On 5/6/2021 1:18 PM, WM wrote:
> Jim Burns schrieb
> am Donnerstag, 6. Mai 2021 um 02:07:07 UTC+2:
>> On 5/5/2021 4:57 PM, WM wrote:

>>>> Your position is that my arguments are correct, but
>>>> that the arguments of "set theory" are wrong. Yes?
>>>
>>> I told you several times what set theoristst claim.
>>> Why do you always delete it and then ask the same
>>> question?
>>
>> I ask again whether your position is that "set theory" is
>> incorrect

Me:
>> I ask again whether your position is that "set theory" is
>> incorrect because

You asked. I answered. You misquoted.

> Of course it is incorrect.

Then, if I want to convince you of something,
I should not use "set theory". Do you agree that I shouldn't?

> See the thread "The three worst mistakes of set theory."
> https://groups.google.com/g/sci.math/c/NCjyH4_cArg

William schrieb am Mittwoch, 5. Mai 2021 um 23:52:34 UTC+2:
> On Wednesday, May 5, 2021 at 5:46:50 PM UTC-3, WM wrote:
> > William schrieb am Mittwoch, 5. Mai 2021 um 20:27:34 UTC+2:
> > > On Wednesday, May 5, 2021 at 3:20:21 PM UTC-3, WM wrote:
> > > > William schrieb am Mittwoch, 5. Mai 2021 um 19:55:48 UTC+2:
> > > > > On Wednesday, May 5, 2021 at 2:44:08 PM UTC-3, WM wrote:
> > >
> > > > > > But if all elements have gone and if only one element can go per step, then there must be finite terms
> > > > > Nope, Counterintuitive certainly, but it is not a contradiction.
> > > > The usual reproach that I have often encounterd
> > > And correctly. A contradiction is not something you do not like.
> > Why do you delete the mathematics?
> It is repetitive and unneeded. We both agree that for "actual infinity" every E(n) is infinite and the intersection of all the E(n) does not contain an n that is subject to induction, so if all natural numbers are subject to induction the intersection is empty. You say this is a contradiction I say it is a result that you do not like.

It contradicts mathematics. It contradicts the bijection without which set theory is void.

> Note you have not found an infinite set from which you remove one element and it becomes empty. Each of the E(n) is infinite and infinitely close to the start of the process. (Yes, even when you have all the E(n) you only have sets that are infinitely close to the start)

So it is. If there are all sets, which are necessary if the intersection shall be empty, then there must be far more than can be addressed.

> Note you have not found an infinite set from which you remove one element and it becomes empty.

Such a set cannot be found. There are two alternative explanations: Either there is no set |N and no endsegment. Or there are all endsegments each of which deletes precisely one natnumber. Then all natnumbers become deleted in a linear step-by-step process or one-to-one mapping. Then all finite endsegments must be stepped through. But we cannot find them. The reason is that dark nunbers cannot be identified and they even have no order. That means there is no last endsegmnet before the empty set. Or it exists and contains the grossone, but it is not accessible.

> Each of the E(n) is infinite and infinitely close to the start of the process.

And so are all n that are subject to induction. By induction and set theory we prove that every n that is subject to induction has aleph_0 successors, which are not all subject to induction because they all could become used up, so all would have gone. But if they were accessible, a last one would have gone.

Note that there are two very different kinds of argument in set theory. For all intersections of finitely many endsegments we assume that beyond the last n there are infinitely many natnumbers. We can count them as far as we like: n, n+1, n+2, ...

For the intersection of an infinite set of endsegments however there is an indirect proof onl,y: Every n must have gone because there is an endsegment E(n+1) not containing it. This is correct but all n addressed in this way belong to a tiny initial segment. The owerwhelming majority is dark. For instance all addressable n belong to less than the 10^-1000000 th part of all natnumbers. If there are all, then there is a second half. But all this has never been understood by set theorists.

Regards, WM

Jim Burns schrieb am Donnerstag, 6. Mai 2021 um 19:32:08 UTC+2:
> On 5/6/2021 1:18 PM, WM wrote:

> >> I ask again whether your position is that "set theory" is
> >> incorrect because

The "because" is explained here:

> > See the thread "The three worst mistakes of set theory."
> > https://groups.google.com/g/sci.math/c/NCjyH4_cArg

Regards, WM

On Thursday, May 6, 2021 at 2:16:29 PM UTC-3, WM wrote:
> When we move from 1 to 0, we have necessarily passed a last [point].

Nope, there is no first real number greater than 0. So there is no last real number when going from 1 to 0.

--
William Hughes

On Thursday, May 6, 2021 at 2:36:43 PM UTC-3, WM wrote:
> Every n must have gone because there is an endsegment E(n+1) not containing it. This is correct but all n addressed in this way belong to a tiny initial segment.

There is no single "tiny initial segment" that works for all n. You need an infinite number of "tiny initial segments". When you put all of these segments together you get the set of natural numbers. At no step does a "tiny initial segment" become something other than a "tiny initial seqment". (This is the same counterintuitive result in a different form)

--
William Hughes

William schrieb am Donnerstag, 6. Mai 2021 um 20:10:44 UTC+2:
> On Thursday, May 6, 2021 at 2:16:29 PM UTC-3, WM wrote:
> > When we move from 1 to 0, we have necessarily passed a last [point].

We necessarily have passed a last interval (0, 1/n] where the 1/n depends on several circumstances.
>
> Nope, there is no first real number greater than 0. So there is no last real number when going from 1 to 0.
>
The last 1/n that we can realize or address depends on our abilities. Such abilities can be increased. The smaller points 1/n will not be *created* by our abilities. They had been already existing as dark points. And there is no gap or empty interval next to 0. Therefore there are all dark points which may become addressed and also the infinitely many which never will become addressed.

Regards, WM

On 5/6/2021 1:08 PM, WM wrote:
> Jim Burns schrieb am Donnerstag, 6. Mai 2021 um 19:32:08 UTC+2:
>> On 5/6/2021 1:18 PM, WM wrote:
>
>>>> I ask again whether your position is that "set theory" is
>>>> incorrect because
>
> The "because" is explained here:
>
>>> See the thread "The three worst mistakes of set theory."
>>> https://groups.google.com/g/sci.math/c/NCjyH4_cArg
>
> Regards, WM
>

WM OP;

"1) An inclusion-monotonic sequence of sets is a sequence where every
set S(n+1) contaiins only elements which are in the preceding set.S(n).
All elements of all sets are in the first set S(1). "

So all elements of all sets are in set S(1), and s(2) must have all
elements of s(1) + a few more elements(?).

no!

then those new elements would have to be included in S(1) by definition

so

1. s(1) = s(2) = s(3)= ... =s(n) = s(n+1) =... = s(oo)
2. sets s(n) for n>2 and higher are redundant and meaningless
3. WMs statement 1) above is meaningless as it is not a sequence of sets.

Looks like a BIG Mistake in WMs Set Theory alright. Fail.

On Thursday, May 6, 2021 at 3:56:27 PM UTC-3, WM wrote:
> William schrieb am Donnerstag, 6. Mai 2021 um 20:10:44 UTC+2:
> > On Thursday, May 6, 2021 at 2:16:29 PM UTC-3, WM wrote:
> > > When we move from 1 to 0, we have necessarily passed a last [point].
> Nope, there is no first real number greater than 0. So there is no last real number when going from 1 to 0.
> We necessarily have passed a last interval (0, 1/n] where the 1/n depends on several circumstances
No 1/n is a real number, There is no last real number when going from 0 to 1.

We may have passed the last addressed 1/n, but there is no last 1/n which may become addressed (trivial induction on n).

--
William Hughes

William schrieb am Donnerstag, 6. Mai 2021 um 22:05:42 UTC+2:
> On Thursday, May 6, 2021 at 3:56:27 PM UTC-3, WM wrote:
> > William schrieb am Donnerstag, 6. Mai 2021 um 20:10:44 UTC+2:
> > > On Thursday, May 6, 2021 at 2:16:29 PM UTC-3, WM wrote:
> > > > When we move from 1 to 0, we have necessarily passed a last [point]..
> > Nope, there is no first real number greater than 0. So there is no last real number when going from 1 to 0.
> > We necessarily have passed a last interval (0, 1/n] where the 1/n depends on several circumstances
> No 1/n is a real number, There is no last real number when going from 0 to 1.
>
> We may have passed the last addressed 1/n, but there is no last 1/n which may become addressed (trivial induction on n).

That is trivial because all 1/n which can become addressed belong to a potentially infinite collection (Peano). The question is only whether the interval (0, 1] is filled with points or whether the not yet addressed points will have to be created in the course of addressing them.

Regards, WM

On Thursday, May 6, 2021 at 5:17:32 PM UTC-3, WM wrote:
> William schrieb am Donnerstag, 6. Mai 2021 um 22:05:42 UTC+2:
> > On Thursday, May 6, 2021 at 3:56:27 PM UTC-3, WM wrote:
> > > William schrieb am Donnerstag, 6. Mai 2021 um 20:10:44 UTC+2:
> > > > On Thursday, May 6, 2021 at 2:16:29 PM UTC-3, WM wrote:
> > > > > When we move from 1 to 0, we have necessarily passed a last [point].
> > > Nope, there is no first real number greater than 0. So there is no last real number when going from 1 to 0.
> > > We necessarily have passed a last interval (0, 1/n] where the 1/n depends on several circumstances
> > No 1/n is a real number, There is no last real number when going from 0 to 1.
> >
> > We may have passed the last addressed 1/n, but there is no last 1/n which may become addressed (trivial induction on n).
> That is trivial because all 1/n which can become addressed belong to a potentially infinite collection (Peano). The question is only whether the interval (0, 1] is filled with points or whether the not yet addressed points will have to be created in the course of addressing them.

In standard mathematics (which you call actual infinity) all real numbers in (0,1] are assumed to exist. In your "potential" infinity real numbers have to be created and do not exist before they are created. In the former there is no existing real number which is closest to 0. I the later there is an existing real number which is closest to 0, (1/M where M is the greatest integer, note M can change) but this real number is greater than 0.

--
William Hughes

William schrieb am Donnerstag, 6. Mai 2021 um 20:36:52 UTC+2:
> On Thursday, May 6, 2021 at 2:36:43 PM UTC-3, WM wrote:
>
> > Every n must have gone because there is an endsegment E(n+1) not containing it. This is correct but all n addressed in this way belong to a tiny initial segment.
> There is no single "tiny initial segment" that works for all n. You need an infinite number of "tiny initial segments".

It is a potentially infinite but nevertheless, compared to ℵo, tiny always finite collection. The bird an the mountain.

> When you put all of these segments together you get the set of natural numbers.

Not the set required for set theory. The tiny initial segment of indices will not suffice to index the digits of a real number or the lines of a Cantor-list.

> At no step does a "tiny initial segment" become something other than a "tiny initial seqment". (This is the same counterintuitive result in a different form)

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

They are just enough to index the same amount of natural and rational numbers in order to let these sets appear equinumerous.

But that is not actual infinity: ω - n = ω.

Regards, WM

William schrieb am Donnerstag, 6. Mai 2021 um 22:36:42 UTC+2:
> On Thursday, May 6, 2021 at 5:17:32 PM UTC-3, WM wrote:

> In standard mathematics (which you call actual infinity) all real numbers in (0,1] are assumed to exist.

And further it is assumed that each one can be addressed individually.

And for each individually addressed number a closer number can be addressed, found, or created?

What is it? If it is not created, but each one is existing and can be chosen, why then not choosing that immediately before 0. There is no gap next to zero. There are points which can be chosen collectively, because there will none remain immediately before 0.

Regards, WM