Le 05/04/2024 à 20:07, Moebius a écrit :
>
> Hint@Mückenheim: {2, 4, 6, 8, 10, ..., 2ω} = {2n : n e IN} u {2ω}.
>
we can use the ordinal axis as Cantor has described it
0, 1, 2, 3, ..., ω, ω + 1, ..., ω + k, ..., ω + ω (= ω2), ω2 + 1,
...
and multgiply 0, 1, 2, 3, ..., ω, by 2. What is the fate of the distance
between ℕ and ω? Does it grow to the complete interval ω + 1, ..., ω
+ k, ..., ω + ω?
Regards, WM

Le 06/04/2024 à 15:40, Richard Damon a écrit :
> On 4/6/24 9:26 AM, WM wrote:
>> Le 05/04/2024 à 12:57, FromTheRafters a écrit :
>>> WM explained on 4/4/2024 :
>>
>>>> Explain why first bijecting n and n/1 should destroy an existing
>>>> bijection!
>>>
>>> You still seem to think that sets change. If you mean 'n' is an
>>> element of the naturals then of course N bijects with the naturals as
>>> embedded in Q.
>>
>> Of course. But if someone doubts it, I could directly map the naturals
>> n/1 to the fractions with the result that there is no bijection.
>
> No, not "No Bijection", but that mapping isn't a bijection.

That mapping is Cantor's proposal. But for every other mapping, the O's
would also remain. All O's! It is th lossless exchange which proves it.
>>
>> No, that is disproved by the remaining Os.
>
> Which only shows that this one mapping doesn't work.
>
It is Cantor's famaous mapping, more than a century believed to be a
bijection.

> And, when you try it within one set, as opposed to between two sets,

If it operates, it must operate within one set too.

Regards, WM

On 4/6/24 9:55 AM, WM wrote:
> Le 06/04/2024 à 15:40, Richard Damon a écrit :
>> On 4/6/24 9:26 AM, WM wrote:
>>> Le 05/04/2024 à 12:57, FromTheRafters a écrit :
>>>> WM explained on 4/4/2024 :
>>>
>>>>> Explain why first bijecting n and n/1 should destroy an existing
>>>>> bijection!
>>>>
>>>> You still seem to think that sets change. If you mean 'n' is an
>>>> element of the naturals then of course N bijects with the naturals
>>>> as embedded in Q.
>>>
>>> Of course. But if someone doubts it, I could directly map the
>>> naturals n/1 to the fractions with the result that there is no
>>> bijection.
>>
>> No, not "No Bijection", but that mapping isn't a bijection.
>
> That mapping is Cantor's proposal. But for every other mapping, the O's
> would also remain. All O's! It is th lossless exchange which proves it.

Cantor's proposal is between members of two distinct sets.

>>>
>>> No, that is disproved by the remaining Os.
>>
>> Which only shows that this one mapping doesn't work.
>>
> It is Cantor's famaous mapping, more than a century believed to be a
> bijection.

But HIS does work, when you do it right.

>
>> And, when you try it within one set, as opposed to between two sets,
>
> If it operates, it must operate within one set too.

Why?

IT is a mapping betweens elements of two defined sets.

If you don't have those two sets, you are following the mapping.

You are just proving your stupidity.

>
> Regards, WM
>
>
>
>

On 04/06/2024 06:31 AM, Jim Burns wrote:
> On 4/5/2024 10:37 PM, Ross Finlayson wrote:
>> On 04/05/2024 11:04 AM, Jim Burns wrote:
>
>>> [...]
>>
>> I usually write that n/d with
>> n as "numerator" and d as "denominator".
>
> Of course you do. I stand corrected.
>
>> It's not identified which f(n) is 1/root2,
>> only that it exists.
>> There's that for each r in [0,1],
>> exists n s.t. f(n) = r,
>> and f^-1(r) = n, mostly infinite.
>
> I don't see how ⅟√​̅2 is known to exist in [0,1]ᴿꟳ
> from the definition
> [0,1]ᴿꟳ =
> lim[d→∞, d∈ℕ⁺] lim[n→d, n∈ℕ] n/d =
> lim[d→∞, d∈ℕ⁺] {n/d: 0≤n≤d, n∈ℕ}
>
> Or, more generally, what supports a claim that
> [0,1]ᴿꟳ = [0,1]ᶜᵒⁿᵗⁱⁿᵘᵘᵐ
>
>
> I gave an example of what would support
> such a claim, from a different definition.
> [0,1]ⁿᵒᵗᐧᴿꟳ =
> { x = lub bnes ⊆ ℚ: 0≤x≤1}
>
> lub least.upper.bound
> bnes bounded.non.empty.set
>
> I'm not telling you to do it my way, but,
> if not my way, then how?
>
>> There's not much said except that
>> d goes to infinity, and n goes to d.
>> (Thus that it's not just zero.)
>
> I think it's essential to your project
> that more be said.
>
> I have inserted what I can, as best I can,
> of what you might mean.
>
> For example, for some reason, you really like
> to call a connected domain "continuous",
> an adjective I expect to see applied to
> a function. Not a deal.breaker.
>
> You have the opportunity to correct my insertions
> where I have misunderstood you.
>
> I read "goes to" as "ranges up to"
> which is why I write
> lim[d→∞, d∈ℕ] lim[n→d, n∈ℕ] n/d =
> lim[d→∞, d∈ℕ] {n/d: 0≤n≤d, n∈ℕ}
>
> And there's the rub.
>
> I can expand the limit in an often.used way
> lim[d→∞, d∈ℕ] {d/n: 0≤n≤d, n∈ℕ} =
> ⋂[d∈ℕ] ⋃[d′≥d:d′∈ℕ] {n/d′: 0≤n≤d′, n∈ℕ}
>
> However,
> ⋂[d∈ℕ+] ⋃[d′≥d:d′∈ℕ] {n/d′: 0≤n≤d′, n∈ℕ} =
> [0,1]ʳᵃᵗⁱᵒⁿᵃˡ
>
> [0,1]ʳᵃᵗⁱᵒⁿᵃˡ ≠ [0,1]ᶜᵒⁿᵗⁱⁿᵘᵘᵐ
> [0,1]ʳᵃᵗⁱᵒⁿᵃˡ ∌ ⅟√​̅2
>
> But
> [0,1]ᴿꟳ =
> lim[d→∞, d∈ℕ⁺] lim[n→d, n∈ℕ] n/d =
> lim[d→∞, d∈ℕ⁺] {n/d: 0≤n≤d, n∈ℕ} =
> ⋂[d∈ℕ+] ⋃[d′≥d:d′∈ℕ] {n/d′: 0≤n≤d′, n∈ℕ} =
> [0,1]ʳᵃᵗⁱᵒⁿᵃˡ
>
> Either
> that's how "limit" is defined here and
> [0,1]ᴿꟳ ≠ [0,1]ᶜᵒⁿᵗⁱⁿᵘᵘᵐ
> or
> "limit" is defined some other way and
> you should say what that way is.
>
>

It's pretty simple, a growing and constantly growing,
filling of [0,1], to be full.

This would be most simple as a function of a real-valued
variable x, f(x) = x, from zero to one.

Yet, starting with only natural integers, that are whole
quantities and units that range from 0, to 1, through infinity,
it's after division is defined, that these are growing and
constantly growing, the fractions filling [0,1], until,
in the limit, in the infinite limit, it's full.

So, the most natural usual sort of act of line-drawing,
is the idea that there's a space [0,1] and it's filled from
zero to one through each value, as beads-on-a-string,
contiguous, consecutive, what result the continuous,
in the infinite limit or continuum limit, then it results
that that's all there is to it.

Now, we know "points don't make a line" or "lines
don't make a point", yet, already today, the model
of the lines is of point-sets, that a line is the points
that comprise it, vis-a-vis Euclid's theory, where,
they're entirely different types in their relations.

So, the idea to make a line from points, that points
make a line, sort of follows from line-drawing, then
that in the infinitely-many dimensions not just one,
that it's a natural spiral space-filling curve as a
natural continuum, of points and spaces, making
for a geometry of points and spaces, altogether,
with regards to a very simple and fundamental
primitive theory, that results a geometry, that
results also Euclidean geometry.

Most people who've studied calculus are familiar
with the differential or d, where delta is a finite
difference, d is no more than an infinitesimal difference.

Then, this ran(f) the iota-values, is the entire apparatus,
which arrives at why it's not a Cartesian function, and
not-a-real-function in the standard sense, so that this
way, it lives as an object of the overall theory of all
the objects of mathematics, mutually intercompatible,
if not mutually interchangeable, with the other and
that being the standard usual model of a continuous
domain, the complete ordered field.

Thusly, line-reals as a continuous domain, and field-reals
as a continuous domain, have a very simple and direct
formalism, why they are so fundamental in the objects
of mathematics, connecting the discrete and continuous.

Now, more directly to your point, what you seem to
arrive at is "[0,1]^{dense though discontinuous}",
that completeness the least-upper-bound property,
is defined as that sup {f(m)} is f(max(m)+1), so that
there's least-upper-bound, so it's complete, thus it
results instead "[0,1]^{dense and not discontinuous}".

Of course, it would have to be arrived at why that's
not incompatible with the results of uncountability,
and it's so that how it is so compatible, is that via
inspection it falls out of the results of uncountability
as a counterexample and not contradicted, and lives
in the world of function theory, as not connected by
the usual Cantor-Schroeder-Bernstein theorem of
the transitive bijectivity as applying to Cartesian functions
particularly, then there's also for the set-theoretic
powerset result a notion of ubiquitous ordinals
about set theory vis-a-vis ordering theory, just
noting that it's overall a coherent and consistent thing.

I sort of defined it in terms of limit this way
about 25 or 30 years ago here, it's defined the same
quite thoroughly and throughout all my conversations since,
it's an object of mathematics so doesn't it exactly matter
what I say about it, only that it is what it is.

So, relating and connecting the discrete and continuous,
is about the most central and primary thing that this is.

Thanks for your reply, warm regards.

On 4/6/2024 9:44 AM, WM wrote:
> Le 05/04/2024 à 19:03, Jim Burns a écrit :
>> On 4/5/2024 5:06 AM, WM wrote:

>>> Dark ordinals reach till ω.
>>> Agreed?
>>
>> ⟦0,ξ⟧ which reaches 'til ω  both
>> is a Mückenheim.set and is not a Mückenheim.set.
>> Agreed?
>
> There are no Mückenheim sets.

It looks like you mean to say
there are no not.Mückenheim sets.

Mückenheim.ness is a property which you use
to prove bijections aren't bijections.
Paraphrasing:
| Look!
| This is a not.bijection.
| Therefore,
| that bijection is a not.bijection.
| Because darkᵂᴹ numbers.

You (WM) implicitly use a claim that
all sets are Mückenheim
That implicit claim is what you call logicᵂᴹ.

Mückenheim sets aren't a problem
for you or for us.
We call them finiteⁿᵒᵗᐧᵂᴹ.

You dislike not.Mückenheim sets,
which, de gustibus non disputandum est,
is not a problem, either.

If you don't want to talk about not.Mückenheim sets,
you can not.talk about not.Mückenheim sets.

Your problem is that you think that, by both
disliking and talking about not.Mückenheim sets,
not.Mückenheim sets become Mückenheim sets.

Sadly for you, fortunately for not.you,
arithmetic is not impressed by your résumé.

> But we can use the ordinal axis
> as Cantor has described it
> 0, 1, 2, 3, ...,
> ω, ω + 1, ..., ω + k, ...,
> ω + ω (= ω2), ω2 + 1, ..

The Mückenheim ordinals are confined to
the first row, before ω

We can talk about the others.
We can not.talk about the others.
Either way, the others aren't Mückenheim.

> and multiply 0, 1, 2, 3, ..., ω, by 2.

Two times anything on the first row
is something on the first row.

Being on the first row is determined by
the Mückenheim property.

If ⟨1,...,k⟩ has the Mückenheim property
then ⟨1,...,k,k+1,...,k+k⟩ also has it
and is also on the first row, before ω

One way to describe ordinal k as Mückenheim ==
as first.row number k == as k before ω is that
non.0 k and each non.0 before k
has an immediate predecessor.

k+k ​̄⟶ k+k-1 ​̄⟶ ... ​̄⟶ k+1 ​̄⟶ k ​̄⟶ ... ​̄⟶ 0

Each non.0 Mückenheim ordinal (before ω) has
an immediate predecessor.
If ω also had an immediate predecessor,
ω would be on the first row,
instead of beginning the second row.

>>> The difference between ⟦0,m+1⟧ and ⟦0,ω⟧ is
>>> how large?
>>
>>> Is it ω for every m?
>>
>> ∀k,m ∈ ⟦0,ω⦆: k+m ∈ ⟦0,ω⦆
>> So, yes.
>
> Then it is variable, not a fixed number.
> Actually infinite sets are constant.
> Potentially infinite collections are variable.

⟦0,ω⦆ doesn't change.
∀k,m ∈ ⟦0,ω⦆: k+m ∈ ⟦0,ω⦆

Perhaps there is a problem with
your assertion that all sets are Mückenheim.

Le 06/04/2024 à 15:58, Richard Damon a écrit :
> On 4/6/24 9:55 AM, WM wrote:

>> That mapping is Cantor's proposal. But for every other mapping, the O's
>> would also remain. All O's! It is th lossless exchange which proves it.
>
> Cantor's proposal is between members of two distinct sets.

No. He does not specify that. And there is no reason to do so, except that
it can be used to contradict the ridiculous nonsense that there are as
many fractions as prime numbers.
+ >> It is Cantor's famaous mapping, more than a century believed to be a
>> bijection.
>
> But HIS does work, when you do it right.

No, his bijection works only for potential infinity applying the "...". I
show that his mapping does not work for the complete actually infinite
sets. He uses "and so on". Why does any intelligent mind believe that? I
show that the remainder will never decrease. There is no belief required.
It is provable fact.
>
>> If it operates, it must operate within one set too.
>
> Why?

Because there are as many naturals in ℕ as in ℚ. Precisely as many.
But only my approach shows that they are less than the fractions.

Regards, WM

Le 06/04/2024 à 17:49, Jim Burns a écrit :
> On 4/6/2024 9:44 AM, WM wrote:

>> But we can use the ordinal axis
>> as Cantor has described it
>> 0, 1, 2, 3, ...,
>> ω, ω + 1, ..., ω + k, ...,
>> ω + ω (= ω2), ω2 + 1, ..
>> and multiply 0, 1, 2, 3, ..., ω, by 2.
>
> Two times anything on the first row
> is something on the first row.

That is contradicted by this argument: Two times all numbers of 1, 2, 3,
..., ω will double the distance between ℕ and ω to that between 2ℕ
and 2ω. That distance is less than the distance between ω and ω2.
Everything before this is a doubled natural number and simultaneously a
transfinite number.

Regards, WM

On 4/6/24 3:40 PM, WM wrote:
> Le 06/04/2024 à 15:58, Richard Damon a écrit :
>> On 4/6/24 9:55 AM, WM wrote:
>
>>> That mapping is Cantor's proposal. But for every other mapping, the
>>> O's would also remain. All O's! It is th lossless exchange which
>>> proves it.
>>
>> Cantor's proposal is between members of two distinct sets.
>
> No. He does not specify that. And there is no reason to do so, except
> that it can be used to contradict the ridiculous nonsense that there are
> as many fractions as prime numbers.y

But he DOES, as he talks about the two SETS of numbers that are matched up.

And yes, the size of the set of all fractions is EXACTLY of the same
size as the set of all Prime Numbers, and that size is Aleph_0.

That you can't understand that is YOUR problem, because your mind just
can't understand things bigger than it.

> +
>>> It is Cantor's famaous mapping, more than a century believed to be a
>>> bijection.
>>
>> But HIS does work, when you do it right.
>
> No, his bijection works only for potential infinity applying the "...".
> I show that his mapping does not work for the complete actually infinite
> sets. He uses "and so on". Why does any intelligent mind believe that? I
> show that the remainder will never decrease. There is no belief
> required. It is provable fact.

Excpet that he doesn't need to use "..." as he can just list the formula
that maps any k to n,d and back, and show that each n.d generates a
unique k, and each k list a unique n,d. That is the bijection.

Only if you want to try to LIST all the members of the mapping, do you
need to use ..., and that is because the set is INFINITE in length, and
thus unlistable.

You logic is just INVALID when applied to infinite sets, and totally
blows up and becomes inconsistant when you try to do so.

You seem to be stuck in "Dead Man Walking" mode.

>>
>>> If it operates, it must operate within one set too.
>>
>> Why?
>
> Because there are as many naturals in ℕ as in ℚ. Precisely as many. But
> only my approach shows that they are less than the fractions.

Since Q is the set of ALL RATIONAL numbers, what "fraction" isn't a
Rational Number?

Note, the fact that you can map the Natural numbers N to a subset of Q
(the elements that have the same value as a natural number) doesn't mean
that their can not be also a mapping from N to ALL of Q.

You are just showing you don't understand that nature of infinite sets,
because you brain just can't handle something bigger than it.

>
> Regards, WM
>
>

Le 06/04/2024 à 22:03, Richard Damon a écrit :
> On 4/6/24 3:40 PM, WM wrote:
>> Le 06/04/2024 à 15:58, Richard Damon a écrit :
>>> On 4/6/24 9:55 AM, WM wrote:
>>
>>>> That mapping is Cantor's proposal. But for every other mapping, the
>>>> O's would also remain. All O's! It is th lossless exchange which
>>>> proves it.
>>>
>>> Cantor's proposal is between members of two distinct sets.
>>
>> No. He does not specify that. And there is no reason to do so, except
>> that it can be used to contradict the ridiculous nonsense that there are
>> as many fractions as prime numbers.y
>
> But he DOES, as he talks about the two SETS of numbers that are matched up.

One set and its subset. Dedekind: A system S is said to be /infinite/ if
it is similar to a real part of itself. To consider them as two sets does
not change the numbers of elements.
>
> And yes, the size of the set of all fractions is EXACTLY of the same
> size as the set of all Prime Numbers, and that size is Aleph_0.

Wrong. Proof: All prime numbers p are fractions p/1 ∈ ℚ, but 1/2 is
not prime.

Regards, WM

On 4/7/24 4:32 AM, WM wrote:
> Le 06/04/2024 à 22:03, Richard Damon a écrit :
>> On 4/6/24 3:40 PM, WM wrote:
>>> Le 06/04/2024 à 15:58, Richard Damon a écrit :
>>>> On 4/6/24 9:55 AM, WM wrote:
>>>
>>>>> That mapping is Cantor's proposal. But for every other mapping, the
>>>>> O's would also remain. All O's! It is th lossless exchange which
>>>>> proves it.
>>>>
>>>> Cantor's proposal is between members of two distinct sets.
>>>
>>> No. He does not specify that. And there is no reason to do so, except
>>> that it can be used to contradict the ridiculous nonsense that there
>>> are as many fractions as prime numbers.y
>>
>> But he DOES, as he talks about the two SETS of numbers that are
>> matched up.
>
> One set and its subset. Dedekind: A system S is said to be /infinite/ if
> it is similar to a real part of itself. To consider them as two sets
> does not change the numbers of elements.

But does affect your logic of pairing.

>>
>> And yes, the size of the set of all fractions is EXACTLY of the same
>> size as the set of all Prime Numbers, and that size is Aleph_0.
>
> Wrong. Proof: All prime numbers p are fractions p/1 ∈ ℚ, but 1/2 is not
> prime.
>

So, With infinite sets, a proper subset CAN be the same size as its parent.

You are just PROVING you don't understand how infinity works, because
your brain just can't handle it.

Your logic system has gone BOOM and taken your intelect with it.

> Regards, WM

Le 07/04/2024 à 13:16, Richard Damon a écrit :
> On 4/7/24 4:32 AM, WM wrote:
>> Le 06/04/2024 à 22:03, Richard Damon a écrit :
>>> On 4/6/24 3:40 PM, WM wrote:
>>>> Le 06/04/2024 à 15:58, Richard Damon a écrit :
>>>>> On 4/6/24 9:55 AM, WM wrote:
>>>>
>>>>>> That mapping is Cantor's proposal. But for every other mapping, the
>>>>>> O's would also remain. All O's! It is th lossless exchange which
>>>>>> proves it.
>>>>>
>>>>> Cantor's proposal is between members of two distinct sets.
>>>>
>>>> No. He does not specify that. And there is no reason to do so, except
>>>> that it can be used to contradict the ridiculous nonsense that there
>>>> are as many fractions as prime numbers.y
>>>
>>> But he DOES, as he talks about the two SETS of numbers that are
>>> matched up.
>>
>> One set and its subset. Dedekind: A system S is said to be /infinite/ if
>> it is similar to a real part of itself. To consider them as two sets
>> does not change the numbers of elements.
>
>
> But does affect your logic of pairing.

No. Since there are precisely as many natnumbers n as natnumber fractions
n/1, nothing is affected. The only effect is that the Os can be proven to
remain the same number in every step. This is true in all mappings but
more easily seen in mine.

> So, With infinite sets, a proper subset CAN be the same size as its parent.

Impossible.
>
> You are just PROVING you don't understand how infinity works,

I understand that a crowd of fools has been tricked by Cantor.

Regards, WM

On 4/7/24 9:23 AM, WM wrote:
> Le 07/04/2024 à 13:16, Richard Damon a écrit :
>> On 4/7/24 4:32 AM, WM wrote:
>>> Le 06/04/2024 à 22:03, Richard Damon a écrit :
>>>> On 4/6/24 3:40 PM, WM wrote:
>>>>> Le 06/04/2024 à 15:58, Richard Damon a écrit :
>>>>>> On 4/6/24 9:55 AM, WM wrote:
>>>>>
>>>>>>> That mapping is Cantor's proposal. But for every other mapping,
>>>>>>> the O's would also remain. All O's! It is th lossless exchange
>>>>>>> which proves it.
>>>>>>
>>>>>> Cantor's proposal is between members of two distinct sets.
>>>>>
>>>>> No. He does not specify that. And there is no reason to do so,
>>>>> except that it can be used to contradict the ridiculous nonsense
>>>>> that there are as many fractions as prime numbers.y
>>>>
>>>> But he DOES, as he talks about the two SETS of numbers that are
>>>> matched up.
>>>
>>> One set and its subset. Dedekind: A system S is said to be /infinite/
>>> if it is similar to a real part of itself. To consider them as two
>>> sets does not change the numbers of elements.
>>
>>
>> But does affect your logic of pairing.
>
> No. Since there are precisely as many natnumbers n as natnumber
> fractions n/1, nothing is affected. The only effect is that the Os can
> be proven to remain the same number in every step. This is true in all
> mappings but more easily seen in mine.

Yes, the size of the Natual Numbers is the same size as the Size of the
Rational Fractions that have a denomenator equal to 1.

This doesn't mean it can't ALSO be the same size as the full field of n/d.

Your inability to understand that is just your own problem.

>
>> So, With infinite sets, a proper subset CAN be the same size as its
>> parent.
>
> Impossible.

Nope, PROVEN.

Since the DEFINITION of "Same Size" is the ability to make a 1-to-1
mapping between the sets.

Do you want to claim that two sets that you can match EVERY DISTINCT
element of one to a UNIQUE DISTINCT ELEMENT of the other are NOT the
same size?

and we can build such a mapping between the set of natural Numbers (N)
with the set of even Numbers (E).

Since for ALL elements n, a member of the Natural Numbers, there exists
an element e, a member of tghe Even Nubers, such that the value of e is
twice the value of n (e = 2n)

EVERY element of N is mapped to a DISTINCT element of E.

Try to find an exception.

If E is smaller than N, then BY DEFINITION, when building a bijection,
two different values of n must map to the same value of E, but that
never happens.

But E is ALSO a proper subset of the Natural Numbers, as we can also
build E by removing all the "odd" values from the Natural Numbers.

In fact, it turns out that the size of the set of Natural Numbers is
exactly the same size of ANY non-finite subset of any set based on some
bounded length tuple of natural numbers (like the rationals). They are
all "Countably Infinite" with a size of Aleph_0.

>>
>> You are just PROVING you don't understand how infinity works,
>
> I understand that a crowd of fools has been tricked by Cantor.

NOPE, you have FOOLED YOURSELF by beleiving your own lies.

Your problem is your instance on using "finite" logic on infinite sets,
which makes your world blow itself up in contradictions.

Which seems to have blown your brains out of your head.

>
> Regards, WM

On 4/6/2024 3:49 PM, WM wrote:
> Le 06/04/2024 à 17:49, Jim Burns a écrit :
>> On 4/6/2024 9:44 AM, WM wrote:

>>> But we can use the ordinal axis
>>> as Cantor has described it
>>> 0, 1, 2, 3, ...,
>>> ω, ω + 1, ..., ω + k, ...,
>>> ω + ω (= ω2), ω2 + 1, ..
>>> and multiply 0, 1, 2, 3, ..., ω, by 2.
>>
>> Two times anything on the first row
>> is something on the first row.
>
> That is contradicted by

....your anti.arithmetism.

tl;dr
The successor operation is closed in
the natural numbers.
Natural.number.addition is closed in
the natural numbers.
Natural.number.multiplication is closed in
the natural numbers.

Those claims are theorems.
In order to be theorems,
it is important to know the meaning of
successor, natural number, addition and multiplication.

If you want to see the proofs, ask.

Consider the successor.operation ⁺¹ which is
non.0, 1.to.1, and closed in the successor.havers.
i⁺¹≠0 ∧ ¬∃h≠i:h⁺¹=i⁺¹ ∧ ∃k=i⁺¹⁺¹ ∧ 1=0⁺¹

For some successor.havers k
a set ⟦0,k⦆ exists such that
predecessor k⁻¹ of k and
predecessor i⁻¹ of each i≠0 in ⟦0,k⦆ and
0 are in ⟦0,k⦆
∀i ∈ ⟦0,k⦆: i⁺¹ ∈ ⦅0,k⟧

The natural numbers ℕ are
0 and successor.havers with ⟦0,k⦆ like that
k ∈ ℕ ⟺
k=0 ∨ ∃⟦0,k⦆: ∀i ∈ ⟦0,k⦆: i⁺¹ ∈ ⦅0,k⟧

That successor operation ⁺¹ is closed in ℕ
If
∃⟦0,k⦆: ∀i ∈ ⟦0,k⦆: i⁺¹ ∈ ⦅0,k⟧
then
∃⟦0,k⁺¹⦆: ∀i ∈ ⟦0,k⁺¹⦆: i⁺¹ ∈ ⦅0,k⁺¹⟧
Thus,
if k ∈ ℕ
then k⁺¹ ∈ ℕ

Addition in ℕ is defined so that
k+m=n ⟺
a sequence fₖ₊ₘ₌ₙ: ⟦0,m⟧ ⟶ 𝔸 exists of addition.facts
starting with ⟨k+0=k⟩, ending with ⟨k+m=n⟩
and ⟨k+i=j⟩ ⇔ ⟨k+i⁺¹=j⁺¹⟩
∀k,m,n ∈ ℕ:
k+m=n ⟺
∃fₖ₊ₘ₌ₙ: ⟦0,m⟧ ⟶ 𝔸:
fₖ₊ₘ₌ₙ(0)=⟨k+0=k⟩ ∧ fₖ₊ₘ₌ₙ(m)=⟨k+m=n⟩ ∧
∀i ∈ ⟦0,m⦆:
fₖ₊ₘ₌ₙ(i)=⟨k+i=j⟩ ⇔ fₖ₊ₘ₌ₙ(i⁺¹)=⟨k+i⁺¹=j⁺¹⟩

Addition in ℕ is closed in ℕ
If
∀i ∈ ⟦0,k⦆: i⁺¹ ∈ ⦅0,k⟧ and
∀i ∈ ⟦0,m⦆: i⁺¹ ∈ ⦅0,m⟧
then
∀i ∈ ⟦0,k+m⦆: i⁺¹ ∈ ⦅0,k+m⟧
Thus,
if k,m ∈ ℕ
then k+m ∈ ℕ

Multiplication in ℕ is defined so that
k⋅m=n ⟺
a sequence fₖ.ₘ₌ₙ: ⟦0,m⟧ ⟶ 𝕄 exists of multiplication.facts
starting with ⟨k⋅0=0⟩, ending with ⟨k⋅m=n⟩
and ⟨k⋅i=j⟩ ⇔ ⟨k⋅i⁺¹=j+k⟩
∀k,m,n ∈ ℕ:
k⋅m=n ⟺
∃fₖ.ₘ₌ₙ: ⟦0,m⟧ ⟶ 𝕄:
fₖ.ₘ₌ₙ(0)=⟨k⋅0=0⟩ ∧ fₖ.ₘ₌ₙ(m)=⟨k⋅m=n⟩ ∧
∀i ∈ ⟦0,m⦆:
fₖ.ₘ₌ₙ(i)=⟨k+i=j⟩ ⇔ fₖ.ₘ₌ₙ(i⁺¹)=⟨k+i⁺¹=j+k⟩

Multiplication in ℕ is closed in ℕ
If
∀i ∈ ⟦0,k⦆: i⁺¹ ∈ ⦅0,k⟧ and
∀i ∈ ⟦0,m⦆: i⁺¹ ∈ ⦅0,m⟧
then
∀i ∈ ⟦0,k⋅m⦆: i⁺¹ ∈ ⦅0,k⋅m⟧
Thus,
if k,m ∈ ℕ
then k⋅m ∈ ℕ

> That is contradicted by this argument:
> Two times all numbers of 1, 2, 3, .., ω
> will double the distance between ℕ and ω
> to that between 2ℕ and 2ω.

No, it won't.

Le 07/04/2024 à 19:56, Richard Damon a écrit :
> On 4/7/24 9:23 AM, WM wrote:

>>> So, With infinite sets, a proper subset CAN be the same size as its
>>> parent.
>>
>> Impossible.
>
> Nope, PROVEN.

Proven impossble with my matrix,
>
> Since the DEFINITION of "Same Size" is the ability to make a 1-to-1
> mapping between the sets.
>
> Do you want to claim that two sets that you can match EVERY DISTINCT
> element of one to a UNIQUE DISTINCT ELEMENT of the other are NOT the
> same size?
>
> and we can build such a mapping between the set of natural Numbers (N)
> with the set of even Numbers (E).

Only handwaving by "and so on"

> Since for ALL elements n, a member of the Natural Numbers, there exists
> an element e, a member of tghe Even Nubers, such that the value of e is
> twice the value of n (e = 2n)
>
> EVERY element of N is mapped to a DISTINCT element of E.
>
> Try to find an exception

In all cases there are infinitely many exceptions.
∀n ∈ ℕ_applied: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

Regards, WM

Le 07/04/2024 à 21:47, Jim Burns a écrit :
> On 4/6/2024 3:49 PM, WM wrote:
>> Le 06/04/2024 à 17:49, Jim Burns a écrit :
>>> On 4/6/2024 9:44 AM, WM wrote:
>
>>>> But we can use the ordinal axis
>>>> as Cantor has described it
>>>> 0, 1, 2, 3, ...,
>>>> ω, ω + 1, ..., ω + k, ...,
>>>> ω + ω (= ω2), ω2 + 1, ..
>>>> and multiply 0, 1, 2, 3, ..., ω, by 2.
>>>
>>> Two times anything on the first row
>>> is something on the first row.
>>
>> That is contradicted by
>
> ...your anti.arithmetism.

If all products of the sequence 1, 2, 3, ..., remain below ω but ω ss
mapped to ω2, then the distance between ℕ and ω is increased from 0 to
infinity by multiplying.
>
> The successor operation is closed in
> the natural numbers.

For visible numbers only.

> Natural.number.addition is closed in
> the natural numbers.
> Natural.number.multiplication is closed in
> the natural numbers.
>
> Those claims are theorems.
> In order to be theorems,
> it is important to know the meaning of
> successor, natural number, addition and multiplication.
>
> If you want to see the proofs, ask.

The proofs are valid for visible numbers only. But most numbers are
invisible as is shown by the fact that no numbers fits between ℕ and ω.

>> Two times all numbers of 1, 2, 3, .., ω
>> will double the distance between ℕ and ω
>> to that between 2ℕ and 2ω.
>
> No, it won't.

*2 doubles every structure. According to my logic.

Regards, WM

On 4/8/24 9:44 AM, WM wrote:
> Le 07/04/2024 à 19:56, Richard Damon a écrit :
>> On 4/7/24 9:23 AM, WM wrote:
>
>>>> So, With infinite sets, a proper subset CAN be the same size as its
>>>> parent.
>>>
>>> Impossible.
>>
>> Nope, PROVEN.
>
> Proven impossble with my matrix,

Nope, since you matrix doesn't follow the required form.

All you have done is proven that YOUR logic can yield different
contractory results depending on which valid path you follow.

That means that YOUR logic system is proven INCONSISTENT, and thus BLOWN
UP.

>>
>> Since the DEFINITION of "Same Size" is the ability to make a 1-to-1
>> mapping between the sets.
>>
>> Do you want to claim that two sets that you can match EVERY DISTINCT
>> element of one to a UNIQUE DISTINCT ELEMENT of the other are NOT the
>> same size?
>>
>> and we can build such a mapping between the set of natural Numbers (N)
>> with the set of even Numbers (E).
>
> Only handwaving by "and so on"

Nope.

>
>> Since for ALL elements n, a member of the Natural Numbers, there
>> exists an element e, a member of tghe Even Nubers, such that the value
>> of e is twice the value of n (e = 2n)
>>
>> EVERY element of N is mapped to a DISTINCT element of E.
>>
>> Try to find an exception
>
> In all cases there are infinitely many exceptions.
> ∀n ∈ ℕ_applied: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

I didn't say "N_applied", I said N.

Your problem is that YOUR logic system can't actually have N, but you
still talk as if it does.

Thus, your system is BLOWN UP.

N is the FULL SET of Natural Numbers, all countable infinite number of
them all the way to the unreachable (by finite operations) end.

If your logic system can't handle that, you LIE every time you mention
that set.

>
> Regards, WM

On 4/8/2024 9:55 AM, WM wrote:
> Le 07/04/2024 à 21:47, Jim Burns a écrit :

>> The successor operation is closed in
>> the natural numbers.
>
> For visible numbers only.

Visibleᵂᴹ or darkᵂᴹ,
k is a natural number :⟺
k=0 ∨ ∃⟦0,k⦆: ∀i ∈ ⟦0,k⦆: i⁺¹ ∈ ⦅0,k⟧

Visibleᵂᴹ or darkᵂᴹ,
if
∃⟦0,k⦆: ∀i ∈ ⟦0,k⦆: i⁺¹ ∈ ⦅0,k⟧
then
∃⟦0,k⁺¹⦆: ∀i ∈ ⟦0,k⁺¹⦆: i⁺¹ ∈ ⦅0,k⁺¹⟧

⟦0,k⁺¹⦆ = ⟦0,k⦆∪{k}

Visibleᵂᴹ or darkᵂᴹ,
if
k is a natural number,
then
then k⁺¹ is a natural number.

Visibleᵂᴹ or darkᵂᴹ,
the successor operation is closed in
the natural numbers.

Am 09.04.2024 um 03:54 schrieb Jim Burns:
> On 4/8/2024 9:55 AM, WM wrote:
>> Le 07/04/2024 à 21:47, Jim Burns a écrit :
>
>>> The successor operation is closed in
>>> the natural numbers.
>>
>> For visible numbers only.
>
> Visibleᵂᴹ or darkᵂᴹ,
> k is a natural number  :⟺

k e IN. (I hope that you don't mind that slight simplification).

> Visibleᵂᴹ or darkᵂᴹ,
> the successor operation is closed in
> the natural numbers.

As is stated as (i) a Peano axiom (where we do not differentiate between
visibleᵂᴹ and darkᵂᴹ natural numbers)

An e IN: s(n) e IN

or (ii) can be PROVED in set theory, where s(x) := x u {x} and IN := the
intersection of all "successor sets". [Hence IN is a successor set
itself, which means: 0 e IN and An e IN: s(n) e IN.]

Again this holds for ALL natural numbers (no matter if they are
visibleᵂᴹ or darkᵂᴹ).

Le 09/04/2024 à 01:22, Richard Damon a écrit :
> On 4/8/24 9:44 AM, WM wrote:
>> Le 07/04/2024 à 19:56, Richard Damon a écrit :
>>> On 4/7/24 9:23 AM, WM wrote:
>>
>>>>> So, With infinite sets, a proper subset CAN be the same size as its
>>>>> parent.
>>>>
>>>> Impossible.
>>>
>>> Nope, PROVEN.
>>
>> Proven impossble with my matrix,
>
> Nope, since you matrix doesn't follow the required form.

It does precilsely.

>>> and we can build such a mapping between the set of natural Numbers (N)
>>> with the set of even Numbers (E).
>>
>> Only handwaving by "and so on"

>> In all cases there are infinitely many exceptions.
>> ∀n ∈ ℕ_applied: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
>
> I didn't say "N_applied", I said N.

But what you can use belongs to ℕ_applied. Otherwise show a natural
number that completes the bijection, i.e., which has not infinitely many
pairings on front.

Regards, WM

Le 09/04/2024 à 01:54, Jim Burns a écrit :
> On 4/8/2024 9:55 AM, WM wrote:
>> Le 07/04/2024 à 21:47, Jim Burns a écrit :
>
>>> The successor operation is closed in
>>> the natural numbers.
>>
>> For visible numbers only.
>
> Visibleᵂᴹ or darkᵂᴹ,
> k is a natural number :⟺
> k=0 ∨ ∃⟦0,k⦆: ∀i ∈ ⟦0,k⦆: i⁺¹ ∈ ⦅0,k⟧

Not correct if there are all natural numbers such that no further one
exists below ω. Multiplication by 2 creates numbers beyond ω, or there
would be numbers immune to multiplication.

Regards, WM

WM drivels horrendous bullshit:

> Multiplication by 2 creates numbers beyond ω

This statement is totally idiotic, as usual.

Am 09.04.2024 um 14:47 schrieb Tom Bola:
> WM drivels horrendous bullshit:
>
>> Multiplication by 2 creates numbers beyond ω
>
> This statement is totally idiotic, as usual.

Ach?

(Wer lutscht denn da?)

Am 09.04.2024 um 14:22 schrieb WM:

> there would be numbers immune to multiplication.

There are such numbers: 0 * n = 0 for all n e IN.

Moebius schrieb:

> Am 09.04.2024 um 14:47 schrieb Tom Bola:
>> WM drivels horrendous bullshit:
>>
>>> Multiplication by 2 creates numbers beyond ω
>>
>> This statement is totally idiotic, as usual.
>
> Ach?
>
> (Wer lutscht denn da?)

Ja, ein einzelnes herausgehobenes Statement mit harscher Kritik (und keine
ganze Abhandlung wegen der Langeweile so wie du oft) im Rahmen einer Phase,
wenn das alle täten, würde WM womöglich die Laune zu posten verlieren...

Am 09.04.2024 um 17:15 schrieb Tom Bola:
> Moebius schrieb:
>
>> Am 09.04.2024 um 14:47 schrieb Tom Bola:
>>> WM drivels horrendous bullshit:
>>>
>>>> Multiplication by 2 creates numbers beyond ω
>>>
>>> This statement is totally idiotic, as usual.
>>
>> Ach?
>>
>> (Wer lutscht denn da?)
>
> Ja, ein einzelnes herausgehobenes Statement mit harscher Kritik (und keine
> ganze Abhandlung wegen der Langeweile so wie du oft) im Rahmen einer Phase,
> wenn das alle täten, würde WM womöglich die Laune zu posten verlieren...

Ja, da gebe ich Dir Recht. :-)