayson]

It's interesting. One wonders how it's about
quantification, also the terms under iteration,
and separately, collection.

I.e., where the predicate's true for one or all,
and for example one and not all or all and not one.

Thanks, this is a bit past syllogism to the
contrapositive.

[Mostowski Collapse]

My argument in favor of modale logic. Maybe
this could explaun WMs psychosis. Ordinary
people would do the following:

forall x e N A(x)

With modal logic we could formulate WMs idee
fixe, namely not to use actual infinity N,
but use initial segments N' only:

each x e N' A(x)

The two would say then same, if the quantifier
"each" behaves modally, for example if we would
use these two definitions:

forall x e N A(x) :<=> -exists x e N -A(x)

each x e N' A(x) :<=> -<>exists x e N' -A(x)

<> does not denote necessity [], but it would
denote possibility. This would pretty much sum up
WMs psychosis.

But what is good for? WM would move infinity to the
background, in that in the surface, he would not
anymore talk about infinitely many elements if N,

but instead he would nevertheless have an infinity,
namely there are many initial segments N'. WM
repeatedly quantifies over N':

He writes on de.sci.mathematik:

WM haluziniert:
in **einem** einzigen endlichen Anfangsabschnitt

That somehow belongs into the loony bin. WM replaces
quantification over:

1
2
3
4
...

By quantification over initial segments, and then its
elements:

{1}
{1,2}
{1,2,3}
{1,2,3,4}
...

I am only a little bit interested in this loony bin stuff,
because the modal operator <> would hide the plurality
of N'. So maybe the pyschosis of WM can be explained

that he accepts the plurality of N', namely the possibility
<> of an arbitrary N', but rejects the direct plurality
of elements from N. Possibly because he is under

the impression that this brings him closer to his
salvation, not to use the evil N. But if one would
analyze the modal logic more closely, one would

see that infinity has been shifted into the frame
of possible worlds. Techically not much happened.
Salvation is still in danger, since all FISONs

instead of N is used. This would be all clever,
if something interesting would come out of this
venture. But WM only creates nonsense, like

for example dark numbers. LoL

[Jim Burns]

On 7/28/2020 3:27 AM, WM wrote:
> Am Dienstag, 28. Juli 2020 00:34:20 UTC+2
> schrieb Jim Burns:

>> every digit belongs to a finite initial segment of digits.
>
> Therefore it represents a rational number.

Every digit belongs to a finite initial segment of
_an infinite sequence_ of digits. Therefore, it "represents"
either a rational _or an irrational_ number.

I borrowed your phrasing, but this is also incorrect.
It's _all_ of the digits and which places they are which
represent a number, whether rational or irrational.

Compare this to:
Look at any atom.
No one is able to say whether that atom is part of a
living creature or not _by looking only at that atom_
This does not "prove" that there are no living creatures.

Here is the non-arbitrary difference between the two arguments
that you're promoting as some kind of arbitrariness.
-- A single digit is enough to know two numbers are not equal.
-- No finite initial segment _of an infinite digit sequence_
is enough to know if the number is rational or irrational.

[...]

>> You're not very good at irony, are you?
>
> That is not my profession.

I would say "Don't quit your day job" if your day job weren't
teaching nonsense to your students. As it is, though, hell, yeah,
quit your day job.

[George Greene]

On Sunday, July 26, 2020 at 4:17:00 AM UTC-4, WM wrote:
> Every initial sequence

Nobody gives a shit. The quantifier does not say anything about sequence
of anything. It ranges OVER ALL of the domain, NOT in any particular order --
you really should think of the ranging as happening IN PARALLEL RATHER THAN
serially, in ANY order -- NO order is relevant. That is why an extensional
class theory provides the classic examples of these domains -- that is why
the model theory of these theories is usually a class theory. That is why
the usual model-construction language for FOL is ZFC.

> of digits D_n = d1d2...dn of the antidiagonal D differs from the first n

No, WE are NOT reasoning from initial segments -- we take ONE arbitrary n and
insist that it differs AT THAT n. FROM THAT ONE n, it FOLLOWS LOGICALLY that
it differs for 1) each, 2) every, and 3) ALL n.
All of these ARE ENTIRELY synonymous AND this inference is SOUND, as long as
"n" (not n) wasn't a term that previously appeared in the language.

[WM]

Am Dienstag, 28. Juli 2020 19:00:13 UTC+2 schrieb Jim Burns:
> On 7/28/2020 3:27 AM, WM wrote:
> > Am Dienstag, 28. Juli 2020 00:34:20 UTC+2
> > schrieb Jim Burns:
>
> >> every digit belongs to a finite initial segment of digits.
> >
> > Therefore it represents a rational number.
>
> Every digit belongs to a finite initial segment of
> _an infinite sequence_ of digits. Therefore, it "represents"
> either a rational _or an irrational_ number.

No. A digit d_n represents the fraction d_n/10^n. Even infinitely many digits have no digit at omega. Therefore all digits represent fractions.

>
> I borrowed your phrasing, but this is also incorrect.
> It's _all_ of the digits and which places they are which
> represent a number, whether rational or irrational.
>
> Compare this to:
> Look at any atom.
> No one is able to say whether that atom is part of a
> living creature or not _by looking only at that atom_
> This does not "prove" that there are no living creatures.

But it proves that atoms are atoms.

>
> Here is the non-arbitrary difference between the two arguments
> that you're promoting as some kind of arbitrariness.
> -- A single digit is enough to know two numbers are not equal.
> -- No finite initial segment _of an infinite digit sequence_
> is enough to know if the number is rational or irrational.

If you give me a digit sequence, then I know that you give a fraction.

Irrational numbers can only be given by algorithms (which produce digit sequences which in turn approximate the irrational number - but never hit them).

Regards, WM

[WM]

Am Dienstag, 28. Juli 2020 20:02:40 UTC+2 schrieb George Greene:
> On Sunday, July 26, 2020 at 4:17:00 AM UTC-4, WM wrote:
> > Every initial sequence

> > of digits D_n = d1d2...dn of the antidiagonal D differs from the first n
>
> No, WE are NOT reasoning from initial segments

You cannot escape the finite because every checked digit d_n sits at a finite place n.

> -- we take ONE arbitrary n and
> insist that it differs AT THAT n.

Every position where it could sit is finite. You cannot go further because there is no digit at omega.

> FROM THAT ONE n, it FOLLOWS LOGICALLY that
> it differs for 1) each, 2) every, and 3) ALL n.

It follows as well logically that each and every and all d_n do not define an irrational number (and each and every and all n belong to a finite set).

Regards, WM

[Ross A. Finlayson]

Burse finds his way out of a box, it's laudable.

[Alan Smaill]

WM <wolfgang.m...@hs-augsburg.de> writes:

> Am Dienstag, 28. Juli 2020 20:02:40 UTC+2 schrieb George Greene:
>> On Sunday, July 26, 2020 at 4:17:00 AM UTC-4, WM wrote:
>> > Every initial sequence
>> > of digits D_n = d1d2...dn of the antidiagonal D differs from the first n
>>
>> No, WE are NOT reasoning from initial segments
>
> You cannot escape the finite because every checked digit d_n sits at a
> finite place n.

Likewise you cannot show that sqrt(2) is irrational,
because each checked pair of naturals (n,m) sits at a finite
place in the listing of pairs.

Yet you accept that sqrt(2) is irrational.

WM is inconsistent.

--
AS

[Ross A. Finlayson]

I'd encourage you to consider that for questions of universals
in the universal quantifier, that while it is indeed about
domains and ranges and ranging in the domain, that these logical
case do introduce what would be the correct form for a formalism,
a disambiguation among the logical cases that correctly see them
hold in their tersest form.

One way to consider it is that it's a relativization, that writing
different labels for the universal quantifier, do make for that
what is inaccessible to some ground model is well-defined for
what otherwise is logically incomplete.

Kind of like set theory and part theory having eventually the
same theorems and even eventually (in their course) being the
same objects, here is that eventually the theorems about these
logical cases is written in particulars these different direct
interpretations of each/any/every/all.

[Jim Burns]

On 7/29/2020 10:02 AM, WM wrote:
> Am Dienstag, 28. Juli 2020 19:00:13 UTC+2
> schrieb Jim Burns:

>> Every digit belongs to a finite initial segment of
>> _an infinite sequence_ of digits. Therefore, it "represents"
>> either a rational _or an irrational_ number.
>
> No. A digit d_n represents the fraction d_n/10^n.
> Even infinitely many digits have no digit at omega.
> Therefore all digits represent fractions.

Infinite sequences of digits represent unique number line points
without having digits at infinite places. _How_ they're represented
is not how you _think_ they're represented.

Consider the infinite sequence of finite initial segments.
D_0 = 3.
D_1 = 3.1
D_2 = 3.14
D_3 = 3.141
D_4 = 3.1415
D_5 = 3.14159
...

Each finite initial segment is an entry at a finite index.

Each entry D_k draws a circle C_k around which points
the _whole_ infinite sequence might represent.
C_k has a width 1/10^k
Each circle draws a circle 90% tighter than the previous circle.
All the circles contain more than one point.

However, there is at most one point that is in all the circles.

Suppose that x and y are both in C_k.
d = abs(x-y)
1/10^k >= d >= 0

If both x and y are in _all_ circles, then
forall (finite) k, 1/10^k >= d

But this is only true if d = 0,
and, if d = abs(x-y) = 0, then x = y

Therefore, you can't have x and y in all C_k and ~(x = y).

But this result doesn't apply to any finite initial segment.

>> I borrowed your phrasing, but this is also incorrect.
>> It's _all_ of the digits and which places they are which
>> represent a number, whether rational or irrational.
>>
>> Compare this to:
>> Look at any atom.
>> No one is able to say whether that atom is part of a
>> living creature or not _by looking only at that atom_
>> This does not "prove" that there are no living creatures.
>
> But it proves that atoms are atoms.

I'll concede that atoms are atoms and that rationals are
rationals.
Will you concede that infinite sequences of digits represent
unique points on the number line?

[Mostowski Collapse]

It would explain WMs psychosis, but not show any
inconsistency of set theory.

It would help to pinpoint, where WMs psychosis
creates an alternative reality.

"Psychosis is an abnormal condition of the mind
that results in difficulties determining what
is real and what is not real."
https://en.wikipedia.org/wiki/Psychosis

[Mostowski Collapse]

Lets check this out, about WMs psychosis:

1. Elevated Mood
==> Euphoric, inappropriate laughter, singing
Writes transfinity PDF full of shit

2. Increased Motor Activity or Energy
==> Absent
Spends all day in his hammock

3. Sexual Interest
==> Mildly or possibly increased
Dreams of Berliner Mädchengymnasium

4. Sleep
==> Sleeping less than normal amount by up to one hour
Posts erarly morning 06:00 about p/q --> 2^p*3^q

5. Irritability
==> Frequently irritable; short, curt
Most easily trollable crank

6. Speech: Rate & Amount
==> Push; consistently increased rate and amount;
Pretty inhibited spammer frequenty copy pasting his own nonsense

7. Language: Thought Disorder
==> Distractible; loses goal of thought; change
topics frequently; racing thoughts
Doesn't stay on topic often changes topic

8. Content
==> Grandiose or paranoid ideas; ideas of reference
Does not exercise mathematics, rather humanities against
his anti-hero Cantor

9. Disruptive or Aggressive Behavior
==> Sarcastic; loud at times, guarded
Calls his audience poor-spirited, think he satisfies some thirsting

10. Appearance
==> Appropriate dress and grooming
Dresses like a teacher, likes outdoors

11. Insight
==> Denies any behavior change
Is totally immune to mathematics

https://psychology-tools.com/test/young-mania-rating-scale

[WM]

Am Mittwoch, 29. Juli 2020 16:55:02 UTC+2 schrieb Alan Smaill:
> WM <wolfgang.m...@hs-augsburg.de> writes:
>
> > Am Dienstag, 28. Juli 2020 20:02:40 UTC+2 schrieb George Greene:
> >> On Sunday, July 26, 2020 at 4:17:00 AM UTC-4, WM wrote:
> >> > Every initial sequence
> >> > of digits D_n = d1d2...dn of the antidiagonal D differs from the first n
> >>
> >> No, WE are NOT reasoning from initial segments
> >
> > You cannot escape the finite because every checked digit d_n sits at a
> > finite place n.
>
> Likewise you cannot show that sqrt(2) is irrational,
> because each checked pair of naturals (n,m) sits at a finite
> place in the listing of pairs.

The irrationality of sqrt2 can be proven by contradiction:
2q^2 = p^2 ==> p is even ==> q is even.
No digit required.

Regards, WM

[WM]

Am Mittwoch, 29. Juli 2020 19:24:13 UTC+2 schrieb Jim Burns:
> On 7/29/2020 10:02 AM, WM wrote:
> > A digit d_n represents the fraction d_n/10^n.
> > Even infinitely many digits have no digit at omega.
> > Therefore all digits represent fractions.
>
> Infinite sequences of digits represent unique number line points
> without having digits at infinite places. _How_ they're represented
> is not how you _think_ they're represented.
>
> Consider the infinite sequence of finite initial segments.
> D_0 = 3.
> D_1 = 3.1
> D_2 = 3.14
> D_3 = 3.141
> D_4 = 3.1415
> D_5 = 3.14159
> ...
>
> Each finite initial segment is an entry at a finite index.
>
> Each entry D_k draws a circle C_k around which points
> the _whole_ infinite sequence might represent.
> C_k has a width 1/10^k
> Each circle draws a circle 90% tighter than the previous circle.
> All the circles contain more than one point.
>
> However, there is at most one point that is in all the circles.

You never complete all circles.

>
> Suppose that x and y are both in C_k.
> d = abs(x-y)
> 1/10^k >= d >= 0
>
> If both x and y are in _all_ circles,

But there are not all circles.

> I'll concede that atoms are atoms and that rationals are
> rationals.
> Will you concede that infinite sequences of digits represent
> unique points on the number line?

Why should I support obvious nonsense?

Regards, WM

[Mostowski Collapse]

Well you used infinitely many pairs
p,q. But you wrote on de.sci.mathematik:

"Ich gebe zu, noch nie bis ins Unendliche
gezählt zu haben. Das ist aber auch nicht ratsam."
https://groups.google.com/d/msg/de.sci.mathematik/3cnmDRpVprQ/A6OGzRF9BAAJ

That sqrt2 is irrational means for all p,q
sqrt=/=p,q. In Augsburg Crank institute
such proof is never possible, since

every demonstrated pair p,q leaves the
overwhelming majority of pairs without
demonstration. From this it is not

concluded that all pairs are demonstrated.

[Mostowski Collapse]

Corr.:

That sqrt2 is irrational means for all p,q

sqrt=/=p/q. In Augsburg Crank institute

such proof is never possible, since

BTW: Credits Alan Smaill

[Alan Smaill]

I'm afraid you are in serious contradiction with yourself
(nothing new there).

In the potentially infinite collection
{ (2,1), (3,2), (3,1), (4,3),(4,2),(4,1), (5,4) .. }
you cannot escape the finite, because every checked
pair sits at a finite place n.

QED.

> Regards, WM

--
Alan Smaill

[WM]

Am Mittwoch, 29. Juli 2020 21:20:02 UTC+2 schrieb Alan Smaill:
> WM <wolfgang.m...@hs-augsburg.de> writes:
>

> > The irrationality of sqrt2 can be proven by contradiction:
> > 2q^2 = p^2 ==> p is even ==> q is even.
> > No digit required.
>

> In the potentially infinite collection
> { (2,1), (3,2), (3,1), (4,3),(4,2),(4,1), (5,4) .. }
> you cannot escape the finite, because every checked
> pair sits at a finite place n.

I do not use numbers here but the property that the factor 2 makes all numbers being even.

Further I *can* use proofs by complete induction, for instance, in order to prove that no digit sequence represents an irrational number.

Regards, WM

[Ross A. Finlayson]

For a "more complete" set theory, where after Goedel
there's incompleteness of the ordinary set theory for
consistency, it's simply more complete for a broader
vision.

To help explain transfer principle and a primitive,
or primary, machinery as what implements it
symbolically in a notation, for the quantifier
(and the implicit infinitely many quantifiers
in front of it) and the variable (for what it is),
such notions as refining the universal quantifier
make some sense for those whose goals in foundations
include consistency and completeness.

(And concreteness, in constancy.)

[Mostowski Collapse]

Its rather hiding the foundation. If you delegate
infinity to the possibility operator <>.

This is the opposite goal of set theory. Set theory
wants to excavate the foundation.

[Mostowski Collapse]

WMs math does not give more completness, he himself
halucinates dark numbers. His con artistry resembles:

https://en.wikipedia.org/wiki/The_Pont_Neuf_Wrapped

[Ross A. Finlayson]

It's a property of predicates corresponding to their domains,
basically as what is extracted from tagging predicates to
tagging the quantifier about them.

[Mostowski Collapse]

Yes and no. When you excavate foundation, you
don't use predicates anymore. The domain of a
predicate is the domain of discourse, in
set theory its V, not a set anymore,

the "universal set" V is not a set.
See Russels paradox:

https://en.wikipedia.org/wiki/Russell's_paradox

You would abandone the FOL terminology "predicate",
and maybe start talk about set-like "relations".
The good thing for "predicates" there is even
an alternative word. But then it the past Dan-O-Matik

and Crank "Me" could even not make the step from
FOL "functions" to set-like "functions". Maybe if
there were a second name things would be more
easier. Lets call the set-like functions "maps",

and the FOL functions, which also act on V,
the "universal set", which is not a set,
"functions". What does mathematics deal with
"functions" or "maps"?

[Mostowski Collapse]

You would then have a language:

Foundation Mathematics
-----------------------------
Predicates Relations
Functions Maps

So to some extend foundation could
also talk about itself, or how
it relates to mathematics.

With large cardinal axioms you
can map Predicates and Functions
back to Relations and Maps,

and then view foundation as a
mathematical object itself,
applying foundation to foundation.

or don't use a large cardinal axiom
and be happy with classes as some
kind of objectivization of certain

predicates and functions.

[Mostowski Collapse]

You might find some "tagging" in
type theory, the competitor for set
theory. Go figure out what is

foundation and what is mathematics
in type theory. Can you?

[Me]

On Thursday, July 30, 2020 at 5:16:54 PM UTC+2, Mostowski Collapse wrote:

> and Crank "Me" could even not make the step from
> FOL "functions" to set-like "functions".

Shut up, you silly crank! The one who is not able to grasp the difference is YOU.

Holy shit!

Still taking drugs, man?

[Mostowski Collapse]

LoL

[Jim Burns]

On 7/29/2020 10:02 AM, WM wrote:

> Irrational numbers can only be given by algorithms

There aren't enough algorithms for all the points that exist
on the number line.

What should we mean by "all the points"?

This is an essential step in reasoning _validly_ about points
(reasoning by true statements about a point which could be any point).
We describe the points _validly_ in order to reason validly about them.

(i)
There are no positive-length gaps in the points.

I don't expect you to dispute (i). There exist _at least_ the
points given by algorithms, and there are no positive-length
gaps in those. More points won't create gaps.

(ii)
There are no single-point gaps in the points.

If we _cut_ the number line C (for "continuum")
(separate its points into two collections L and H,
such that x e L & y e H -> x < y )

then either there IS a point z that _determines_ L/{z} and H/{z}
( L/{z} = { u e C | u < z }, H/{z} = { u e C | z < u }
or there ISN'T such a point z.

If there ISN'T such a cut-determining point z for L and H,
I call that a _single-point gap_ between L and H.

Part (ii) says, for every cut L,H of the number line, there
IS a point that determines it, that there ISN'T a gap.

----
Suppose that, for the number line,
(i) there are no positive-length gaps, and
(ii) there are no single-point gaps.

Suppose that there are a countable infinity of algorithms
which give points in the number line.

Assign a natural number to each of these algorithmic points.
a[1], a[2], a[3], ...

There is at least one point without an algorithm, without
any natural number assigned to it.

Proof:
There exists an infinite sequence of intervals
I[1], I[2], I[3], ...

with algorithmic endpoints,
-- I[k] = ( a[p], a[q] )
nested
-- k > j -> I[k] sub I[j] ),
where, for each interval I[k] = ( a[p], a[q] ),
any algorithmic point in I[k] has a higher index
than its endpoints
-- ( a[r] e ( a[p], a[q] ) ) -> ( r > p & r > q )

If a[r] was an algorithmic point in _all_ of the intervals I[k],
its index r would need to be higher than _all_ of the
indexes of _all_ the I[k]. But there are infinitely many
indexes. r cannot be a natural number index.
Therefore, there is no _algorithmic_ point a[r] in all
the intervals.

However, the sequence I[k] determines a cut L,H, and,
by (ii), there must exist a point z that determines L,H.
This point z is in all of the intervals I[k].
Being in all of the intervals, z is not an algorithmic point.

Therefore, there exists at least one point in the number line
not given by any algorithm.

> (which produce digit sequences which in turn approximate
> the irrational number - but never hit them).

An infinite digit sequence does not determine a point by
hitting the point.

An infinite digit sequence _rules out_ every point except one
by one or another finite initial segment of digits. The unique
point which is NOT ruled out by any finite initial segment
is the unique point which IS represented by the infinite digit
sequence.

[WM]

Am Donnerstag, 30. Juli 2020 19:27:17 UTC+2 schrieb Jim Burns:
> On 7/29/2020 10:02 AM, WM wrote:
>
> > Irrational numbers can only be given by algorithms
>
> There aren't enough algorithms for all the points that exist
> on the number line.

That is wrong. Uncountability is nonsense.

> An infinite digit sequence does not determine a point by
> hitting the point.

Therefore infinitely many points remain.

>
> An infinite digit sequence _rules out_ every point except one

No, that would require to finish the sequence. But an infinite sequence has infinitely many digits following every digit.

Regards, WM

[Ross A. Finlayson]

Is not a usual philosophy a philosophy, technical,
with "Being" and "Nothing" and "Time" in Kant,
Hegel, and Heidegger?

There's to leave out "Leviathan" but it's baggage.

I.e. it's for the technical parts, Kant, Hegel, and
Heidegger, Being, Nothing, and Time, and Kant
with Thing-Itself also.

[Ross A. Finlayson]

It's been a function theory here a few years.

(The set theory.)

"That there's a pure set theory" is a fundamental
theorem of mathematics.

Please to be keeping in mind the type theory and
the category theory as quite most usual theories,
and that there are pure theories of them are
fundamental theorems of mathematics.

[Jim Burns]

On 7/30/2020 5:24 PM, WM wrote:
> Am Donnerstag, 30. Juli 2020 19:27:17 UTC+2 schrieb Jim Burns:

>> An infinite digit sequence does not determine a point by
>> hitting the point.
>
> Therefore infinitely many points remain.

An infinite digit sequence determines a single point, but
it doesn't do this by hitting that point.

>> An infinite digit sequence _rules out_ every point except one
>
> No, that would require to finish the sequence.

Consider the infinite digit sequence
3(.)1 4 1 5 9 2 6 5 3 5 ...

After 1 digit (3.), points < 2 and > 5 are ruled out.
The sequence is not finished.
After 2 digits (3.1), points < 3.0 and > 3.3 are ruled out.
The sequence is not finished.
After 3 digits (3.14), points < 3.13 and > 3.16 are ruled out.
The sequence is not finished.
After 4 digits (3.141), points < 3.140 and > 3.143 are ruled out.
The sequence is not finished.

Every point _except one_ is ruled out by one or another of
the finite initial segments of 3(.)1 4 1 5 9 2 6 5 3 5 ...
Being finite, none of them finish the sequence.

How do we know there is only one which is not ruled out?

forall d > 0, exists k e N such that 1/10^k < d

For any two distinct points x and y, abs(x-y) = d,
there is some k such that either x or y is ruled out by
a finite initial segment of length k.

We declare the value of the infinite digit sequence to be
that one point which is not ruled out by any finite initial segment.

> But an infinite sequence has infinitely many digits
> following every digit.

There is an infinite sequence following 3(.)1 4
At the same time, 3.13 and 3.16 are ruled out,
and remain ruled out. So?

Your argument seems to be that you FEEL like it should be wrong,
and things that make you FEEL that way are "matheology".

[Jim Burns]

On 7/30/2020 5:24 PM, WM wrote:

> Am Donnerstag, 30. Juli 2020 19:27:17 UTC+2
> schrieb Jim Burns:
>> On 7/29/2020 10:02 AM, WM wrote:

>>> Irrational numbers can only be given by algorithms
>>
>> There aren't enough algorithms for all the points that exist
>> on the number line.
>
> That is wrong. Uncountability is nonsense.

We reason about _all_ the points-given-by-algorithms and
about _all_ the points-in-the-number-line the same way that
we reason about _all_ the green frogs (Lithobates clamitans).

We make _valid_ claims true of each of them.
("A green frog is a frog. A green frog is green.")

And we apply truth-preserving rules to reason our way
to new claims, which, because of how we arrived at them,
we know are also true of each of them.
("Green frogs are not writing desks.")

We _do not_ hunt through all the streams and ponds of the world,
checking for green-frog/writing-desks. You seem to think that's
the only way we can know things.

The new claims aren't necessarily even a little interesting
or surprising. But they are well-justified.

In some instances, the new claims ARE interesting, surprising,
even (dare I say it?) ) _counter-intuitive_
The same justification applies in the counter-intuitive
instances as applies in the deadly-boring instances.
They do not prove set theory is inconsistent.
They prove intuition is less than all-knowing.

[Hagen Schwaß]

Am Sonntag, 26. Juli 2020 10:17:00 UTC+2 schrieb WM:

In finite sets I can use every=all. Finally I can use an expression between a finite set A and an infinte set B saying every x in A : x something in B.

[Hagen Schwaß]

Am Samstag, 1. August 2020 12:11:10 UTC+2 schrieb Hagen Schwaß:
> Am Sonntag, 26. Juli 2020 10:17:00 UTC+2 schrieb WM:
>
> In finite sets I can use every=all. Finally I can use an expression between a finite set A and an infinte set B saying every x in A : x something in B.

And further I can use every=all within an infinite set itself: for every x in A : x something in A

[WM]

Am Freitag, 31. Juli 2020 18:29:05 UTC+2 schrieb Jim Burns:

> An infinite digit sequence determines a single point, but
> it doesn't do this by hitting that point.

But it hits infinitely many points. The sequence for pi for instance hits 3, 3.1, 3.14, ... and so on. It hits aleph_0 fractions. Every additional digit hits and determines one additional fraction - but not more. When should this stop?

>
> >> An infinite digit sequence _rules out_ every point except one
> >
> > No, that would require to finish the sequence.
>
> Consider the infinite digit sequence
> 3(.)1 4 1 5 9 2 6 5 3 5 ...
>
> After 1 digit (3.), points < 2 and > 5 are ruled out.
> The sequence is not finished.
> After 2 digits (3.1), points < 3.0 and > 3.3 are ruled out.
> The sequence is not finished.
> After 3 digits (3.14), points < 3.13 and > 3.16 are ruled out.
> The sequence is not finished.
> After 4 digits (3.141), points < 3.140 and > 3.143 are ruled out.
> The sequence is not finished.
>
> Every point _except one_ is ruled out

Never. That would require a last digit - and would determine a fraction.

>
> How do we know there is only one which is not ruled out?
>
> forall d > 0, exists k e N such that 1/10^k < d
>
> For any two distinct points x and y, abs(x-y) = d,
> there is some k such that either x or y is ruled out by
> a finite initial segment of length k.
>
> We declare the value of the infinite digit sequence to be
> that one point which is not ruled out by any finite initial segment.

Any finite initial segment does not rule out uncountably many points.

>
> > But an infinite sequence has infinitely many digits
> > following every digit.
>
> There is an infinite sequence following 3(.)1 4
> At the same time, 3.13 and 3.16 are ruled out,
> and remain ruled out. So?
>
> Your argument seems to be that you FEEL like it should be wrong,

You can be proved wrong. For every n in |N the interval 1/10^n contains 2^aleph_0 real numbers.

Even if you feel that infinitely many natnumbers should contain and infinite number, your feeling is wrong. Therefore the interval is infinitely often decreased but it remains an uncountable set - if uncountable sets exist.

Regards, WM

[WM]

Am Freitag, 31. Juli 2020 19:19:42 UTC+2 schrieb Jim Burns:

> We reason about _all_ the points-given-by-algorithms and
> about _all_ the points-in-the-number-line the same way that
> we reason about _all_ the green frogs (Lithobates clamitans).
>
> We make _valid_ claims true of each of them.

Who has proved that bijections between infinite sets show their equinumerosity? On the contrary, it is easy to prove that they do not. The even natural numbers and the natural numbers are provably not equinumerous. Therefore the "bijection" is not really a bijection but a delusion. That is fact.

> In some instances, the new claims ARE interesting, surprising,
> even (dare I say it?) ) _counter-intuitive_
> The same justification applies in the counter-intuitive
> instances as applies in the deadly-boring instances.
> They do not prove set theory is inconsistent.
> They prove intuition is less than all-knowing.

What you call intuition is logical thinking. What you call logic is wrong.

In a recent appraisal of Russell's contribution to mathematical logic he {{Gödel}} says that the paradoxes reveal "the amazing fact that our logical intuitions are self-contradictory". I confess that in this respect I remain steadfastly on the side of Brouwer who blames the paradoxes not on some transcendental logical intuition which deceives us but on an error inadvertendly commited in the passage from finite to infinite sets. [p. 234] [H. Weyl: "Philosophy of mathematics and natural science", Princeton Univ. Press (2009)]

Me too.

Regards, WM

[Ross A. Finlayson]

It's usually attributed to Galileo.

Two sets with a bijection between them
have a same size by that definition.
There are other definitions,
usually comparative density in the
naturals or rationals.

Half of the integers are even, via density.

Of course it's even been so that "half of
the integers are even" since times of Virgil.

[Jim Burns]

On 8/1/2020 10:15 AM, WM wrote:
> Am Freitag, 31. Juli 2020 18:29:05 UTC+2
> schrieb Jim Burns:

>> An infinite digit sequence determines a single point, but
>> it doesn't do this by hitting that point.
>
> But it hits infinitely many points.

Hitting a point is not how an infinite digit sequence
determines a point. How many different ways have I said this?

It hits infinitely many points -- *points that the infinite*
*digit sequence does not represent*

So, it hits infinitely many points. *So what* ?

> The sequence for pi for instance hits 3, 3.1, 3.14, ...
> and so on.
> It hits aleph_0 fractions. Every additional digit hits and
> determines one additional fraction - but not more.
> When should this stop?

A better question is "Why should we care?"
These aleph_0 fractions are not what the infinite digit
sequence represents.

>>>> An infinite digit sequence _rules out_ every point except one
>>>
>>> No, that would require to finish the sequence.
>>
>> Consider the infinite digit sequence
>> 3(.)1 4 1 5 9 2 6 5 3 5 ...
>>
>> After 1 digit (3.), points < 2 and > 5 are ruled out.
>> The sequence is not finished.
>> After 2 digits (3.1), points < 3.0 and > 3.3 are ruled out.
>> The sequence is not finished.
>> After 3 digits (3.14), points < 3.13 and > 3.16 are ruled out.
>> The sequence is not finished.
>> After 4 digits (3.141), points < 3.140 and > 3.143 are ruled out.
>> The sequence is not finished.
>>
>> Every point _except one_ is ruled out
>
> Never. That would require a last digit

You FEEL -- despite being corrected for YEARS now --
that an infinite digit sequence represents a point by
*hitting* the point.

That's not how an infinite digit sequence represents a point.
Too bad about your FEELINGS, but there it is.

[Jim Burns]

On 8/1/2020 10:15 AM, WM wrote:

> Am Freitag, 31. Juli 2020 18:29:05 UTC+2
> schrieb Jim Burns:

>> How do we know there is only one which is not ruled out?
>>
>> forall d > 0, exists k e N such that 1/10^k < d
>>
>> For any two distinct points x and y, abs(x-y) = d,
>> there is some k such that either x or y is ruled out by
>> a finite initial segment of length k.
>>
>> We declare the value of the infinite digit sequence to be
>> that one point which is not ruled out by any finite initial
>> segment.
>
> Any finite initial segment does not rule out uncountably many
> points.

Each finite initial segment rules out certain _segments_ of
the number line.

Raising the question of how many points are ruled out is a distraction.

For one finite initial segment D_k of digits of length k, where
D_k represents p, open segments (-inf, p) and ( p+1/10^k, +inf )
are ruled out.

We declare the value of the infinite digit sequence to be

that _one_ point which is not ruled out by any finite initial
segment.

It's not _two_ points "not ruled out by any finite initial segment"
because _two_ points x and y must be some positive distance d apart,
and, for any positive distance d, there is some _finite_ initial
segment D_k of length k such that, 1/10^k < d,
If x is "not ruled out by any finite initial segment,
then x is not ruled out by D_k.
If x is not ruled out by D_k, then x is in [ p, p+1/10^k ].
If x is in [ p, p+1/10^k ], then y is not in [ p, p+1/10^k ].
If y is not in [ p, p+1/10^k], then y is ruled out by D_k.
If y is ruled out by D_k,
then y is NOT "not ruled out by any finite initial segment".

For all different x and y, x and y are NOT BOTH "not ruled out

by any finite initial segment".

Because

forall d > 0, exists k e N such that 1/10^k < d

>>> But an infinite sequence has infinitely many digits
>>> following every digit.
>>
>> There is an infinite sequence following 3(.)1 4
>> At the same time, 3.13 and 3.16 are ruled out,
>> and remain ruled out. So?
>>
>> Your argument seems to be that you FEEL like it should be wrong,
>
> You can be proved wrong. For every n in |N the interval 1/10^n
> contains 2^aleph_0 real numbers.

Irrelevant.

forall d > 0, exists k e N such that 1/10^k < d

This is how we know there aren't _two_ points "not ruled out

by any finite initial segment".

> Even if you feel that infinitely many natnumbers should contain
> and infinite number, your feeling is wrong.

For any two points x and y which are a positive distance d apart,
there is a *FINITE* natural k such that 1/10^k < d.

I have been explicit throughout that I'm NOT referring to
any "infinite" natural number k.

> Therefore the interval is infinitely often decreased
> but it remains an uncountable set

"The" interval is each interval in (for example) the sequence
[ 3., 4. ], [ 3.1, 3.2 ], [ 3.14, 3.15 ], ...

Each interval is decreased from the previous interval.
Each interval is an uncountable set.
*No more than one point is in all intervals*

We declare the value of the infinite digit sequence to be

that _one_ point which is not ruled out by any finite initial
segment.

> - if uncountable sets exist.

This is a different question from
"Is there more than one point not ruled out by any
finite initial segment of digits?"

[Jim Burns]

On 8/1/2020 10:29 AM, WM wrote:
> Am Freitag, 31. Juli 2020 19:19:42 UTC+2
> schrieb Jim Burns:

>> We reason about _all_ the points-given-by-algorithms and
>> about _all_ the points-in-the-number-line the same way that
>> we reason about _all_ the green frogs (Lithobates clamitans).
>>
>> We make _valid_ claims true of each of them.
>
> Who has proved that bijections between infinite sets show
> their equinumerosity?

You don't know what a proof is.
You don't know what a definition is.
How is someone supposed to communicate with you?

----
For the moment, forget the words "bijection" and "equinumerous".

Consider two sets B and C such that
forall x in B, exists _unique_ y in C
forall y in C, exists _unique_ x in B

What would you _like_ to call that situation?
Is whatever name you chose going to suggest that
B has more elements than C?
Why, given that forall x in B, exists _unique_ y in C ?
That C has more elements than B?
Why, given that forall y in C, exists _unique_ x in B ?

In fact, though, we say that B and C are equinumerous,
because "equinumerous" is an apt description of B and C.

----
The "paradox" (not-a-paradox) in all this is that some sets
are equinumerous with some of their proper subsets.

We all know that. We all _have known_ that.
We call these sets "Dedekind infinite".
Being a Dedekind set only contradicts your FEELINGS that
there shouldn't be any Dedekind infinite sets.
Too bad about your FEELINGS.
How many decades have you been at this?
Isn't it past time for you to suck it up and move on?

> On the contrary, it is easy to prove that they do not.
> The even natural numbers and the natural numbers are provably
> not equinumerous.

Oh, Lordy.
No.
"They are equinumerous" *MEANS* "There is a bijection".
Get help.

You picked a nice example, though.

Consider the set of natural numbers, labeled in binary:
{ 1, 10, 11, 100, 101, ... }
Append a '0' to the right of each label:
{ 10, 100, 110, 1000, 1010, ... }

Does appending a '0' to the right of each label mean
there are now half as many elements in that set?

I ask because that set with appended '0' is the set of
even natural numbers, in binary.
{ 10, 100, 110, 1000, 1010, ... }

And, *DON'T BLAME THAT ON SET THEORY*. That's all yours.

[WM]

Am Samstag, 1. August 2020 19:41:36 UTC+2 schrieb Jim Burns:
> On 8/1/2020 10:15 AM, WM wrote:

> > The sequence for pi for instance hits 3, 3.1, 3.14, ...
> > and so on.
> > It hits aleph_0 fractions. Every additional digit hits and
> > determines one additional fraction - but not more.
> > When should this stop?
>
> A better question is "Why should we care?"

You should care because a digit cannot define or represent two numbers.

> >>
> >> Every point _except one_ is ruled out
> >
> > Never. That would require a last digit
>
> You FEEL -- despite being corrected for YEARS now --
> that an infinite digit sequence represents a point by
> *hitting* the point.

That is so for every finite digit sequence in mathematics. An infinite sequence is merely a never ending sequel where each member obeys mathematics.

You feel that it should be different. But that's not the case.

>
> We declare the value of the infinite digit sequence to be
> that one point which is not ruled out by any finite initial
> segment.

There remain infinitely many such points after any finite initial segment, all points of the interval 1/10^n.

Regards, WM

[WM]

Am Samstag, 1. August 2020 22:05:29 UTC+2 schrieb Jim Burns:
> On 8/1/2020 10:29 AM, WM wrote:

> > Who has proved that bijections between infinite sets show
> > their equinumerosity?
>
> You don't know what a proof is.
> You don't know what a definition is.
> How is someone supposed to communicate with you?

So you don't know who. But I know: Nobody.

>
> Consider two sets B and C such that
> forall x in B, exists _unique_ y in C
> forall y in C, exists _unique_ x in B
>
> What would you _like_ to call that situation?

I call it the claim of a fool or a fraud: In mathematics "for all" means that a last one can be shown. It does not mean that no exception can be shown.

> In fact, though, we say that B and C are equinumerous,
> because "equinumerous" is an apt description of B and C.

It is obvious that the prime numbers and the natural numbers are not equinumerous. So there is a blatant lie.

>
> ----
> The "paradox" (not-a-paradox) in all this is that some sets
> are equinumerous with some of their proper subsets.

The paradox is that there are people who believe these blatant lies.

>
> We all know that. We all _have known_ that.
> We call these sets "Dedekind infinite".

You all suffer from a malfunction of your brain. Otherwise this sitiuation could not continue.

> Being a Dedekind set only contradicts your FEELINGS that
> there shouldn't be any Dedekind infinite sets.

It contradict the provable fact that there are not Dedekind infinite complete sets. There are Dedekind mappings between potentially infinite sets. Remember what he said:

"Every time when there is a cut (A1, A2) which is not produced by a rational number, we create a new, an irrational number  which we consider to be completely defined by this cut (A1, A2);" [R. Dedekind: "Stetigkeit und irrationale Zahlen", 6th ed., Vieweg, Braunschweig (1960) p. 13]

He considered only potential infinity.

Regards, WM

[Mostowski Collapse]

WM halucinates: "The even natural numbers and

the natural numbers are provably not equinumerous."

Well they are equinumerous by this definition:

"In mathematics, two sets or classes A and B are
equinumerous if there exists a one-to-one correspondence
(a bijection) between them"
https://en.wikipedia.org/wiki/Equinumerosity

Here is a bijection:

f : N -> E
f(x)=2*x

[Jim Burns]

On 8/2/2020 10:13 AM, WM wrote:
> Am Samstag, 1. August 2020 19:41:36 UTC+2
> schrieb Jim Burns:
>> On 8/1/2020 10:15 AM, WM wrote:

>>>> Every point _except one_ is ruled out
>>>
>>> Never. That would require a last digit

Each finite initial segment has a last digit, and each finite
initial segment rules out _some_ of the points in the line
(as definitely NOT represented by the whole infinite sequence).

All of the finite initial sequences together rule out
_all but one_ of the points. The one point NOT ruled out
is the one point that the infinite sequence represents.

>> You FEEL -- despite being corrected for YEARS now --
>> that an infinite digit sequence represents a point by
>> *hitting* the point.
>
> That is so for every finite digit sequence in mathematics.
> An infinite sequence is merely a never ending sequel
> where each member obeys mathematics.

What you _mean_ by
"obeys mathematics"
is
"obeys your (WM's) FEELINGS".

Infinite sequences do not represent points the same way
finite sequences represent points, despite your FEELINGS.

We declare the value of the infinite digit sequence to be

that _one_ point which is not ruled out by any finite initial
segment.

[Jim Burns]

On 8/2/2020 10:29 AM, WM wrote:
> Am Samstag, 1. August 2020 22:05:29 UTC+2
> schrieb Jim Burns:

>> Consider two sets B and C such that
>> forall x in B, exists _unique_ y in C
>> forall y in C, exists _unique_ x in B
>>
>> What would you _like_ to call that situation?
>
> I call it the claim of a fool or a fraud:
> In mathematics "for all" means that a last one can be shown.
> It does not mean that no exception can be shown.

"For all" means that no exception _exists_ in the domain,
shown or not shown, showable or not showable.

Did you see where cosmologists have determined that the whole
cosmos is _at least_ 250 times the size of our observable piece
of it? Our mathematics describes the part which _exists_ beyond
what we can observe even in principle.

https://www.technologyreview.com/2011/02/01/197279/cosmos-at-least-250x-bigger-than-visible-universe-say-cosmologists/

>> Being a Dedekind set only contradicts your FEELINGS that
>> there shouldn't be any Dedekind infinite sets.
>
> It contradict the provable fact

...he says, proving he doesn't know what a proof is...

> It contradict the provable fact that there are not Dedekind
> infinite complete sets. There are Dedekind mappings between
> potentially infinite sets. Remember what he said:
>
> "Every time when there is a cut (A1, A2) which is not produced
> by a rational number, we create a new, an irrational number 
> which we consider to be completely defined by this cut (A1, A2);"
> [R. Dedekind: "Stetigkeit und irrationale Zahlen", 6th ed.,
> Vieweg, Braunschweig (1960) p. 13]
>
> He considered only potential infinity.

There are uncountably many Dedekind cuts, as I'm sure
Richard Dedekind knew. It has essentially the same proof as
for uncountably many real numbers, since Dedekind cuts
*are* the real numbers in every important sense.

Did you have a point here?

[Ross A. Finlayson]

Keep in mind that Dedekind cuts are defined _after_
defining the complete ordered field, not just rationals.

Other notions of cuts include Veronese and Peano.
The usual notion of infinitesimals as "bullets"
(iota-values) are in a sense also cuts.

(And look like line reals that a modern definition of
line continuity besides (Dedekind's) field continuity.)

They're topology's, and it's not so much however
many of them are as that they are there.

Dedekind cuts are a model of rational bounds of reals
by their usual order of the rationals, laws of arithmetic.
But, they're not "the equivalence classes of
sequences that are Cauchy", in the development:
the required sense. I.e., _after_ defining the
complete order field and Least Upper Bound axiom,
_then_ Dedekind cuts are about rational bounds,
but, they get gaplessness, which the rationals don't,
from Least Upper Bound and the development.


I.e. proofs of Dedekind's cuts being uncountable and
of the cardinality of the continuum and though only
defined by rationals that the open and closed about
them sew up, are of course only after Cauchy/Weierstrass
and Least Upper Bound.

Of course, various definitions of "continuous" in function
make the rationals or a function defined on them look
"continuous" (or meeting the definition, that arbitrarily
small differences in domain have arbitrarily small differences
in range), but conscientiously 24/7, the rationals
are countable or continuous, not both.

The line reals are countable and continuous - only one function.

The signal reals or after the rationals that simply enough
"doubling the density" of the rationals arrives at covers
for all points, besides neighborhoods, again one wouldn't
conscientiously let stray the required formal surrounds
for soundness and constancy in definition.

(This is where uncountable is after limit ordinals anyways,
that basically between the discrete so well-defined by all
the rationals and the continuous thus well-defined by
completing the signal, is limit over that, too,
that it's a special case according to the definition
of "function" of the "numbers" in the "set theory".)

The field reals are quite usual about the modular and algebra.

Most continously-variable processes are parameterized by time,
which according to a clock-principle is uniform and constant,
and thus that "geometry is motion".

One wonders such notions might be lost on the retro-finitist,
but, the conscientious formalist keeps them.

I.e. that mathematics is a truth not a game,
though that of course it's sublime.

I have here three definitions of "continuous" after various
corresponding notions of "discrete", line, signal, and field
continuity and for models of real numbers in set theory
(the model theory the function theory the number theory,
the descriptive set theory). This is called "repleteness".

[Jim Burns]

On 8/2/2020 1:47 PM, Ross A. Finlayson wrote:
> On Sunday, August 2, 2020 at 10:15:23 AM UTC-7,
> Jim Burns wrote:
>> On 8/2/2020 10:29 AM, WM wrote:

>>> He considered only potential infinity.
>>
>> There are uncountably many Dedekind cuts, as I'm sure
>> Richard Dedekind knew. It has essentially the same proof as
>> for uncountably many real numbers, since Dedekind cuts
>> *are* the real numbers in every important sense.
>>
>> Did you have a point here?
>
> Keep in mind that Dedekind cuts are defined _after_
> defining the complete ordered field, not just rationals.

No, this is incorrect.

The cuts on the rationals are a model of the real numbers.
So are the cuts on points given by algorithms.
So are the cuts on finite decimals.

A Dedekind cut L,H on, for example, the finite decimals is able to
represent _any_ point x in the line, even though the finite decimals
are clearly NOT all of the points in the line.

How L,H is able to do this is by determining all the points
in the line _WHICH ARE NOT_ x. L,H is a photographic negative
of x, if anyone remembers "photographs" and "film" anymore.

The points y in L are all less than this unique x
_AND ALSO_ all the points u in the line < any point y in L
And similarly for points in H.

Thus, even though the point in L and the points in H are NOT
all the points in the line, they still DETERMINE all the
points in the line _that are not x_

(We should take care not to count twice what's essentially the
same cut, as L,H and L',H' with x in L and x in H', but we can
pick one in the definition of cut.)

By defining addition, subtraction, multiplication and division
on the _cuts_ to correspond to operations on the points _between_
the cuts, we get our model of _all_ the points in the line and
their operations. Even though we only use the finite decimals in
our model.

> _then_ Dedekind cuts are about rational bounds,
> but, they get gaplessness, which the rationals don't,
> from Least Upper Bound and the development.

"Least upper bound" is one way to express gaplessness.

Dedekind cuts get gaplessness/least-upper-bound as a property
of being Dedekind cuts.

The least upper bound of a bounded, non-empty collection of
Dedekind cuts is the _set union_ of the cuts in that collection.
Provably.

I suppose we could say that this collection gets the existence
of a least upper bound from the existence of a set union.
Well, that, and the definition of "Dedekind cut".

I think it's a very pretty result.

[WM]

Am Sonntag, 2. August 2020 17:16:30 UTC+2 schrieb Mostowski Collapse:
> WM: "The even natural numbers and

> the natural numbers are provably not equinumerous."
>
> Well they are equinumerous by this definition:

This definition is in contradiction to mathematics.

>
> "In mathematics, two sets or classes A and B are
> equinumerous if there exists a one-to-one correspondence
> (a bijection) between them"
> https://en.wikipedia.org/wiki/Equinumerosity

That shows that there is not a bijection between even natural numbers and natural numbers. There is not even a bijection between |N and |N U [0}.

>
> Here is a bijection:
>
> f : N -> E
> f(x)=2*x

It is wrong. Proof by mathematics: For every large finite set we have about twice as many naturals than even naturals. Therefore quinumerosity can be excluded. Hence no bijection possible.

Regards, WM

[WM]

Am Sonntag, 2. August 2020 18:25:28 UTC+2 schrieb Jim Burns:

> All of the finite initial sequences together rule out
> _all but one_ of the points. The one point NOT ruled out
> is the one point that the infinite sequence represents.

You get boring. Try to learn some maths: Every digits leaves infinitely many possibilities open, namely an interval including uncountably many reals.

> We declare the value of the infinite digit sequence to be
> that _one_ point which is not ruled out by any finite initial
> segment.

Every digit belongs to a finite sequence and is followed by infinitely many, each of which is followed by infinitely many. If you are unable to understand that, then try to define a real number by a digit sequence without a finite algorithm.

Regards, WM

[WM]

Am Sonntag, 2. August 2020 19:15:23 UTC+2 schrieb Jim Burns:


> "For all" means that no exception _exists_ in the domain,
> shown or not shown, showable or not showable.

Your "for all" means that you believe that no exception exists. But that is irrelevant because it cannot be proved for infinite sets and is even wrong.

>
> There are uncountably many Dedekind cuts, as I'm sure
> Richard Dedekind knew.

Hardly. Every Dedekind cut has a finite definition. At his times all mathematicians knew that there are only finite definitions.

> Did you have a point here?

Dedekind created real numbers. Not even aleph_0 could be created during the lifetime of the universe.

Regards, WM

[Ralf Bader]

You are so incredibly stupid.

[Jim Burns]

On 8/2/2020 4:04 PM, WM wrote:
> Am Sonntag, 2. August 2020 18:25:28 UTC+2
> schrieb Jim Burns:

>> All of the finite initial sequences together rule out
>> _all but one_ of the points. The one point NOT ruled out
>> is the one point that the infinite sequence represents.
>
> You get boring.

If you decide to make more interesting errors, or even
merely different boring ones, I certainly won't complain.

> Try to learn some maths: Every digits leaves infinitely many
> possibilities open, namely an interval including uncountably
> many reals.

Every individual finite initial segment _is not_ the infinite
sequence.

All of the finite initial sequences together rule out
_all but one_ of the points. The one point NOT ruled out
is the one point that the infinite sequence represents.

>> We declare the value of the infinite digit sequence to be
>> that _one_ point which is not ruled out by any finite initial
>> segment.
>
> Every digit belongs to a finite sequence and is followed by
> infinitely many, each of which is followed by infinitely many.
> If you are unable to understand that, then try to define a
> real number by a digit sequence without a finite algorithm.

There are more real numbers than algorithms.

There is a real number x different from each real number f(k)
of an infinite sequence of real numbers, a number such that
x is different from f(1) in their 1_st digits,
x is different from f(2) in their 2_nd digits,
x is different from f(3) in their 3_rd digits,
x is different from f(4) in their 4_th digits,
x is different from f(5) in their 5_th digits,
...

[Jim Burns]

On 8/2/2020 4:13 PM, WM wrote:
> Am Sonntag, 2. August 2020 19:15:23 UTC+2
> schrieb Jim Burns:

>> "For all" means that no exception _exists_ in the domain,
>> shown or not shown, showable or not showable.
>
> Your "for all" means that you believe that no exception exists.

The kind of reasoning that uses "for all" _starts_ with
statements that are true _for all_ individuals in a domain.

Very often, these statements (axioms) describe these individuals.
We might say "FOR ALL natural numbers, there exists a successor".

Since these statements (axioms) describe what we intend to
reason about, why, YES, there are no exceptions.
For example, there are no exceptions to the statement about
natural numbers and successors, because any supposed natural
number that DIDN'T have a successor would not be what we were
talking about.

We advance to other statements that ALSO have no exceptions
in the domain, by use of inference rules that preserve truth.
If, for all natural numbers k, the successor of k is not 0,
then (etc etc etc) the successor of k is not k. Theorems.

Theorems can turn out to be very surprising, on occasion.
This DOES NOT mean there's a contradiction somewhere.

> But that is irrelevant because it cannot be proved for
> infinite sets and is even wrong.

You also want a _proof_ of the definition of
"equinumerous".

[Mostowski Collapse]

WM halucinated: "Every Dedekind cut has a finite definition."

Nope. You are confusing Dedekind cuts constructed,
for example from rational numbers.

But a Dedekind cut doesn't need a definition at all.

"Jedesmal nun, wenn ein Schnitt (Al, A2) **vorliegt**,
welcher durch keine rationale Zahl hervorgebracht wird,
so erschaffen wir eine neue, eine irrationale Zahl α,
welche wir als durch diesen Schnitt (Al , A2) vollständig
definiert ansehen"
https://people.math.ethz.ch/~halorenz/4students/Begleitseminar/Dedekind.pdf

The verb **be present** also indicates and
actual infinite view of Dedekind cuts.

[WM]

Am Montag, 3. August 2020 09:02:35 UTC+2 schrieb Mostowski Collapse:


> But a Dedekind cut doesn't need a definition at all.

In matheology most objects need no definition but only belief of the parishioners. But here we are in sci.logic

>
> "Jedesmal nun, wenn ein Schnitt (Al, A2) **vorliegt**,
> welcher durch keine rationale Zahl hervorgebracht wird,
> so erschaffen wir eine neue, eine irrationale Zahl α,
> welche wir als durch diesen Schnitt (Al , A2) vollständig
> definiert ansehen"
> https://people.math.ethz.ch/~halorenz/4students/Begleitseminar/Dedekind.pdf
>
> The verb **be present** also indicates and
> actual infinite view of Dedekind cuts.

"erschaffen" means constructing. The cut cannot "vorliegen" without having been constructet or without being agreed by the community of matheologians.

Regards, WM

[WM]

Am Montag, 3. August 2020 01:58:43 UTC+2 schrieb Jim Burns:
> On 8/2/2020 4:04 PM, WM wrote:


> > Every digits leaves infinitely many
> > possibilities open, namely an interval including uncountably
> > many reals.
>
> Every individual finite initial segment _is not_ the infinite
> sequence.

But it is all that can be given and known.

>
> All of the finite initial sequences together rule out
> _all but one_ of the points.

Try logic: For every n: finite initial segment n is not the infinite sequence.
If every n fails, then every n fails. But more than every n cannot be given or known.

> The one point NOT ruled out
> is the one point that the infinite sequence represents.

Give me a sequence (without finite algorithm) where only one point is not ruled out.

>
> There are more real numbers than algorithms.
>
> There is a real number x different from each real number f(k)
> of an infinite sequence of real numbers, a number such that
> x is different from f(1) in their 1_st digits,
> x is different from f(2) in their 2_nd digits,
> x is different from f(3) in their 3_rd digits,
> x is different from f(4) in their 4_th digits,
> x is different from f(5) in their 5_th digits,
> ...

Up to every k, x is a rational number.

Regards, WM

[WM]

Am Montag, 3. August 2020 02:31:55 UTC+2 schrieb Jim Burns:


> You also want a _proof_ of the definition of
> "equinumerous".

I can prove that your definition is wrong and that it even cannot be satisfied according to your belief

It is wrong that there are as many primes as naturals. Nobody would believe that a person counting all primes counts all natnumbers.

But you believe that Cantor enumerates all fractions. Therefore you must believe that he enumerates all fractions q_n + 100 with q_n a fraction enumerated in the interval (0, 1]. Do you really believe that? Then you can as well believe that the set of primes contains all natnumbers.

Regards, WM

[Mostowski Collapse]

"erschaffen" comes after "vorlegen". Erschaffen means
that a Dedekind Cut implies a irrationale Zahl α.

So basically he constructs irrational numbers from
actual infinity. Namley from the sets A1,A2 in a

Dedekind cut (A1,A2).

[Mostowski Collapse]

The construction is not effective, even if you
would only look at FOL definable cuts (A1,A2)

and (B1,B2) you might not decide α = β. You can
formulate Alan Turings Halteproblem as

real number equality.

[George Greene]

On Monday, July 27, 2020 at 5:34:24 PM UTC-4, WM wrote:
> Am Montag, 27. Juli 2020 20:34:09 UTC+2 schrieb Jim Burns:

> > *Not* an initial finite sequence of digits, a *single digit* d_n
> > of D is enough to prove that D -- the whole D -- does not equal
> > the n_th entry L(n).
>
> But that

Bzzt. Antecedent failure. You have NO IDEA what you mean by "that" here.
WHAT is impossible to prove??
> is impossible to prove,

That a single digit at a single n is enough to prove that D does not equal
L(n) is true BY DEFINITION -- THAT does not NEED to be proved -- THAT is
what "=" MEANS. So your "that" must be referring to some other sentence,
then. Might you possibly have meant that the individual case -- that a single
digit d_n is different from the nth digit of L(n) -- is impossilbe to
prove? Why would that be impossible to prove?? THAT TOO is true BY THE
DEFINITION OF D!

> because every checked digit belongs to an initial sequence
> of digits D_n = d1d2...dn.

SO *FUCKING* what?? Every natural number belongs to the infinite sequence
1,2,3,..., etc., but that does NOT stop anyone from proving that 7 is prime!
The fact that a number occurs in some larger collection has no bearing
whatsoever on anything that anybody is proving about it individually.

Your use of "because" is just incompetent.

Order and sequence ARE NOT relevant to this question.

Cantor's theorem is true OF ALL sets and sets UNLIKE lists usually DON'T
have ORDER! I repeat, all this "checking" (of "checked" digits) you are talking about can be thought of as occurring IN PARALLEL, NOT SERIALLY -- THERE IS NO order. The case for N (which has an inherent order) IS NOT different from the
case FOR ANY OTHER set. Whether the set is finite or infinite DOESN'T MATTER --
that does NOT come up in the proof. Whether the set can be ordered or not DOESN'T MNATTER -- that doesn't come up in the proof of Cantor's theorem either.


This is finally reaching the point where the mental illness is going to have to be ascribed to US for not giving up.

[George Greene]

On Monday, July 27, 2020 at 5:34:24 PM UTC-4, WM wrote:

> You should at least try to understand that
> every n belongs to a finite initial segment.

OF COURSE IT DOES. Why are you trying to ask people to understand
things that are basic to the definition? YOU are the one who doesn't understand!

> More is not possible to check.

NOBODY IS TRYING to check "more"! THE MAXIMUM
number of digits or places THAT ANYbody is trying to "check" is
*ONE*!!

NO MATTER HOW FAR OUT the digit may be, no matter HOW LONG the initial-segment-that-it-ends may be, IT ITSELF is still only ONE digit and comparing it to
the nth digit of L(n) IS TRIVIAL!

[Jim Burns]

On 8/3/2020 8:57 AM, WM wrote:
> Am Montag, 3. August 2020 01:58:43 UTC+2
> schrieb Jim Burns:
>> On 8/2/2020 4:04 PM, WM wrote:

>>> Every digits leaves infinitely many
>>> possibilities open, namely an interval including
>>> uncountably many reals.
>>
>> Every individual finite initial segment _is not_ the
>> infinite sequence.
>
> But it is all that can be given and known.

This is no longer an objection that an infinite sequence of
digits is *ambiguous* (infinitely many left after each digit).

Instead, this most recent objection is that infinite digit
sequences *don't exist* You base it on your own requirement
that it be given or known in order to exist.

Have you (quietly) conceded that your earlier objection is
not correct? The way you're changing the subject makes it look
that way.

So far as this most recent objection about giving and knowing,
keep in mind what I had to say about green frogs and how we can
reason _in the real world_ about a green frog without giving or
knowing that green frog.

But I don't want to lose sight of your old claim that infinite
digit sequences are ambiguous, when we're so close (it seems) to
agreeing that they're not ambiguous -- assuming they exist.

>> There are more real numbers than algorithms.
>>
>> There is a real number x different from each real number f(k)
>> of an infinite sequence of real numbers, a number such that
>> x is different from f(1) in their 1_st digits,
>> x is different from f(2) in their 2_nd digits,
>> x is different from f(3) in their 3_rd digits,
>> x is different from f(4) in their 4_th digits,
>> x is different from f(5) in their 5_th digits,
>> ...
> Up to every k, x is a rational number.

"Up to every k", x is not that number represented by the
first k digits. Your "x is a rational number" is nonsense.

----
For all k, x is not f(k).
From
x#k = ( f(k)#k + 5 ) mod 10
we know
abs(x - f(k)) >= 4/10^k

Every assignment of natural numbers to the reals leaves
at least one real not assigned any natural number.

There are more real numbers than natural numbers.

[Jim Burns]

On 8/3/2020 9:05 AM, WM wrote:

> It is wrong that there are as many primes as naturals.
> Nobody would believe that a person counting all primes
> counts all natnumbers.

The naturals and the primes are not-all-self-injections-onto.

There exists a bijection between the naturals and the primes.
Such a bijection is a self-injection which is not onto.
This is entirely unremarkable.

> But you believe that Cantor enumerates all fractions.

Give p/q the index k, where
k = (p+q-1)*p+q-2)/2 + p

So, yes.

> Therefore you must believe that he enumerates all fractions
> q_n + 100 with q_n a fraction enumerated in the interval (0, 1].
> Do you really believe that?

There are enough naturals to assign one of them to each fraction.
So, yes,
there are enough naturals to assign one of them to _some_ fractions.

Am I to understand that your FEELINGS tell you differntly?

> Then you can as well believe that the set of primes contains
> all natnumbers.

It's too bad about your STILL not knowing what a bijection is.

[Ross A. Finlayson]

Crank!

This is like "ZF minus empty set", all left out.

It seems that there's already Least Upper Bound as
an axiom, which includes irrationals and in fact is
a necessary property after "rationals exist".

So, as equivalence classes, of all the sequences,
that are Cauchy or converge, for that real number
in its value in a modular form, the sets under the
rationals or Dedekind Cuts: make advantage of the
property of being Dedekind-complete, the rationals,
as what the reals are Dedekind-complete and gapless.

Treating the rationals as the real is fair in approximation
and more or less is the rooted definition of equivalence
classes of sequences that are Cauchy, decimal numbers
or modular as what result from the field axioms,
including the completeness axiomatized by "Least Upper
Bound" and for lim sup and lim inf (supremum and infimum).

Not that those are so direct - LUB and lim sup and lim inf,
being equivalent properties, what though they're often
enough assumed together (building and making the bounds
in the bounds).

Excuse me, that's crankish.

Dedekind cuts are gapless what _after_ LUB, which for
example one can assure you using mathematics, can
then be used for regular derivations on "trust".

(That Dedekind cuts with including the rationals and
the irrationals in their neighborhood make them "values"
of partitions in rationals as partitions in reals.)

(Sufficient to represent the range.)

There's already the complete ordered field that
Dedekind hangs his cuts on (about the rationals).

It's even as simple then a model to make the rationals,
what are their values in simple integer models of the
rationals, that these have enough. (Not just using
sets bounded on one side by rationals as real numbers,
but using rationals as real numbers.)

(Which by their real value they are.)

[Jim Burns]

On 8/3/2020 11:48 PM, Ross A. Finlayson wrote:
> On Sunday, August 2, 2020 at 12:37:07 PM UTC-7,
> Jim Burns wrote:

>> Dedekind cuts get gaplessness/least-upper-bound as a property
>> of being Dedekind cuts.
>>
>> The least upper bound of a bounded, non-empty collection of
>> Dedekind cuts is the _set union_ of the cuts in that collection.
>> Provably.
>>
>> I suppose we could say that this collection gets the existence
>> of a least upper bound from the existence of a set union.
>> Well, that, and the definition of "Dedekind cut".
>>
>> I think it's a very pretty result.
>
> Crank!

This is a standard treatment of constructing the reals.

(I need to emphasize that what is generally meant by
"constructing the reals" is NOT what WM wants it to mean.
Given our assumptions, we prove that things exist which behave
like the reals. It is not a requirement in order for them
to exist that any of us define each of them or give each of
them or have any contact in any sense with each of them.
They _exist_ That's the claim.)

> This is like "ZF minus empty set", all left out.
>
> It seems that there's already Least Upper Bound as
> an axiom, which includes irrationals and in fact is
> a necessary property after "rationals exist".

There isn't "already" a least upper bound property.
What we're doing is PROVING that _something_ exists
that has the least upper bound property.
In particular, in this case, we prove that Dedekind cuts
(of a collection of things WITHOUT the least upper bound
property) HAVE the least upper bound property.

If we can do that (Spoiler: We can do that), then we can
declare such and such has the least upper bound property.
If we are challenged on whether such a thing exists,
we can answer,
"If the rationals plus a bit of set theory are consistent,
then these Dedekind cuts exist and have the least upper
bound property."

> So, as equivalence classes, of all the sequences,
> that are Cauchy or converge,

Equivalence classes of Cauchy sequences are a different
construction of the real numbers. They also prove that
_something_ exists which has the least upper bound property.
Having more than one way to prove something is not a bad thing.

[...]

> Excuse me, that's crankish.
>
> Dedekind cuts are gapless what _after_ LUB, which for
> example one can assure you using mathematics, can
> then be used for regular derivations on "trust".

(i)
"Trust" is not a thing, in this context.

(ii)
The rationals do not have the least upper bound property.
The Dedekind cuts of the rationals DO have the least upper
bound property.

The least upper bound of a bounded, non-empty collection B of
Dedekind cuts is the set union UB of B.

It is surely more than coincidence that UB is the smallest set
(by inclusion) which contains (by inclusion) every set in B.
Here, B can be _any_ set, it's what UB _means_ so, B can
also be some collection of Dedekind cuts, too.

That's pretty much game, set, and match. The rest of the proof
shows that the union of an arbitrary bounded, non-empty
collection of Dedekind cuts is also a Dedekind cut.

[...]

[WM]

Am Montag, 3. August 2020 15:53:11 UTC+2 schrieb Mostowski Collapse:

> "erschaffen" comes after "vorlegen".

erschaffen means erschaffen (create). People who believe that there are uncountable Dedekind cuts can also claim that there are uncountable countables.

Regards, WM

[WM]

Am Montag, 3. August 2020 18:41:19 UTC+2 schrieb George Greene:


> SO *FUCKING* what?? Every natural number belongs to the infinite sequence
> 1,2,3,..., etc., but that does NOT stop anyone from proving that 7 is prime!

But it stops everyone from naming every natural number.

The set of definable natural numbers ℕ_def is so small compared to the whole set, that

∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

You can write |ℕ \ ℕ| = 0. But you cannot define all natural numbers.

> The fact that a number occurs in some larger collection has no bearing
> whatsoever on anything that anybody is proving about it individually.
>
> Your use of "because" is just incompetent.

Why don't you try it out? Name all natural numbers individually which are required to get |ℕ \ {1, 2, 3, ...}| = 0. Why do you fail? If you can answer this question, you will understand set theory. But you will probably be even too stupid to ask the question let alone to understand it.

Regards, WM

[WM]

Am Montag, 3. August 2020 18:41:19 UTC+2 schrieb George Greene:

> I repeat, all this "checking" (of "checked" digits) you are talking about can be thought of as occurring IN PARALLEL, NOT SERIALLY -- THERE IS NO order.

I repeat: That is the behaviour of tricksters and of frauds. There is no mathematics, no logic, no honesty in it.

Counting is a serial procedure.

Regards, WM

[WM]

Am Montag, 3. August 2020 18:46:42 UTC+2 schrieb George Greene:
> On Monday, July 27, 2020 at 5:34:24 PM UTC-4, WM wrote:
>
> > You should at least try to understand that
> > every n belongs to a finite initial segment.
>
> OF COURSE IT DOES. Why are you trying to ask people to understand
> things that are basic to the definition?

Because up to every digit at finite position a rational number is defined.

> > More is not possible to check.
>
> NOBODY IS TRYING to check "more"! THE MAXIMUM
> number of digits or places THAT ANYbody is trying to "check" is
> *ONE*!!
>
> NO MATTER HOW FAR OUT the digit may be, no matter HOW LONG the initial-segment-that-it-ends may be, IT ITSELF is still only ONE digit and comparing it to
> the nth digit of L(n) IS TRIVIAL!

It is as trivial to see that all relevant things of the Cantor list are rational. Every digit sequence from the decimal point to the diagonal digit is finite and thus rational. The diagonal up to every such digit is finite and thus rational. More can neither be checked not be created - in particular nothing irrational.

Regards, WM

[Python]

Crank Wolfgang Mueckenheim, aka WM wrote:
> ... of tricksters and of frauds. There is no mathematics, no logic, no
honesty in it.

Very nice and accurate way to describe your whole life, Crank Wolfgang
Mueckenheim, from Hochschule Augbsburg.

[WM]

Am Montag, 3. August 2020 19:18:20 UTC+2 schrieb Jim Burns:

> Instead, this most recent objection is that infinite digit
> sequences *don't exist* You base it on your own requirement
> that it be given or known in order to exist.

If you believe that they exist, then give one.

>
> But I don't want to lose sight of your old claim that infinite
> digit sequences are ambiguous, when we're so close (it seems) to
> agreeing that they're not ambiguous -- assuming they exist.

They are ambiguous because after every digit infinitely many follow. But if you can give an infinite digit sequence that is not ambiguous, then please give it.

Regards, WM

[WM]

Am Montag, 3. August 2020 19:55:15 UTC+2 schrieb Jim Burns:
> On 8/3/2020 9:05 AM, WM wrote:
>
> > It is wrong that there are as many primes as naturals.
> > Nobody would believe that a person counting all primes
> > counts all natnumbers.

> > But you believe that Cantor enumerates all fractions.
>
> Give p/q the index k, where
> k = (p+q-1)*p+q-2)/2 + p
>
> So, yes.
>
> > Therefore you must believe that he enumerates all fractions
> > q_n + 100 with q_n a fraction enumerated in the interval (0, 1].
> > Do you really believe that?
>
> There are enough naturals to assign one of them to each fraction.
> So, yes,
> there are enough naturals to assign one of them to _some_ fractions.
>
> Am I to understand that your FEELINGS tell you differntly?

It is mathematics. Up to every index the enumerated fractions in (100, 101] are far less than in (0, 1]. We can prove that this remains so even in the limit. Do you plead that mathematics should be replaced by belief in this instance?

Regards, WM

[Jim Burns]

On 8/4/2020 5:17 PM, WM wrote:
> Am Montag, 3. August 2020 19:18:20 UTC+2
> schrieb Jim Burns:

>> But I don't want to lose sight of your old claim that infinite
>> digit sequences are ambiguous, when we're so close (it seems) to
>> agreeing that they're not ambiguous -- assuming they exist.
>
> They are ambiguous because after every digit infinitely many
> follow.

They are _unambiguous_ because every two different points have
two different infinite digit sequences. And different _infinite_
sequences of digits must have different _finite_ initial
segments of length k, for some finite k.

Two points x and y with _all_ the the same k-length finite initial
segments of digits must be a distance apart _smaller than any_
_non-zero distance_ That's only true if their distance apart
is zero and so x = y.

For each infinite digit sequence, no more than one point.

> But if you can give an infinite digit sequence that is not
> ambiguous, then please give it.

We reason about about what _exists_
The claims apply to what we give and also
to what we don't give but which exists anyway.

But why follow us, all of a sudden?
Go ahead and give an _ambiguous_ infinite digit sequence.

[Jim Burns]

On 8/4/2020 5:21 PM, WM wrote:
> Am Montag, 3. August 2020 19:55:15 UTC+2
> schrieb Jim Burns:
>> On 8/3/2020 9:05 AM, WM wrote:

>>> It is wrong that there are as many primes as naturals.
>>> Nobody would believe that a person counting all primes
>>> counts all natnumbers.
>
>>> But you believe that Cantor enumerates all fractions.
>>
>> Give p/q the index k, where
>> k = (p+q-1)*p+q-2)/2 + p
>>
>> So, yes.
>>
>>> Therefore you must believe that he enumerates all fractions
>>> q_n + 100 with q_n a fraction enumerated in the interval (0, 1].
>>> Do you really believe that?
>>
>> There are enough naturals to assign one of them to each fraction.
>> So, yes,
>> there are enough naturals to assign one of them to _some_ fractions.
>>
>> Am I to understand that your FEELINGS tell you differntly?
>
> It is mathematics.

"Mathematics" is the name you give to your FEELINGS, so you
answer my question "yes".

The one rule of your "mathematics" is: it is impossible for
your FEELINGS to be wrong.

> Up to every index the enumerated fractions in (100, 101] are
> far less than in (0, 1]. We can prove that this remains so
> even in the limit. Do you plead that mathematics should be
> replaced by belief in this instance?

ALL the fractions in (0, 1] can be bijected with ALL the
fractions in (100, 101].

Also,
ALL the fractions in (0, 1] can be bijected with LESS THAN ALL
the fractions in (100, 101]. That's NOT a contradiction,
whatever your FEELINGS tell you.

[Mostowski Collapse]

Yes "construct from", he says "**so** erschaffen".

So basically he constructs irrational numbers from

actual infinity. Namley from the actual infinte

sets A1,A2 in a Dedekind cut (A1,A2).

Whats wrong with you?

[WM]

Am Mittwoch, 5. August 2020 02:41:31 UTC+2 schrieb Jim Burns:
> On 8/4/2020 5:17 PM, WM wrote:


> > But if you can give an infinite digit sequence that is not
> > ambiguous, then please give it.
>
> We reason about about what _exists_

How can you know what exists if you cannot find, prove, apply it?

> The claims apply to what we give and also
> to what we don't give but which exists anyway.

Where does the objects of your belief exist?

>
> Go ahead and give an _ambiguous_ infinite digit sequence.

That is easy. I always have to stop at a finite digit. But even if I never stop I will never complete the sequence, never get a last digit, never get a unique real number (which moreover would be a fraction). Same with you, even if you believe otherwise.

Regards, WM

[WM]

Am Mittwoch, 5. August 2020 03:22:02 UTC+2 schrieb Jim Burns:
> On 8/4/2020 5:21 PM, WM wrote:


> > It is mathematics.
>
> "Mathematics" is the name you give to your FEELINGS, so you
> answer my question "yes".

Mathematics is the proof for all indices and the derivation of the limit. The sequence a, a, a, ... hs limit a in mathematics. No feeling involved.

> The one rule of your "mathematics" is: it is impossible for
> your FEELINGS to be wrong.

It is impossible for the sequence 1, 1, 1, ... to have limit 2.

>
> > Up to every index the enumerated fractions in (100, 101] are
> > far less than in (0, 1]. We can prove that this remains so
> > even in the limit. Do you plead that mathematics should be
> > replaced by belief in this instance?
>
> ALL the fractions in (0, 1] can be bijected with ALL the
> fractions in (100, 101].

Of course. All the prime numbers can be bijected with all algebraic numbers. Nevertheless the prime numbers are not the algebraic numbers.

>
> Also,
> ALL the fractions in (0, 1] can be bijected with LESS THAN ALL
> the fractions in (100, 101]. That's NOT a contradiction,

Of course not. (By the way that shows what unseless nonsense your alleged "bijections" are.) But that is not the point. The point is that we can prove using first-semester mathematics (nevertheless first-class mathematics, pun intended) that Cantor enumerates *in the limit*, using all finite indices, less than 1 % fractions q_k + 100 of all q_k he enumerates.

> whatever your FEELINGS tell you.

Chuckle. Repeat analysis and finding limits. Perhaps my book may of help for you: [W. Mückenheim: "Mathematik für die ersten Semester", 4th ed., De Gruyter, Berlin (2015)] Note: 4th edition! Many students have learnt mathematics without feelings from it

Regards, WM

[WM]

Am Mittwoch, 5. August 2020 08:49:58 UTC+2 schrieb Mostowski Collapse:
> Yes "construct from", he says "**so** erschaffen".
> So basically he constructs irrational numbers from
> actual infinity.

There they would exist already, created by God.

Regards, WM

[Ross A. Finlayson]

Yeah - you're expected to learn all that yourself and then know.

The "set of reals are the equivalence classes of sequences that are
Cauchy", after Least Upper Bound property, is the "standard" set
of reals. The Dedekind cuts, under rationals as rationals are dense
in the reals, is a reduced, simplified construction, for a value apparatus.

"Trust" is not a thing, in this context: it's subsumed by "truth".

Which I trust....