On Monday, January 17, 2022 at 6:59:37 AM UTC-4, WM wrote:
> William schrieb am Montag, 17. Januar 2022 um 02:30:35 UTC+1:
> > On Sunday, January 16, 2022 at 5:24:29 PM UTC-4, WM wrote:
> > > William schrieb am Sonntag, 16. Januar 2022 um 18:51:48 UTC+1:
> > > > On Sunday, January 16, 2022 at 4:45:55 AM UTC-4, WM wrote:
> > > > > William schrieb am Sonntag, 16. Januar 2022 um 02:50:30 UTC+1:
> > > > >
> > > > > > N_P does not change. A potentially infinite object changes. N_P is not potentially infinite.
> > > > > Construct the largest number you can.
> > > > It is an element of N_P. And it was an element of N_P before you "constructed" it.
> > > Who told you
> > It follows from the fact that N_P does not change. You cannot add an element to it. You can show by induction that N_P does not contain *any* "dark" elements. This includes elements that have not been used,
> By induction we can show that almost all elements are dark

nope. Induction shows the opposite. 1 is not dark, If n is not dark then n+1 is not dark. Thus all element of N_P, including those that will never have been "used", have the property that they are not dark.

> and will never have been used:

"used" is one of your many, many, many. ways of saying "cannot be written down". The fact that an element has the property "cannot be written down" does not mean it cannot have other properties. E.g. it has the property that it is the largest element of a FISON.

--
William Hughes

On Monday, January 17, 2022 at 6:59:37 AM UTC-4, WM wrote:
> William schrieb am Montag, 17. Januar 2022 um 02:30:35 UTC+1:
> > On Sunday, January 16, 2022 at 5:24:29 PM UTC-4, WM wrote:
> > > William schrieb am Sonntag, 16. Januar 2022 um 18:51:48 UTC+1:
> > > > On Sunday, January 16, 2022 at 4:45:55 AM UTC-4, WM wrote:
> > > > > William schrieb am Sonntag, 16. Januar 2022 um 02:50:30 UTC+1:
> > > > >
> > > > > > N_P does not change. A potentially infinite object changes. N_P is not potentially infinite.
> > > > > Construct the largest number you can.
> > > > It is an element of N_P. And it was an element of N_P before you "constructed" it.
> > > Who told you
> > It follows from the fact that N_P does not change. You cannot add an element to it. You can show by induction that N_P does not contain *any* "dark" elements. This includes elements that have not been used,
> By induction we can show that almost all elements are dark

Nope, Induction shows the opposite. 1 is not dark. If n is not dark then n+1 is not dark. Thus every element of N_P, including elements that are never "used", have the property that they are not dark.

> and will never have been used:

"used" is another of the many ways you have of saying "can be written down" Before you try to prove it at again let me specify that almost all elements of N_P will never be "used". The fact that an element of N_P has the property that it "cannot be written down" does not mean it does not have other properties, E,g is the largest element of a FISON.

--
William Hughes

..

Jim Burns schrieb am Montag, 17. Januar 2022 um 18:37:16 UTC+1:

> An example showing why a quantifier shift is not reliable:

Here I show you a proof that cannot be refuted by shouting "quantifier shift". Indexing the fractions

1/1, 1/2, 1/3, 1/4, ...
2/1, 2/2, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ...
....

by natural numbers according to
k = (m + n - 1)(m + n - 2)/2 + m (*)
is called Cantor's first diagonl argument.

We put the indexes X in the first column and mark all not yet indexed fractions by O.

XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
....

Then every index k will be attached to a fraction m/n according to (*). The O's will be removed from the target element but will appear where the index has taken from. I call this change of places between X and O a transposition. Here is the configuration according to (*) for k = 8:

XXXXOOOOO...
XXXOOOOOO...
XOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
XOOOOOOOO...
....

Note that every transposed O is dropped above the X to be applied next. Therefore it is impossible that an O disappears from the board as long as X's are available. Indexing all fractions of the matrix would require all O's to disappear. That is impossible as long as indexes X are in the first column below the dropped O's.

No try to find a way to revert the positions of O's and X's in the first column or to get rid of the O's in another way.

Regards, WM

William schrieb am Montag, 17. Januar 2022 um 19:15:09 UTC+1:
> On Monday, January 17, 2022 at 6:59:37 AM UTC-4, WM wrote:
> > William schrieb am Montag, 17. Januar 2022 um 02:30:35 UTC+1:
> > > On Sunday, January 16, 2022 at 5:24:29 PM UTC-4, WM wrote:
> > > > William schrieb am Sonntag, 16. Januar 2022 um 18:51:48 UTC+1:
> > > > > On Sunday, January 16, 2022 at 4:45:55 AM UTC-4, WM wrote:
> > > > > > William schrieb am Sonntag, 16. Januar 2022 um 02:50:30 UTC+1:
> > > > > >
> > > > > > > N_P does not change. A potentially infinite object changes. N_P is not potentially infinite.
> > > > > > Construct the largest number you can.
> > > > > It is an element of N_P. And it was an element of N_P before you "constructed" it.
> > > > Who told you
> > > It follows from the fact that N_P does not change. You cannot add an element to it. You can show by induction that N_P does not contain *any* "dark" elements. This includes elements that have not been used,
> > By induction we can show that almost all elements are dark
> Nope, Induction shows the opposite. 1 is not dark. If n is not dark then n+1 is not dark. Thus every element of N_P, including elements that are never "used", have the property that they are not dark.

No, that is only provable for n+1 if n has been named.

> > and will never have been used:
> "used" is another of the many ways you have of saying "can be written down"

Yes.

> Before you try to prove it at again let me specify that almost all elements of N_P will never be "used".

That is true.

> The fact that an element of N_P has the property that it "cannot be written down" does not mean it does not have other properties, E,g is the largest element of a FISON.

Maybe. But we cannot know that FISON. It remains dark to us. Otherwise the element is visible, defined, and used.

Regards, WM

On 1/17/2022 1:31 PM, WM wrote:
> Jim Burns schrieb am Montag, 17. Januar 2022 um 18:37:16 UTC+1:
>
>> An example showing why a quantifier shift is not reliable:
>
> Here I show you a proof that cannot be refuted by shouting "quantifier shift". Indexing the fractions
>
> 1/1, 1/2, 1/3, 1/4, ...
> 2/1, 2/2, 2/3, 2/4, ...
> 3/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ...
> ...
>
> by natural numbers according to
> k = (m + n - 1)(m + n - 2)/2 + m (*)
> is called Cantor's first diagonl argument.

wrong. when m=1 and n=1 k should = 1, but you have 0 Fail.

also, in your equation m and n are symetrical, which means when they are switched you get the same k. Fail.

this is not enumeration at all. It is fake math.

>
> We put the indexes X in the first column and mark all not yet indexed fractions by O.

this is not Cantors enumeration at all, which diagonally threads

<snip rest of misleading garf>

On 1/17/2022 1:34 PM, WM wrote:
> William schrieb am Montag, 17. Januar 2022 um 19:15:09 UTC+1:
>> On Monday, January 17, 2022 at 6:59:37 AM UTC-4, WM wrote:
>>> William schrieb am Montag, 17. Januar 2022 um 02:30:35 UTC+1:
>>>> On Sunday, January 16, 2022 at 5:24:29 PM UTC-4, WM wrote:
>>>>> William schrieb am Sonntag, 16. Januar 2022 um 18:51:48 UTC+1:
>>>>>> On Sunday, January 16, 2022 at 4:45:55 AM UTC-4, WM wrote:
>>>>>>> William schrieb am Sonntag, 16. Januar 2022 um 02:50:30 UTC+1:
>>>>>>>
>>>>>>>> N_P does not change. A potentially infinite object changes. N_P is not potentially infinite.
>>>>>>> Construct the largest number you can.
>>>>>> It is an element of N_P. And it was an element of N_P before you "constructed" it.
>>>>> Who told you
>>>> It follows from the fact that N_P does not change. You cannot add an element to it. You can show by induction that N_P does not contain *any* "dark" elements. This includes elements that have not been used,
>>> By induction we can show that almost all elements are dark
>> Nope, Induction shows the opposite. 1 is not dark. If n is not dark then n+1 is not dark. Thus every element of N_P, including elements that are never "used", have the property that they are not dark.
>
> No, that is only provable for n+1 if n has been named.

No.

>
>>> and will never have been used:
>> "used" is another of the many ways you have of saying "can be written down"
>
> Yes.
>
>> Before you try to prove it at again let me specify that almost all elements of N_P will never be "used".
>
> That is true.
>
>> The fact that an element of N_P has the property that it "cannot be written down" does not mean it does not have other properties, E,g is the largest element of a FISON.
>
> Maybe. But we cannot know that FISON. It remains dark to us. Otherwise the element is visible, defined, and used.

which is Observer Dependent Math, a farce.

>
> Regards, WM

On Monday, January 17, 2022 at 8:34:47 PM UTC+1, WM wrote:
>
> No, that is only provable for n+1 if n has [a] name[.]

If you say so. Ok, then I will supply all n e IN with names: The name for n (where n is any natural number) is "|...|" where "|..|" consist of n strokes. Hence the name of the first natural number is just "|", that of the second is "||", etc.

We can do this "recursively" too: 1 has the name "|". For all n e IN: n+1 has the name of n appended by "|".

Hence each and every natural number has a name.

And no, I can't personally "batize" them all, one after the other. :-)

Only Chuck Norris could do that!

On Monday, January 17, 2022 at 3:34:47 PM UTC-4, WM wrote:
> William schrieb am Montag, 17. Januar 2022 um 19:15:09 UTC+1:
> > On Monday, January 17, 2022 at 6:59:37 AM UTC-4, WM wrote:
> > > William schrieb am Montag, 17. Januar 2022 um 02:30:35 UTC+1:
> > > > On Sunday, January 16, 2022 at 5:24:29 PM UTC-4, WM wrote:
> > > > > William schrieb am Sonntag, 16. Januar 2022 um 18:51:48 UTC+1:
> > > > > > On Sunday, January 16, 2022 at 4:45:55 AM UTC-4, WM wrote:
> > > > > > > William schrieb am Sonntag, 16. Januar 2022 um 02:50:30 UTC+1:
> > > > > > >
> > > > > > > > N_P does not change. A potentially infinite object changes. N_P is not potentially infinite.
> > > > > > > Construct the largest number you can.
> > > > > > It is an element of N_P. And it was an element of N_P before you "constructed" it.
> > > > > Who told you
> > > > It follows from the fact that N_P does not change. You cannot add an element to it. You can show by induction that N_P does not contain *any* "dark" elements. This includes elements that have not been used,
> > > By induction we can show that almost all elements are dark
> > Nope, Induction shows the opposite. 1 is not dark. If n is not dark then n+1 is not dark. Thus every element of N_P, including elements that are never "used", have the property that they are not dark.
> No, that is only provable for n+1 if n has been named.

Nope. Nope if n is not dark then n+1 is not dark. This does not require that n has or can be written down ("named" is just another of you many ways of saying "written down".)

.... know[n]... visible, defined, used.

You have a lot of ways of say "can be written down"

--
William Hughes

On Monday, 17 January 2022 at 09:31:38 UTC-4, WM wrote:
> horand....@gmail.com schrieb am Montag, 17. Januar 2022 um 13:15:43 UTC+1:
> > On Monday, 17 January 2022 at 06:56:35 UTC-4, WM wrote:
> > > Jim Burns schrieb am Sonntag, 16. Januar 2022 um 23:11:01 UTC+1:
> > > > We have a technique for reasoning about infinitely-many
> > > > individuals.
> > > Yes, and we can reason that most of them will never have been used. That is tantamount to they cannot be used.
> > There is no way to say of any particular number that it will never be used in the future.
> Of course not. Thar one would have been used.
> > Any one of them can be reasoned about,
> Almost all cannot. Proof: Each used one has ℵo succesors, ℵo of which will be successors of every used number.

That is not a proof that any particular number cannot be used, you fucking imbecile. You are up to your old quantifier switching. Man, you are dense!

måndag 17 januari 2022 kl. 11:56:35 UTC+1 skrev WM:
> Jim Burns schrieb am Sonntag, 16. Januar 2022 um 23:11:01 UTC+1:
> > On 1/16/2022 4:29 PM, WM wrote:
>
> > > But not all can be chosen and used.
> > > Almost all will not have been used at the end of times.
> > > That means they cannot have been used.
> > > That means they are dark.
>
> > We have a technique for reasoning about infinitely-many
> > individuals.
> Yes, and we can reason that most of them will never have been used. That is tantamount to they cannot be used.
>
> Regards, WM
it is entirely irrelevant

William schrieb am Montag, 17. Januar 2022 um 21:43:39 UTC+1:

> Nope. Nope if n is not dark then n+1 is not dark. This does not require that n has or can be written down ("named" is just another of you many ways of saying "written down".)
>
> ... know[n]... visible, defined, used.
>
> You have a lot of ways of say "can be written down"

To explain it better I have to go a long way back: In the OP I have shown that the Cantor function k = (m + n - 1)(m + n - 2)/2 + m used for enumerating all positive fractions

1/1, 1/2, 1/3, 1/4, ...
2/1, 2/2, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ...
....

can be represented by diagrams. When we put all indexes X in the first column and apply them as Cantor prescribed, then we get first, as the starting position:

XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
....

where the X's represent the indexes 1, 2, 3, ...

The index 1 remains at fraction 1/1. The first move, by 2, indexes 1/2:

XXOOOOOOO...
OOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO...
....

and so on: 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, ...

After eight steps for instance we get

XXXXOOOOO...
XXXOOOOOO...
XOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
XOOOOOOOO...
....

Important is, when continuing, that no O can disappear as long as any index X blocks the possible drain, i.e., the first column.

X
X X
..
..
..
O O
O ..
..
..
X X
X ..
..
..

As long as indexes are availabale, the presence of all O's indicates that almost all fractions are not indexed. And after the drain has become free, there are no indexes available which could index the remaining matrix elements covered by O's.

This proves there remain many fractions without indexes. But we cannot find them. They are dark.

In this example I have assumed that all indexes are visible and can be used. But why should they, if there are dark fractions like n/1?

Regards, WM

tisdag 18 januari 2022 kl. 11:41:22 UTC+1 skrev WM:
> William schrieb am Montag, 17. Januar 2022 um 21:43:39 UTC+1:
>
> > Nope. Nope if n is not dark then n+1 is not dark. This does not require that n has or can be written down ("named" is just another of you many ways of saying "written down".)
> >
> > ... know[n]... visible, defined, used.
> >
> > You have a lot of ways of say "can be written down"
> To explain it better I have to go a long way back: In the OP I have shown that the Cantor function k = (m + n - 1)(m + n - 2)/2 + m used for enumerating all positive fractions
> 1/1, 1/2, 1/3, 1/4, ...
> 2/1, 2/2, 2/3, 2/4, ...
> 3/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ...
> ...
> can be represented by diagrams. When we put all indexes X in the first column and apply them as Cantor prescribed, then we get first, as the starting position:
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> ...
> where the X's represent the indexes 1, 2, 3, ...
>
> The index 1 remains at fraction 1/1. The first move, by 2, indexes 1/2:
>
> XXOOOOOOO...
> OOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> ...
> and so on: 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, ...
>
> After eight steps for instance we get
> XXXXOOOOO...
> XXXOOOOOO...
> XOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> XOOOOOOOO...
> ...
> Important is, when continuing, that no O can disappear as long as any index X blocks the possible drain, i.e., the first column.
>
> X
> X
> X
> .
> .
> .
> O
> O
> O
> .
> .
> .
> X
> X
> X
> .
> .
> .
>
> As long as indexes are availabale, the presence of all O's indicates that almost all fractions are not indexed. And after the drain has become free, there are no indexes available which could index the remaining matrix elements covered by O's.
>
> This proves there remain many fractions without indexes. But we cannot find them. They are dark.
>
> In this example I have assumed that all indexes are visible and can be used. But why should they, if there are dark fractions like n/1?
>
> Regards, WM
proves nothing of the sort and your stupid attempt at this proves you are an idiot.

The original bijection is all that matters, not your stupid shit.

zelos...@gmail.com schrieb am Dienstag, 18. Januar 2022 um 13:41:26 UTC+1:

> > To explain it better I have to go a long way back: In the OP I have shown that the Cantor function k = (m + n - 1)(m + n - 2)/2 + m used for enumerating all positive fractions

> > As long as indexes are availabale, the presence of all O's indicates that almost all fractions are not indexed. And after the drain has become free, there are no indexes available which could index the remaining matrix elements covered by O's.
> >
> > This proves there remain many fractions without indexes. But we cannot find them. They are dark.
> >
> proves nothing of the sort

Where does my model fail to follow Cantor's prescription k = (m + n - 1)(m + n - 2)/2 + m ?

Regards, WM

On 1/18/2022 4:41 AM, WM wrote:
> William schrieb am Montag, 17. Januar 2022 um 21:43:39 UTC+1:
>
>> Nope. Nope if n is not dark then n+1 is not dark. This does not require that n has or can be written down ("named" is just another of you many ways of saying "written down".)
>>
>> ... know[n]... visible, defined, used.
>>
>> You have a lot of ways of say "can be written down"
>
> To explain it better I have to go a long way back: In the OP I have shown that the Cantor function k = (m + n - 1)(m + n - 2)/2 + m used for enumerating all positive fractions
>
> 1/1, 1/2, 1/3, 1/4, ...
> 2/1, 2/2, 2/3, 2/4, ...
> 3/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ...
> ...
>
> can be represented by diagrams. When we put all indexes X in the first column and apply them as Cantor prescribed, then we get first, as the starting position:
>
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> ...
>
> where the X's represent the indexes 1, 2, 3, ...
>

that is not Cantor's mapping at all. You associate an index to an entire row, Fail.

On 1/18/2022 8:25 AM, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 18. Januar 2022 um 13:41:26 UTC+1:
>
>>> To explain it better I have to go a long way back: In the OP I have shown that the Cantor function k = (m + n - 1)(m + n - 2)/2 + m used for enumerating all positive fractions
>
>>> As long as indexes are availabale, the presence of all O's indicates that almost all fractions are not indexed. And after the drain has become free, there are no indexes available which could index the remaining matrix elements covered by O's.
>>>
>>> This proves there remain many fractions without indexes. But we cannot find them. They are dark.
>>>
>> proves nothing of the sort
>
> Where does my model fail to follow Cantor's prescription k = (m + n - 1)(m + n - 2)/2 + m ?
>
> Regards, WM

you listed the k's as a column associating it directly with m.

An intentional misdirection.

On 1/17/2022 2:31 PM, WM wrote:
> Jim Burns schrieb
> am Montag, 17. Januar 2022 um 18:37:16 UTC+1:

>> An example showing why a quantifier shift is
>> not reliable:
>
> Here I show you a proof that cannot be refuted by
> shouting "quantifier shift".

I did not intend to suggest that quantifier shifts
are the only fallacies you commit.

> Indexing the fractions
> 1/1, 1/2, 1/3, 1/4, ...
> 2/1, 2/2, 2/3, 2/4, ...
> 3/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ...
> ...
>
> by natural numbers according to
> k = (m + n - 1)(m + n - 2)/2 + m (*)
> is called Cantor's first diagonl argument.
>
> We put the indexes X in the first column and
> mark all not yet indexed fractions by O.

Step 0.

> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO...
> ...
>
> Then every index k will be attached to a fraction m/n
> according to (*).

Steps 1 through k.

Steps 1 through k have a counting-order
(total order, each cut is a step, each step is a count)
which begins at 1 and ends at k, whatever is.

Without ending somewhere, they're not the first kind.

> The O's will be removed from the target element but
> will appear where the index has taken from.
> I call this change of places between X and O
> a transposition. Here is the configuration according
> to (*) for k = 8:

Steps 1 through 8 have a counting-order
(total order, each cut is a step, each step is a count)
which begins at 1 and ends at 8.

Without ending somewhere, they're not the first kind.

> XXXXOOOOO...
> XXXOOOOOO...
> XOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> XOOOOOOOO...
> ...
>
> Note that every transposed O is dropped above the X
> to be applied next. Therefore it is impossible that
> an O disappears from the board as long as X's are
> available.

As long as X's are available,
the transpositions have a stepping-order
(total order, each cut is a step)
which begins somewhere and ends somewhere.

Without ending somewhere, they're not the first kind.

> Indexing all fractions of the matrix

All fractions of the matrix do NOT have a stepping-
-order (total order, each cut is a step)
which begins somewhere and ends somewhere.

Without ending somewhere, they're not the first kind.

> Indexing all fractions of the matrix would require
> all O's to disappear.

All fractions, the collection of, is the second kind.
d/n gets an x from i/1 where i = (d+n-1)*(d+n-2)/2 + n
All O's disappear.

> That is impossible as long as indexes X are
> in the first column below the dropped O's.
>
> No try to find a way to revert the positions
> of O's and X's in the first column or
> to get rid of the O's in another way.

First kind ≠ second kind.

On Tuesday, January 18, 2022 at 6:41:22 AM UTC-4, WM wrote:
> William schrieb am Montag, 17. Januar 2022 um 21:43:39 UTC+1:
>
> > Nope. Nope if n is not dark then n+1 is not dark. This does not require that n has or can be written down ("named" is just another of you many ways of saying "written down".)
> >
> > ... know[n]... visible, defined, used.
> >
> > You have a lot of ways of say "can be written down"
> To explain it better I have to go a long way back

Nope, I have already said that for the sake of argument I will say the fractions cannot be indexed. If n is not dark it is the largest element of a FISON (which of course may have the property that it cannot be written down.) So n+1 is the largest element of a FISON and is thus not dark. So if n is not dark n+1 is not dark. This is true whether or not you can write n down. By induction we have that no element of N_P is dark.

--
William Hughes

On Tuesday, January 18, 2022 at 3:25:55 PM UTC+1, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 18. Januar 2022 um 13:41:26 UTC+1:
>
> > > To explain it better I have to go a long way back: In the OP I have shown that the Cantor function k = (m + n - 1)(m + n - 2)/2 + m used for enumerating all positive fractions
> > > As long as indexes are availabale, the presence of all O's indicates that almost all fractions are not indexed. And after the drain has become free, there are no indexes available which could index the remaining matrix elements covered by O's.
> > >
> > > This proves there remain many fractions without indexes. But we cannot find them. They are dark.
> > >
> > proves nothing of the sort
> Where does my model fail to follow Cantor's prescription k = (m + n - 1)(m + n - 2)/2 + m ?
>
> Regards, WM

I don't know. I can't bring myself to read any more of your repetitive nonsense.
The function k = (m + n - 1)(m + n - 2)/2 + m is biunique as every child can verify.
That's all there is to the countability of the rationals.

On Tuesday, 18 January 2022 at 14:50:24 UTC-4, JVR wrote:
> On Tuesday, January 18, 2022 at 3:25:55 PM UTC+1, WM wrote:
> > zelos...@gmail.com schrieb am Dienstag, 18. Januar 2022 um 13:41:26 UTC+1:
> >
> > > > To explain it better I have to go a long way back: In the OP I have shown that the Cantor function k = (m + n - 1)(m + n - 2)/2 + m used for enumerating all positive fractions
> > > > As long as indexes are availabale, the presence of all O's indicates that almost all fractions are not indexed. And after the drain has become free, there are no indexes available which could index the remaining matrix elements covered by O's.
> > > >
> > > > This proves there remain many fractions without indexes. But we cannot find them. They are dark.
> > > >
> > > proves nothing of the sort
> > Where does my model fail to follow Cantor's prescription k = (m + n - 1)(m + n - 2)/2 + m ?
> >
> > Regards, WM
> I don't know. I can't bring myself to read any more of your repetitive nonsense.
> The function k = (m + n - 1)(m + n - 2)/2 + m is biunique as every child can verify.
> That's all there is to the countability of the rationals.

It's his usual shit about thinking that somehow all integer values have to be processed to "unclog the brain" and allow the remaining fractional values to be enumerated. It's even clearer where he goes wrong when he uses the X's and O's, because there even he should see that some of the X's move off the first column before the entire column has been processed and that the vertical will not get exhausted. I am not holding my breath, however. Even the three dots he puts at the bottom are smeared there with zero comprehension on his part.

JVR schrieb am Dienstag, 18. Januar 2022 um 19:50:24 UTC+1:
> On Tuesday, January 18, 2022 at 3:25:55 PM UTC+1, WM wrote:
> > zelos...@gmail.com schrieb am Dienstag, 18. Januar 2022 um 13:41:26 UTC+1:
> >
> > > > To explain it better I have to go a long way back: In the OP I have shown that the Cantor function k = (m + n - 1)(m + n - 2)/2 + m used for enumerating all positive fractions
> > > > As long as indexes are availabale, the presence of all O's indicates that almost all fractions are not indexed. And after the drain has become free, there are no indexes available which could index the remaining matrix elements covered by O's.
> > > >
> > > > This proves there remain many fractions without indexes. But we cannot find them. They are dark.
> > > >
> > > proves nothing of the sort
> > Where does my model fail to follow Cantor's prescription k = (m + n - 1)(m + n - 2)/2 + m ?
> >
> I don't know. I can't bring myself to read any more of your repetitive nonsense.
> The function k = (m + n - 1)(m + n - 2)/2 + m is biunique as every child can verify.
> That's all there is to the countability of the rationals.

No, there is a contradiction. No O can leave before all indexes have been issued. No O can leave after all indexes have been issued. (Note that I don't claim that this will ever happen.)

Regards, WM

William schrieb am Dienstag, 18. Januar 2022 um 19:38:46 UTC+1:
> On Tuesday, January 18, 2022 at 6:41:22 AM UTC-4, WM wrote:
> > William schrieb am Montag, 17. Januar 2022 um 21:43:39 UTC+1:
> >
> > > Nope. Nope if n is not dark then n+1 is not dark. This does not require that n has or can be written down ("named" is just another of you many ways of saying "written down".)
> > >
> > > ... know[n]... visible, defined, used.
> > >
> > > You have a lot of ways of say "can be written down"
> > To explain it better I have to go a long way back
> Nope, I have already said that for the sake of argument I will say the fractions cannot be indexed.

But those not indexed are either not existing or not available, i.e., dark. Wouldn't dark fractions hint to dark integers?

> If n is not dark it is the largest element of a FISON (which of course may have the property that it cannot be written down.)

I agree.

> So n+1 is the largest element of a FISON and is thus not dark. So if n is not dark n+1 is not dark. This is true whether or not you can write n down.. By induction we have that no element of N_P is dark.

But ℵo FISONs are not available because of the pigeon hole principle. You cannot distinguish more strings of o's

o
oo
ooo
....

than are o's avaible. There are less than ℵo, i.e., less than more than all.

Regards, WM

Jim Burns schrieb am Dienstag, 18. Januar 2022 um 15:58:27 UTC+1:
> On 1/17/2022 2:31 PM, WM wrote:

> Steps 1 through k have a counting-order
> (total order, each cut is a step, each step is a count)
> which begins at 1 and ends at k, whatever is.
>
> Without ending somewhere, they're not the first kind.

My model need not end (although Cantor claims all indexes are issued). But before ending, the O's cannot leave.

> Without ending somewhere, they're not the first kind.

My model need not end (although Cantor claims all indexes are issued). But before ending, no O can leave.

> As long as X's are available,
> the transpositions have a stepping-order
> (total order, each cut is a step)
> which begins somewhere and ends somewhere.

Of course.
>
> Without ending somewhere, they're not the first kind.

But without ending all O's will remain.

> Without ending somewhere, they're not the first kind.
> > Indexing all fractions of the matrix would require
> > all O's to disappear.
> All fractions, the collection of, is the second kind.
> d/n gets an x from i/1 where i = (d+n-1)*(d+n-2)/2 + n
> All O's disappear.

No, that is provably wrong.

> First kind ≠ second kind.

My model works for all kinds.

Regards, WM

On Tuesday, January 18, 2022 at 3:21:53 PM UTC-4, WM wrote:
> William schrieb am Dienstag, 18. Januar 2022 um 19:38:46 UTC+1:

> > If n is not dark it is the largest element of a FISON (which of course may have the property that it cannot be written down.)
> I agree.
> > So n+1 is the largest element of a FISON and is thus not dark. So if n is not dark n+1 is not dark. This is true whether or not you can write n down. By induction we have that no element of N_P is dark.
> But ℵo FISONs are not available because of the pigeon hole principle. You cannot distinguish more strings of o's
>
> o
> oo
> ooo
> ...
>
> than are o's avaible.

Agreed. There are aleph_0 lines, Thus there are aleph_0 o's available. By induction we can show that every element of the Peano set N_P is the largest element of a FISON.

--
William Hughes

William schrieb am Dienstag, 18. Januar 2022 um 20:51:23 UTC+1:
> On Tuesday, January 18, 2022 at 3:21:53 PM UTC-4, WM wrote:
> > William schrieb am Dienstag, 18. Januar 2022 um 19:38:46 UTC+1:
>
> > > If n is not dark it is the largest element of a FISON (which of course may have the property that it cannot be written down.)
> > I agree.
> > > So n+1 is the largest element of a FISON and is thus not dark. So if n is not dark n+1 is not dark. This is true whether or not you can write n down. By induction we have that no element of N_P is dark.
> > But ℵo FISONs are not available because of the pigeon hole principle. You cannot distinguish more strings of o's
> >
> > o
> > oo
> > ooo
> > ...
> >
> > than are o's avaible.
> Agreed. There are aleph_0 lines, Thus there are aleph_0 o's available.

FISONs have less than aleph_0 o's.

> By induction we can show that every element of the Peano set N_P is the largest element of a FISON.

But aleph_0 = |ℕ| is not reached. ℕ \ U(FISONs) =/= { }.

Regards, WM

On 1/18/2022 1:26 PM, WM wrote:
> Jim Burns schrieb am Dienstag, 18. Januar 2022 um 15:58:27 UTC+1:
>> On 1/17/2022 2:31 PM, WM wrote:
>
>> Steps 1 through k have a counting-order
>> (total order, each cut is a step, each step is a count)
>> which begins at 1 and ends at k, whatever is.
>>
>> Without ending somewhere, they're not the first kind.
>
> My model need not end (although Cantor claims all indexes are issued). But before ending, the O's cannot leave.

your O's will not leave until you pay what you owe them!

>
>> Without ending somewhere, they're not the first kind.
>
> My model need not end (although Cantor claims all indexes are issued). But before ending, no O can leave.

your model is no good at all, because you stuck your index into the matrix as a column. Fail

>
>> As long as X's are available,
>> the transpositions have a stepping-order
>> (total order, each cut is a step)
>> which begins somewhere and ends somewhere.
>
> Of course.
>>
>> Without ending somewhere, they're not the first kind.
>
> But without ending all O's will remain

you stopped again!!

there is no ending an infinite matrix

>
>> Without ending somewhere, they're not the first kind.
>>> Indexing all fractions of the matrix would require
>>> all O's to disappear.
>> All fractions, the collection of, is the second kind.
>> d/n gets an x from i/1 where i = (d+n-1)*(d+n-2)/2 + n
>> All O's disappear.
>
> No, that is provably wrong.

no. It can be proven you have failed again.
you never prove anything.

>
>> First kind ≠ second kind.
>
> My model works for all kinds.

no. your model does not work.

>
> Regards, WM