It happens that WM formulated :
> FromTheRafters schrieb am Mittwoch, 12. Mai 2021 um 14:42:29 UTC+2:
>> on 5/12/2021, WM supposed :
>
>>> (0, 1] \ (0, 1] = { } shows that all elememts of the interval can be
>>> removed.
>> You are not removing anything. You are defining a set which has no
>> elements -- the empty set.
>
> Note that \ is for sets what - is for numbers. We call this operation
> subtracting or removing.

Well, duh! So, you subtract your "set" of real points from an interval,
let's call it "A", then from that you subtract same interval's "set" of
real points let's call it "A" also since they are the same set, and you
got nothing.

Revelation!!

> Subtraction is an arithmetic operation that represents the operation of
> removing objects from a collection. "(Wikipedia)

Sorry, I plum forgot about your rubber sets and serial thinking.

William schrieb am Mittwoch, 12. Mai 2021 um 16:12:53 UTC+2:
> On Wednesday, May 12, 2021 at 10:48:00 AM UTC-3, WM wrote:
> > William schrieb am Mittwoch, 12. Mai 2021 um 15:27:59 UTC+2:
> > > On Wednesday, May 12, 2021 at 9:33:16 AM UTC-3, WM wrote:
> > > > William schrieb am Dienstag, 11. Mai 2021 um 21:10:39 UTC+2:
> > > > > On Tuesday, May 11, 2021 at 3:00:52 PM UTC-3, WM wrote:
> > > > > > William schrieb am Dienstag, 11. Mai 2021 um 15:50:27 UTC+2:
> > > > > > > On Tuesday, May 11, 2021 at 10:00:08 AM UTC-3, WM wrote:
> > > > >
>
> WH: ... a set can be "next" to zero even though no element of the set is "next" to 0.
> WM: That is true, but it would be wrong, if all its elements could be defined
> <snip>
> >To prove this you need to use induction. You cannot prove every element is removable by "removing" elements.
> > You cannot prove it by induction either.
> Of course you can (0 is removable, if x is removable then x+1 is removable). You can also prove that every element is definable. You can also prove that every element is followed by an infinite number of elements.

But you can remove all elements without remainder collectively. Obviously you remove more when exhausting the set than when you leave infinitely many elements.

> You can also prove that no element is dark.

No element that is subject to induction is dark. That is true. Is that a reason to deny the existence of the set "that every element is followed by"?

Regards, WM

On Wednesday, May 12, 2021 at 11:26:12 AM UTC-3, WM wrote:
> William schrieb am Mittwoch, 12. Mai 2021 um 16:12:53 UTC+2:
> > On Wednesday, May 12, 2021 at 10:48:00 AM UTC-3, WM wrote:
> > > William schrieb am Mittwoch, 12. Mai 2021 um 15:27:59 UTC+2:

WH: ... a set can be "next" to zero even though no element of the set is "next" to 0.
WM: That is true, but it would be wrong, if all its elements could be defined

> > <snip>
> > > >To prove this you need to use induction. You cannot prove every element is removable by "removing" elements.
> > > You cannot prove it by induction either.
> > Of course you can (0 is removable, if x is removable then x+1 is removable). You can also prove that every element is definable.

<snip>

> > You can also prove that no element is dark.
> No element that is subject to induction is dark.

And since, by definition, every element of a set subject to induction is subject to induction, we are done

--
William Hughes.

On 5/12/2021 9:12 AM, William wrote:
> On Wednesday, May 12, 2021 at 10:48:00 AM UTC-3, WM wrote:
>> William schrieb am Mittwoch, 12. Mai 2021 um 15:27:59 UTC+2:
>>> On Wednesday, May 12, 2021 at 9:33:16 AM UTC-3, WM wrote:
>>>> William schrieb am Dienstag, 11. Mai 2021 um 21:10:39 UTC+2:
>>>>> On Tuesday, May 11, 2021 at 3:00:52 PM UTC-3, WM wrote:
>>>>>> William schrieb am Dienstag, 11. Mai 2021 um 15:50:27 UTC+2:
>>>>>>> On Tuesday, May 11, 2021 at 10:00:08 AM UTC-3, WM wrote:
>>>>>
>
> WH: ... a set can be "next" to zero even though no element of the set is "next" to 0.

like (0,1] ? my, what HUGE advancements in Math!

> WM: That is true, but it would be wrong, if all its elements could be defined

WM's "defined" is meaningless, it is *observer dependent math*, where
*the math depends upon who is
looking*/thinking/Rapping/Tapping/Buzzing/lights a flashing/horse hoofing/

No IF statement needed.

>
> <snip>
>
>> To prove this you need to use induction. You cannot prove every element is removable by "removing" elements.
>> You cannot prove it by induction either.
>
> Of course you can (0 is removable, if x is removable then x+1 is removable). You can also prove that every element is definable. You can also prove that every element is followed by an infinite number of elements. You can also prove that no element is dark. Induction has many uses.
>

so, where are these Proofs? or are they your sPoofs ?

William schrieb am Mittwoch, 12. Mai 2021 um 16:40:20 UTC+2:
> On Wednesday, May 12, 2021 at 11:26:12 AM UTC-3, WM wrote:

> > > You can also prove that no element is dark.
> > No element that is subject to induction is dark.
> And since, by definition, every element of a set subject to induction is subject to induction, we are done

We are done, if we refrain from finished infinity, exhausting infinite sets.

Regards, WM

On 5/10/2021 6:57 AM, WM wrote:
> zelos...@gmail.com schrieb am Montag, 10. Mai 2021 um 06:37:34 UTC+2:
>> lördag 8 maj 2021 kl. 13:59:36 UTC+2 skrev WM:
>
>> you can well-order any set, it doesn't mean we understand that ordering.
>
> So you agree that you cannot well-order every set i.n an understandable way.
>
>> But your "closest" in normal ordering, does not exist.
>
> If you go through an infinite sequence, term by term then you will never finish unless you have done a last term.

there is no "last term" accept that fact, and deal with it.

> Collectively however, without doing it term by term, you can order the whole sequence.

wrong.

> This shows that there are more terms than can be well-ordered in an understandable way. I call them dark terms.

they are not there

>
> Regards, WM
>

On Wednesday, May 12, 2021 at 1:04:44 PM UTC-3, WM wrote:
> William schrieb am Mittwoch, 12. Mai 2021 um 16:40:20 UTC+2:
> > On Wednesday, May 12, 2021 at 11:26:12 AM UTC-3, WM wrote:

WH: ... a set can be "next" to zero even though no element of the set is "next" to 0.
WM: That is true, but it would be wrong, if all its elements could be defined
WH: You can use induction to prove that every element of S is definable

> > > > You can also prove that no element is dark.
> > > No element that is subject to induction is dark.
> > And since, by definition, every element of a set subject to induction is subject to induction, we are done
> We are done, if we refrain from finished infinity,

"finished infinity" cannot add elements to S that are not part of S by definition. So we ate done if we "refrain from finished infinity" or not.

--
William Hughes

On 5/12/2021 11:04 AM, WM wrote:
> William schrieb am Mittwoch, 12. Mai 2021 um 16:40:20 UTC+2:
>> On Wednesday, May 12, 2021 at 11:26:12 AM UTC-3, WM wrote:
>
>>>> You can also prove that no element is dark.
>>> No element that is subject to induction is dark.
>> And since, by definition, every element of a set subject to induction is subject to induction, we are done
>
> We are done, if we refrain from finished infinity, exhausting infinite sets.
>
> Regards, WM
>

you cannot be done if it takes a pico second to remove an element from
an infinite set, gotta move them all, gotta move them all! (infinite
time required)

oo * one pico second = oo amount of time

On 5/12/2021 8:51 AM, WM wrote:
> Jim Burns schrieb
> am Mittwoch, 12. Mai 2021 um 06:13:27 UTC+2:
>> On 5/11/2021 9:07 AM, WM wrote:

>>> When would your crowd have supplied enough indices
>>> to fix a real number by its digits?
>>
>> You may recall that I came to describing a natural number
>> as countable-to and countable-beyond by way of asking
>> "What is a natural number? What is its purpose?"
>
> Let us first come to the answer of my question.

> Let us first come to the answer of my question.

> That was not my question.

> That was not my question.

> When would your crowd have supplied enough indices to
> fix a real number by its digits?

> When would your crowd have supplied enough indices to
> fix A and B by its digits?

> When would your crowd have supplied enough indices to
> fix A and B by its digits?

In summary, you want what you want when you want it
the way you want it, which excludes those "reasons"
slipped in with the answer.

Your rhetorical strategy appears to be
(i) whine enough to induce others to tell you to f. off
(ii) declare flawless victory.

I can only imagine how proud your students and colleagues at
Hochschule Augsburg are to associate with you,
Wolfgang Mückenheim

>> ----
>> Therefore, _all_ the decimal digits determine _at most_
>> one point in the line.
>
> How much is "all" of a crowd?

P(x) is true of _all_ of a crowd pp iff
P(x) is true _no matter which_ member of pp it is
that x refers to, _without exception_

In this particular instance, _all of a crowd_ refers to
all of the crowd of _indices_ There are claims which are
true of x (one of the crowd of indices) without exception.

x is countable-to.
( A completable 0,...,x exists.

x is countable beyond
( x+1 exists and is not in 0,...,x.

If they're used correctly, in the usual fashion,
all of those countable-to indices are sufficient
to remove ambiguity.

( For any two distinct points, finite decimals on one side
( and finite decimals on the other are closer than the two.
( Therefore, the two cannot both be allowed. For any two.
( Therefore, two are not allowed by the infinite decimal.
( No ambiguity.

Less than all of those indices are not sufficient digits.
Which real number would be ambiguous.

More than the countable-to indices are not necessary.

On 5/11/2021 2:22 PM, William wrote:
> On Tuesday, May 11, 2021 at 3:08:01 PM UTC-3, WM wrote:
>> William schrieb am Dienstag, 11. Mai 2021 um 18:43:54 UTC+2:
>
> WH: ... a set can be "next" to zero even though no element of the set is "next" to 0.
> WM: That is true, but it would be wrong, if all its elements could be defined
>
>>> There are two ways of proving something about every element of an infinite set. On can note that ever element of a set of X is an X and thus has the properties of X (e.g. (0,1] is a set of real numbers so every element of (0,1] is a member of the complete ordered field) or you can use induction (which is described elsewhere). ,
>>>
>>> Note that the "undefinable real numbers" are, despite the name, not real numbers. So (0,1], a set of real numbers, does not contain an "undefinable real number".
>> Every definable real number x has ℵo unit fractions in (0, x). So when you remove
>
> I do not "remove". This does not work. I use one of the two ways to prove something about every element in an infinite set. In this case I am noting that "undefinable real numbers" do not have the properties every element of (0,1] has.
>

how can "undefinable real numbers" be in (0,1] ? they cannot.

For any real interval [a,b], there cannot be "undefined real numbers"
in it. If so, that would put a bound on "undefined real number", which
is false.

that is actually a proof that Dark Numbers can not exist. Any real line
interval would not contain any Dark Numbers.

On 5/10/2021 1:50 PM, WM wrote:
> William schrieb am Montag, 10. Mai 2021 um 20:33:36 UTC+2:
>> On Monday, May 10, 2021 at 3:16:50 PM UTC-3, WM wrote:
>
>>> That is nonsense, because to prove it you have to define individually as many as collectively, i.e. all.
>> You use induction. E.g. All unit fractions are definable. We start by noting that if n is definable, then 1/n is definable. So the problem reduces to showing each of the naturals is definable. We note 0 is definable and if m is definable then m+1 is definable. So, by induction all naturals are definable. So all unit fractions are definable.
>
> Do you wish to remove your agreement to ∀n ∈ ℕ_def: (0, 1] \ [1/n, 1] contains ℵo unit fractions? All unit fractions subject to induction have this property.
>
>> The term "second half" has no meaning for infinite sets.
>
> The term "all with no remainder" has meaning for infinite sets. We can remove all natnural numbers with no remainder collectively. Now do the same individually. You can only remove a tiny initial segment . "Second half" has in fact no meaning because with the exception of a tiny initial segment the complete set is dark and therefore without order.
>
> Regards, WM
>

see? You look away and they all magically turn into Darkies.

What if 3 people were staring at the number 17 on a paper, and one looks
away ?

the number went dark for the person that looked away...

the number went dark for the person who is staring at 17 but thinking
about his new doggie...

WM's "Observer Dependent Math" is baloney...

William schrieb am Mittwoch, 12. Mai 2021 um 19:05:54 UTC+2:
> On Wednesday, May 12, 2021 at 1:04:44 PM UTC-3, WM wrote:
> > William schrieb am Mittwoch, 12. Mai 2021 um 16:40:20 UTC+2:
> > > On Wednesday, May 12, 2021 at 11:26:12 AM UTC-3, WM wrote:
> WH: ... a set can be "next" to zero even though no element of the set is "next" to 0.
> WM: That is true, but it would be wrong, if all its elements could be defined
> WH: You can use induction to prove that every element of S is definable
> > > > > You can also prove that no element is dark.
> > > > No element that is subject to induction is dark.
> > > And since, by definition, every element of a set subject to induction is subject to induction, we are done
> > We are done, if we refrain from finished infinity,
> "finished infinity" cannot add elements to S that are not part of S by definition.

It must and does because otherwise it is inconsistent. Note that when you remove the union of all intervals [1/(n+1), 1/n] from (0, 1] then nothing remains whereas the removal of [1/2, 1/1] and with [1/(n+1), 1/n] also [1/(n+2), 1/(n+1)] will never exhaust the interval (0, 1].

Regards, WM