On 4/14/2022 10:20 AM, William wrote:
> On Thursday, April 14, 2022 at 10:51:12 AM UTC-3, WM wrote:
>> ℕ consists of elements none of which can be subtracted without damaging ℕ.
>
> True. So what? The "damaged N" has the same cardinality as ℕ.
>

If ℕ can be damaged, how do we know if the ℕ we have is good or a damaged one ?

I'm considering the case where WM has to buy more natural numbers off eBay because he ran out of indexs, what if someone sold him a dammaged ℕ ?
Trust but verify, WM should count them all to see if they are all there, or return them.

Gus Gassmann wrote:

> Crackpottery. |ℕ| - 1 = aleph_0 - 1 = aleph_0 = |ℕ|.
>
> But the two statements above make it clear how your demented mind works:
> You think that the sequence of sizes of FISONs goes something like 1, 2,
> 3, ..., |ℕ| - 2, |ℕ| - 1, |ℕ|.
> So then, tell us: which is the fir

FromTheRafters schrieb am Donnerstag, 14. April 2022 um 16:18:22 UTC+2:
> WM was thinking very hard :

> > Because ℕ consists of elements none of which can be subtracted without
> > damaging ℕ.
> You can actually,

It appears so, but it is not so.

> just relabel the elements and

you have one dark element less.

Regards, WM

William schrieb am Donnerstag, 14. April 2022 um 16:36:29 UTC+2:
> A finite set, R, must have a minimum. If S is infinite, it may not have a minimum.

The minimum for endsegments is the empty set. As long as the minimum is not reached, the endsegments contain more. Should more than the minimum be a reason for emptiness? Should the unkown more be the reason for emptiness? Very strange logic!

Regards, WM

William schrieb am Donnerstag, 14. April 2022 um 16:43:55 UTC+2:
> On Thursday, April 14, 2022 at 8:30:41 AM UTC-3, WM wrote:
> > Nope, this is trivially false for a set with no mininum.
> > The minimum is infinite for infinite endsegments.
> An infinite set of endsegments cannot have a minimum.
>
Then all endsegments contain more than nothing. That is not less enough for the empty intersection.

Regards, WM

horand....@gmail.com schrieb am Donnerstag, 14. April 2022 um 16:45:12 UTC+2:
> On Thursday, 14 April 2022 at 09:05:20 UTC-3, WM wrote:
> > horand....@gmail.com schrieb am Donnerstag, 14. April 2022 um 12:35:36 UTC+2:
> [...]
> > The sequence of FISONs goes 1, 2, 3, ... . Above, all is dark.
> Above what???

Above all definable numbers, i.e., above all numbers which are ends of FISONs.

Regards, WM

horand....@gmail.com schrieb am Donnerstag, 14. April 2022 um 16:49:00 UTC+2:
> On Thursday, 14 April 2022 at 10:51:12 UTC-3, WM wrote:
> [...]
> > Yes aleph_0 means actually infinite. But that does not mean |ℕ| - 1 = |ℕ|.
> That is *EXACTLY* what it means.

No, you have no clue.

Regards, WM

On Friday, April 15, 2022 at 3:09:13 PM UTC-3, WM wrote:
> William schrieb am Donnerstag, 14. April 2022 um 16:36:29 UTC+2:
> > A finite set, R, must have a minimum. If S is infinite, it may not have a minimum.
> The minimum for endsegments is the empty set.

Piffle. Every endsegment is infinite. The set of endsegments does not contain the empty set. (The empty set is the GLB, and since the intersection must be contained in the GLB, the intersection is empty)

--
William Hughes

WM has noted that any element in the intersection of all endsegments must be "dark" . Thus, as N_p does not contain a dark element, there is no element of the Peano set of natural numbers in the intersection.

--
William Hughes

On Friday, 15 April 2022 at 15:14:58 UTC-3, WM wrote:
> horand....@gmail.com schrieb am Donnerstag, 14. April 2022 um 16:49:00 UTC+2:
> > On Thursday, 14 April 2022 at 10:51:12 UTC-3, WM wrote:
> > [...]
> > > Yes aleph_0 means actually infinite. But that does not mean |ℕ| - 1 = |ℕ|.
> > That is *EXACTLY* what it means.
> No, you have no clue.

That's rich. coming from the master of ignorance. Kindly fuck off.

On Friday, 15 April 2022 at 15:13:22 UTC-3, WM wrote:
> horand....@gmail.com schrieb am Donnerstag, 14. April 2022 um 16:45:12 UTC+2:
> > On Thursday, 14 April 2022 at 09:05:20 UTC-3, WM wrote:
> > > The sequence of FISONs goes 1, 2, 3, ... . Above, all is dark.
> > Above what???
> Above all definable numbers, i.e., above all numbers which are ends of FISONs.

Aside from the fact that 1, 2, 3, ... is not a sequence, let alone a sequence of FISONs (which I pointed out to you before and which you snipped), there is *NO* natural number n that does not end a FISON, namely FISON(n). The only thing that is dark, as usual, is your mind. Your brain functions flatlined some months ago.
And now, kindly get lost.

On 4/15/2022 1:03 PM, WM wrote:
> FromTheRafters schrieb am Donnerstag, 14. April 2022 um 16:18:22 UTC+2:
>> WM was thinking very hard :
>
>>> Because ℕ consists of elements none of which can be subtracted without
>>> damaging ℕ.
>> You can actually,
>
> It appears so, but it is not so.

then show your work, proof, etc.

can you disprove this ?

|ℕ| - 1 = aleph_0 - 1 = aleph_0 = |ℕ|

>
>> just relabel the elements and
>
> you have one dark element less.

Wrong. You cannot count dark numbers (ants)

>
> Regards, WM

On 4/15/2022 1:14 PM, WM wrote:
> horand....@gmail.com schrieb am Donnerstag, 14. April 2022 um 16:49:00 UTC+2:
>> On Thursday, 14 April 2022 at 10:51:12 UTC-3, WM wrote:
>> [...]
>>> Yes aleph_0 means actually infinite. But that does not mean |ℕ| - 1 = |ℕ|.
>> That is *EXACTLY* what it means.
>
> No, you have no clue.

You are still wrong.

>
> Regards, WM

Why don't you show us your missing dark numbers ?

On 4/15/2022 1:13 PM, WM wrote:
> horand....@gmail.com schrieb am Donnerstag, 14. April 2022 um 16:45:12 UTC+2:
>> On Thursday, 14 April 2022 at 09:05:20 UTC-3, WM wrote:
>>> horand....@gmail.com schrieb am Donnerstag, 14. April 2022 um 12:35:36 UTC+2:
>> [...]
>>> The sequence of FISONs goes 1, 2, 3, ... . Above, all is dark.
>> Above what???
>
> Above all definable numbers, i.e., above all numbers which are ends of FISONs.
>
> Regards, WM

above what ?

do you mean the "greater than" symbol ?

Or do you mean over your head ?

OR up in the air ?

On 4/15/2022 1:10 PM, WM wrote:
> William schrieb am Donnerstag, 14. April 2022 um 16:43:55 UTC+2:
>> On Thursday, April 14, 2022 at 8:30:41 AM UTC-3, WM wrote:
>>> Nope, this is trivially false for a set with no mininum.
>>> The minimum is infinite for infinite endsegments.
>> An infinite set of endsegments cannot have a minimum.
>>
> Then all endsegments contain more than nothing. That is not less enough for the empty intersection.
>
> Regards, WM

All endsegments are infinite. The intersection of all endsegments is empty. (for the 45th time)

William schrieb am Donnerstag, 14. April 2022 um 17:02:44 UTC+2:
> On Thursday, April 14, 2022 at 10:46:39 AM UTC-3, WM wrote:
> > Fritz Feldhase schrieb am Donnerstag, 14. April 2022 um 13:52:35 UTC+2:
> > > The set of all endsegments does not contain a minimal element.
> > The set of infinite endsegments contains elements.
> Correct . However, it does not contain a minimal element. This is sufficient to show that the intersection is empty.

Only in a fool's mind! Large than the minimal element implies a larger intersection.

> One does not need the set to contain an empty element

Absolutely unfounded absurd claim.

Regards, WM

William schrieb am Donnerstag, 14. April 2022 um 17:20:51 UTC+2:
> On Thursday, April 14, 2022 at 10:51:12 AM UTC-3, WM wrote:
> > ℕ consists of elements none of which can be subtracted without damaging ℕ.
> True. So what? The "damaged N" has the same cardinality as ℕ.

There is no cardinality. You can say "the same actual infinity".

Cardinality would require a mapping of all elements. That is impossible for infinite sets. Here is the proof for the simplest case, namely the attempted mapping of naturals and rationals. When trying to collect Cantor's sequence

{1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ...}

in the first column of the matrix of all positive fractions, then every attempt will fail. Here are the first steps:

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, ...
5/1, 5/2, 5/3, 5/4, ...
....

1/1, 2/1, 1/3, 1/4, ... 1/1, 3/1, 1/3, 1/4, ... 1/1, 3/1, 4/1, 1/4, ... 1/1, 3/1, 4/1, 1/4, ...
1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ... 1/2, 5/1, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ... 4/1, 4/2, 4/3, 4/4, ... 1/3, 4/2, 4/3, 4/4, ... 1/3, 4/2, 4/3, 4/4, ...
5/1, 5/2, 5/3, 5/4, ... 5/1, 5/2, 5/3, 5/4, ... 5/1, 5/2, 5/3, 5/4, ... 2/2, 5/2, 5/3, 5/4, ...

Although as many terms as you like of Cantor's sequence will be gathered in the first column, the desired result

1/1, __, __, __, ...
1/2, __, __, __, ...
1/3, __, __, __, ...
2/2, __, __, __, ...
....

will never be accomplished. Never a place of the matrix will become empty.

Regards, WM

WM drivels:

> horand....@gmail.com schrieb
>> WM wrote:
>> [...]
>>> Yes aleph_0 means actually infinite. But that does not mean |ℕ| - 1 = |ℕ|.

>> That is *EXACTLY* what it means.
> No, you have no clue.

LOL
> Regards, WM

ROFL

Piss off, idiot.

WM drivels:

> ...
>
> 1/1, 2/1, 1/3, 1/4, ... 1/1, 3/1, 1/3, 1/4, ... 1/1, 3/1, 4/1, 1/4, ... 1/1, 3/1, 4/1, 1/4, ...
> 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ... 1/2, 5/1, 2/3, 2/4, ...
> 3/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ... 4/1, 4/2, 4/3, 4/4, ... 1/3, 4/2, 4/3, 4/4, ... 1/3, 4/2, 4/3, 4/4, ...
> 5/1, 5/2, 5/3, 5/4, ... 5/1, 5/2, 5/3, 5/4, ... 5/1, 5/2, 5/3, 5/4, ... 2/2, 5/2, 5/3, 5/4, ...
ROFL
> 1/1, __, __, __, ...
> 1/2, __, __, __, ...
> 1/3, __, __, __, ...
> 2/2, __, __, __, ...

ROFL

> Regards, WM

ROFL

Piss off, chum up somewhere else, obnoxious, eek...

On Friday, April 15, 2022 at 5:45:58 PM UTC-3, WM wrote:
> William schrieb am Donnerstag, 14. April 2022 um 17:20:51 UTC+2:
> > On Thursday, April 14, 2022 at 10:51:12 AM UTC-3, WM wrote:
> > > ℕ consists of elements none of which can be subtracted without damaging ℕ.
> > True. So what? The "damaged N" has the same cardinality as ℕ.
> There is no cardinality

Piffle.

> You can say "the same actual infinity".
>
> Cardinality would require a mapping of all elements. That is impossible for infinite sets.

Piffle. The identity function (which exists of any set that exist, e.g. the Peano set of natural numbers) is a bijection

--
William Hughes

On 4/15/2022 3:45 PM, WM wrote:
> William schrieb am Donnerstag, 14. April 2022 um 17:20:51 UTC+2:
>> On Thursday, April 14, 2022 at 10:51:12 AM UTC-3, WM wrote:
>>> ℕ consists of elements none of which can be subtracted without damaging ℕ.
>> True. So what? The "damaged N" has the same cardinality as ℕ.
>
> There is no cardinality. You can say "the same actual infinity".
>

go post in alt.math.fakery.for.dummies

> Regards, WM
>

On 4/15/2022 3:37 PM, WM wrote:
> William schrieb am Donnerstag, 14. April 2022 um 17:02:44 UTC+2:
>> On Thursday, April 14, 2022 at 10:46:39 AM UTC-3, WM wrote:
>>> Fritz Feldhase schrieb am Donnerstag, 14. April 2022 um 13:52:35 UTC+2:
>>>> The set of all endsegments does not contain a minimal element.
>>> The set of infinite endsegments contains elements.
>> Correct . However, it does not contain a minimal element. This is sufficient to show that the intersection is empty.
>
> Only in a fool's mind! Large than the minimal element implies a larger intersection.

Wrong. for any k a natural number, k is not in E(k+1), therefore k is not in the intersection of all endsegments (46th time)

>
>> One does not need the set to contain an empty element
>
> Absolutely unfounded absurd claim.

accept the fact that you are wrong.

>
> Regards, WM

On 3/31/2022 12:58 PM, WM wrote:
> Fritz Feldhase schrieb am Donnerstag, 31. März 2022 um 19:33:23 UTC+2:
>> On Thursday, March 31, 2022 at 7:21:42 PM UTC+2, WM wrote:
>>> Fritz Feldhase schrieb am Donnerstag, 31. März 2022 um 19:02:38 UTC+2:
>>>>>
>>>>> The sequence of all FISONs contains the union of all FISONs.
>>>>>
>>>> "contains"? In what sense?
>>>>
>>>> As "term"?
>>>>
>>> Of course.
>
>>> Why should it be wrong?
>
>> Because the union of all FISONs does not have a largest element, while on the other hand each and every FISON has a largest element (which just is the number of elements in it).
>
> That is not a valid argument. The sequence of all FISONs does not contain a largest element either.

each FISON has a largest element.

>
> The union of all FISONs cannot contain more than the sequence because of the fact that the equence consists of all unions already
> {1} = {1}
> {1} U {1, 2} = {1, 2}
> {1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
> {1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} = {1, 2, 3, 4}
> {1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} U {1, 2, 3, 4, 5} = {1, 2, 3, 4, 5}
> ...
> All that is in the union is also in the sequence as a term because the sequence consists of unions. If there are all terms in the sequence, then the union of all terms is in the sequence.
>
> Regards, WM

petty stuff.

So what is the largest element of the union of all FISONs ?

need a hint ?

William schrieb am Freitag, 15. April 2022 um 23:34:09 UTC+2:
> On Friday, April 15, 2022 at 5:45:58 PM UTC-3, WM wrote:

> > Cardinality would require a mapping of all elements. That is impossible for infinite sets.
> Piffle. The identity function (which exists of any set that exist, e.g. the Peano set of natural numbers) is a bijection

It appears so to simple minds, but there is always a dark part that cannot be mapped. This is proved here:

Cardinality would require a mapping of all elements. That is impossible for infinite sets. Here is the proof for the simplest case, namely the attempted mapping of naturals and rationals. When trying to collect Cantor's sequence

{1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ...}

in the first column of the matrix of all positive fractions, then every attempt will fail. Here are the first steps:

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, ...
5/1, 5/2, 5/3, 5/4, ...
....

1/1, 2/1, 1/3, 1/4, ... 1/1, 3/1, 1/3, 1/4, ... 1/1, 3/1, 4/1, 1/4, ... 1/1, 3/1, 4/1, 1/4, ...
1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ... 1/2, 5/1, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ... 4/1, 4/2, 4/3, 4/4, ... 1/3, 4/2, 4/3, 4/4, ... 1/3, 4/2, 4/3, 4/4, ...
5/1, 5/2, 5/3, 5/4, ... 5/1, 5/2, 5/3, 5/4, ... 5/1, 5/2, 5/3, 5/4, ... 2/2, 5/2, 5/3, 5/4, ...

Although as many terms as you like of Cantor's sequence will be gathered in the first column, the desired result

1/1, __, __, __, ...
1/2, __, __, __, ...
1/3, __, __, __, ...
2/2, __, __, __, ...
....

will never be accomplished. Never a place of the matrix will become empty.

Regards, WM

On 4/16/2022 8:55 AM, WM wrote:
> William schrieb am Freitag, 15. April 2022 um 23:34:09 UTC+2:
>> On Friday, April 15, 2022 at 5:45:58 PM UTC-3, WM wrote:
>
>>> Cardinality would require a mapping of all elements. That is impossible for infinite sets.
>> Piffle. The identity function (which exists of any set that exist, e.g. the Peano set of natural numbers) is a bijection
>
> It appears so to simple minds, but there is always a dark part that cannot be mapped. This is proved here:

that is not a proof, it is a spoof. You do not know math.

>
> Regards, WM
>