Fritz Feldhase schrieb am Montag, 27. Juni 2022 um 18:56:20 UTC+2:
> On Monday, June 27, 2022 at 3:54:59 PM UTC+2, WM wrote:
> > William schrieb am Montag, 27. Juni 2022 um 00:43:38 UTC+2:
> > > On Sunday, June 26, 2022 at 6:45:58 PM UTC-3, WM wrote:
> > > > William schrieb am Sonntag, 26. Juni 2022 um 22:28:44 UTC+2:
> > > > > On Sunday, June 26, 2022 at 5:19:02 PM UTC-3, WM wrote:
> > > > > >
> > > > > > Induction holds for every element that could in principle be accessed and checked.
> > > > > >
> > > > > Piffle, |N is a Peano set. Induction holds for every element of |N
> > > >
> > > > that can be accessed. .
> > > >
> > > Correct, "that can be accessed" means "not dark" which for a Peano set like, |N, is each element.
> > >
> > There is an "each" but not an "all"
>
> Well, in MATHEMATICS we do have the universal quantifier "for all", you know.

∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
|∩{E(k) : k ∈ ℕ_def}| = ℵ₀

Regards, WM

zelos...@gmail.com schrieb am Montag, 27. Juni 2022 um 19:43:09 UTC+2:
> lördag 25 juni 2022 kl. 14:29:56 UTC+2 skrev WM:
> > William schrieb am Freitag, 24. Juni 2022 um 19:06:04 UTC+2:
> > > On Friday, June 24, 2022 at 1:24:52 PM UTC-3, WM wrote:
> > > > William schrieb am Freitag, 24. Juni 2022 um 18:05:45 UTC+2:
> > > at appears to work if not checked
> > > > > |N exists and is a Peano set (axiom of infinity). f(n) exists for every element of |N (induction). Hence the bijection { (n,f(n) | n in |N} exists.
> > > <snip>
> > > >That function does not exist except the first few terms.
> > > Nope. |N is a Peano set thus induction holds for every element of |N
> > Induction holds for every matrix of the infinite sequence
> > 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, ...
> > ... ... ... ...
> >
> >
> > and proves that never the attempted matrix
> >
> > 1/1, __, __, __, ...
> > 1/2, __, __, __, ...
> > 1/3, __, __, __, ...
> > 2/2, __, __, __, ...
> > ...
> >
> > will be reached.
> > > f(n) exist for every element of |N (induction),
> > f(n) in Cantor's function shows that always (i.e., for every index you can check) ℵo terms are not checked.
> >
> Yet that is irrelevant to the functions existence.

Only nonmathematical believers state this opinion.

Regards, WM

Fritz Feldhase schrieb am Montag, 27. Juni 2022 um 19:14:59 UTC+2:
> On Monday, June 27, 2022 at 4:07:33 PM UTC+2, WM wrote:
> > Fritz Feldhase schrieb am Sonntag, 26. Juni 2022 um 17:14:42 UTC+2:
> > >
> > > Are you claiming that induction does not prove that A(n) holds for all n e IN, even though it states that A(n) holds for all n e IN (if so)?
> > >
> > Induction proves that every natural number [...] belongs to a FISON having ℵo successors.
>
> Sure. Still:
>
> Are you claiming that induction does not prove that A(n) holds for all n e IN, even though it states that A(n) holds for all n e IN (if so)?

Induction obviously does not prove that all natural numbers have almost all natural numbers as successors. This proof only hold for definable natural numbers.

Regards, WM

On Tuesday, June 28, 2022 at 1:08:35 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Montag, 27. Juni 2022 um 19:14:59 UTC+2:
> > On Monday, June 27, 2022 at 4:07:33 PM UTC+2, WM wrote:
> > > Fritz Feldhase schrieb am Sonntag, 26. Juni 2022 um 17:14:42 UTC+2:
> > > >
> > > > Are you claiming that induction does not prove that A(n) holds for all n e IN, even though it states that A(n) holds for all n e IN (if so)?
> > > >
> > > Induction proves that every natural number [...] belongs to a FISON having ℵo successors.
> >
> > Sure. Still:
> >
> > Are you claiming that induction does not prove that A(n) holds for all n e IN, even though it states that A(n) holds for all n e IN (if so)?

Still no answer, you silly crank?

You cannot even answer a simple question?

Are you suffering from a brain damage?

> Induction obviously does not prove that all natural numbers have almost all natural numbers as successors.

Of course, it does.

WM expressed precisely :
> William schrieb am Montag, 27. Juni 2022 um 17:22:41 UTC+2:
>> On Monday, June 27, 2022 at 10:54:59 AM UTC-3, WM wrote:
>>> William schrieb am Montag, 27. Juni 2022 um 00:43:38 UTC+2:
>>>> On Sunday, June 26, 2022 at 6:45:58 PM UTC-3, WM wrote:
>>>>> William schrieb am Sonntag, 26. Juni 2022 um 22:28:44 UTC+2:
>>>>>> On Sunday, June 26, 2022 at 5:19:02 PM UTC-3, WM wrote:
>>>>>>
>>>>>>> Induction holds for every element that could in principle be accessed
>>>>>>> and checked.
>>>>>> Piffle, |N is a Peano set. Induction holds for every element of |N
>>>>> that can be accessed. .
>>>>
>>>> Correct, "that can be accessed" means "not dark" which for a Peano set
>>>> like, |N, is each element.
>>> There is an "each" but not an "all".
>> Piffle. The Peano set |N exists (axiom of infinity).
>
> The second even prime number exists (axiom of second even prime number).

Demonstrating once again that you have no clue how this all works.

Sad, really.

Fritz Feldhase schrieb am Dienstag, 28. Juni 2022 um 13:47:24 UTC+2:
> On Tuesday, June 28, 2022 at 1:08:35 PM UTC+2, WM wrote:

> Are you suffering from a brain damage?
> > Induction obviously does not prove that all natural numbers have almost all natural numbers as successors.
> Of course, it does.

Then it would be proving nonsense.

Regards, WM

On Tuesday, June 28, 2022 at 7:58:37 AM UTC-3, WM wrote:

>The Peano-set is not completely accessible because every accessible Peano number has more successors than predecessors. Never >accessed successors are inaccessible.

Piffle. "accessed" is one of your many many many ways of saying "has been written down". Induction can be used to show something is true for n, whether n been written down, has not been written down or cannot in principle be written down. Induction say that if n is an element of the set then P(n) is true. No mention of any other property of n.

--
William Hughes