tisdag 11 maj 2021 kl. 14:02:56 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 11. Mai 2021 um 06:56:01 UTC+2:
> > måndag 10 maj 2021 kl. 19:48:00 UTC+2 skrev WM:
> > > zelos...@gmail.com schrieb am Montag, 10. Mai 2021 um 12:58:54 UTC+2:
> > > > torsdag 6 maj 2021 kl. 13:00:42 UTC+2 skrev WM:
> > >
> > > > > Set theorists claim that LimCardS(n) can differ from CardLimS(n). The explanation of this claim has an easy explanation: LimCardS(n) is the limit of all definable terms of the sequence whereas LimS(n) includes the undefinable dark terms. Therefore CardLimS(n) can differ from LimCardS(n).
> > > > >
> > > > No, Card Lim and Lim Card are different because we can show cases where htey are different.
> > > You can show such cases, but you cannot explain why there is a difference.
> > >
> > Actually we can, one is looking at the cardinality of the limit of set sequence while the other is looking at the limit of the cardinality sequence.
> Example: The endsegments. Limit of cardinality is ℵo. Because all endsegments that you can see are infinite. But cardinality of the limit is zero because ...? No, you cannot see all the finite endsegments which must exist because the strict condition is irrevocable: Only one element lost per term! But you can prove that no element remains forever. So you extrapolate, without viewing all endsegments, that the last endsegment must be empty.
>
> Regards, WM

There are no finite endsegments by the definition.

And yes, because all elements are removed hte limit is the empty set with cardinality 0.

>I see the term 'element' is inauthentic within set theory. When sets contain elements and elements are sets then there is a structural failing.

Why? in set theory everything is a set.

>The notion of the term 'elemental' is to be a primitive thing, whereas the mathematician has been forced to admit a blunderous stage of doubling back and thus adopting a compiler error so low in the hierarchy that rather a lot could collapse. How much is built upon such a faulty footing?

Elemental and Element are not the same fucking word!

Noun

element (plural elements)

One of the simplest or essential parts or principles of which anything consists, or upon which the constitution or fundamental powers of anything are based.

Letters are the elements of written language.

(chemistry) Any one of the simplest chemical substances that cannot be decomposed in a chemical reaction or by any chemical means and made up of atoms all having the same number of protons.
One of the four basic building blocks of matter in theories of ancient philosophers and alchemists: water, earth, fire, and air.
(law) A required aspect or component of a cause of action. A deed is regarded as a violation of law only if each element can be proved.
(set theory) One of the objects in a set.
Any of the teeth of a zip fastener.

A small part of the whole.

an element of doubt;  an element of the picture

(obsolete) The sky.
(plural only, with "the") Atmospheric forces such as strong winds and rains..

exposed to the elements

A place or state of being that an individual or object is best suited to.

to be in one's element

(Christianity, usually in the plural) The bread and wine taken at Holy Communion.
A group of people within a larger group having a particular common characteristic.

You sometimes find the hooligan element at football matches.

A component in electrical equipment, often in the form of a coil, having a high resistance, thereby generating heat when a current is passed through it.

The element in this electric kettle can heat the water in under a minute.

(computing) One of the conceptual objects in a markup language, usually represented in text by tags.

Adjective

elemental (not comparable)

(chemistry) Of, relating to, or being an element (as opposed to a compound).
Basic, fundamental or elementary.

Of the ancient supposed elements of earth, air, fire and water.
(by extension) Of, or relating to a force or nature, especially to severe atmospheric conditions.

On 5/11/2021 6:47 AM, WM wrote:
> Dan Christensen schrieb am Dienstag, 11. Mai 2021 um 01:39:44 UTC+2:
>> On Monday, May 10, 2021 at 4:20:58 PM UTC-4, WM wrote:
>>> Dan Christensen schrieb am Montag, 10. Mai 2021 um 21:22:02 UTC+2:
>>>> On Monday, May 10, 2021 at 1:53:58 PM UTC-4, WM wrote:
>>>
>>>>> The questions is: how can it be reconciled with the fact that only one element is lost per step, i.e., consecutive terms differ merely by one natural number.
>>>> There is nothing to "reconcile."
>>> You can't see it. The questions is: how can it be reconciled with the fact that only one element is lost per step, i.e., consecutive terms differ merely by one natural number.
>>>
>> If you are claiming that this leads to a contradiction in set theory, that would require a proof.
>
> It leads to a contradiction in mathematics. A descending sequence of sets with one element lost per term and an empty end but without finite terms is an inconsistency proof in mathematics. Many set theorists are too blind to understand this. But that is no reason to change mathematical arguments. Either you understand mathematics, or you stay outside.
>
> Regards, WM
>

Please use math to describe what you are saying.
your verbage is too broad, initial conditions are unknown.
In a Proof, all information is there, can you provide one?

If you cannot, no problem, just write down the sets as example;

S_k {1,2,3,...k}
S_k-1 {1,2,3,...k-1}
..
..
..
S_0 {}

is that it ?

if your point is that when k is infinite... how is that inconsistent ?

All you have in above is;

"A descending sequence of sets with one element lost per term and an
empty end but without finite terms"

what does that mean specifically ? (and use math to describe it not
language)

Gus Gassmann schrieb am Dienstag, 11. Mai 2021 um 12:20:24 UTC+2:
> On Monday, 10 May 2021 at 17:24:56 UTC-3, WM wrote:

> > Can a strictly decreasing sequence of sets where every set S(n+1) has precisely one element less than its predecessor S(n) lose every element and yield the empty set without passing through finite sets?
> It is actually very funny to see you talk about mental capacities. It is quite clear that the deficiencies are on your side, and they are selective. You also have an inborn resistance to considering analogies and to learn from them.

I see that there is no answer. Are you too much ashamed to give an answer?
>
> So you have on the one hand an insistence that 0.999... =/= 1, since the left-hand side represents a sequence of partial sums and the right-hand side is the *limit* of that sequence. You're not mainstream on this question, but it is a reasonable way to resolve the inherit ambiguity in the notation 0.999....

Nice to hear.
>
> You must at some point have applied the same reasoning to the sequence {1, 1+1/2, 1+1/2+1/3,...}. Do you understand that the positive real numbers can be completed by adjoining to it the symbol oo, so that divergent sequences of partial sums such as {1, 1+1/2, 1+1/2+1/3,...} can be said to have a limit, namely oo?

oo does not mean completion but represents the potentially infinite. When we say "it is the limit oo" that means only that the sum increases without end. Cantor said 1 + 1/2 + 1/3 + ... = ω. That is a fixed number.
>
> Have you ever thought what happens when you apply that same reasoning to the sequence { {1}, {1} U {2}, {1} U {2} U {3}, ...}?

We get |N, that is all natural numbers, but we do not get ω, the first transfinite number. In the same way we get with 0.999... all finitely indexed nines, but that is not the analytical limit 1.

The set theoretic identification of |N and omega is simply sloppy and worse..

> Have you ever thought *why* Halmos went to the trouble of defining a limit for sequences of sets such as the one above? You *know* what that limit is, but you refuse to accept its significance, and I have never understood why. I mean, the facts are staring you in the face.

It is difficult to distinguish between unions and limits in set theory. The union of all natnumbers is |N. For that we don't need a limit, since we accept that all are there and can simply be unioned.

In set theory we need two different notions. (See "Counting like an old-fashioned station clock" in https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf) Let us use commercial calculation. Unionizing all FISONs we get the set of all natural numbers {1, 2, 3, ...}

{1}
{1, 2}
{1, 2, 3}
....
__________
{1, 2, 3, ...}

But when not only all FISONs but in addition the set of all natural numbers are unionized, then we get the same result

{1}
{1, 2}
{1, 2, 3}
....
{1, 2, 3, ...}
__________
{1, 2, 3, ...}

Strange.

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 11. Mai 2021 um 14:10:21 UTC+2:
> tisdag 11 maj 2021 kl. 14:02:56 UTC+2 skrev WM:
> > zelos...@gmail.com schrieb am Dienstag, 11. Mai 2021 um 06:56:01 UTC+2:
> > > måndag 10 maj 2021 kl. 19:48:00 UTC+2 skrev WM:
> > > > zelos...@gmail.com schrieb am Montag, 10. Mai 2021 um 12:58:54 UTC+2:
> > > > > torsdag 6 maj 2021 kl. 13:00:42 UTC+2 skrev WM:
> > > >
> > > > > > Set theorists claim that LimCardS(n) can differ from CardLimS(n). The explanation of this claim has an easy explanation: LimCardS(n) is the limit of all definable terms of the sequence whereas LimS(n) includes the undefinable dark terms. Therefore CardLimS(n) can differ from LimCardS(n).
> > > > > >
> > > > > No, Card Lim and Lim Card are different because we can show cases where htey are different.
> > > > You can show such cases, but you cannot explain why there is a difference.
> > > >
> > > Actually we can, one is looking at the cardinality of the limit of set sequence while the other is looking at the limit of the cardinality sequence.
> > Example: The endsegments. Limit of cardinality is ℵo. Because all endsegments that you can see are infinite. But cardinality of the limit is zero because ...? No, you cannot see all the finite endsegments which must exist because the strict condition is irrevocable: Only one element lost per term! But you can prove that no element remains forever. So you extrapolate, without viewing all endsegments, that the last endsegment must be empty.
> >
> There are no finite endsegments by the definition.
>
> And yes, because all elements are removed hte limit is the empty set with cardinality 0.

That is impossible by definition, because per term only one natnumber gets lost

∀k ∈ ℕ: E(k+1) = E(k) \ {k}

If all elements are removed but no finite endsegments are passed, then mathematics is violated. Then you can violate every theorem. Then set theory allows you to play with it like mentally disabled would do. Why not violate logic at other places too?

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 11. Mai 2021 um 14:08:28 UTC+2:

> >A descending sequence of sets with one element lost per term and an empty end but without finite terms is an inconsistency proof in mathematics.
> Nope, this is counterintuitive, but not a contradiction.

∀k ∈ ℕ: E(k+1) = E(k) \ {k} is a formula. No intuition.

Regards, WM

>That is impossible by definition, because per term only one natnumber gets lost

Yes but we aren't doing the process so it is irrelevant.

>If all elements are removed but no finite endsegments are passed, then mathematics is violated.

No it isn't because we aren't passing anything, it is not a process.

>∀k ∈ ℕ: E(k+1) = E(k) \ {k} is a formula. No intuition

I didn't say that, the fact we get the empty set is logical, but counterintuitive.

Lim E_n = {x e N: Am e N (x e E_m)}

But given that for any x, m=x+1, we have x /e E_m, then that means x is not part of Lim E_n, ergo, Lim E_n = {}

There is no passing, no finites, nothing of the sort.

On Tuesday, May 11, 2021 at 7:47:54 AM UTC-4, WM wrote:
> Dan Christensen schrieb am Dienstag, 11. Mai 2021 um 01:39:44 UTC+2:
> > On Monday, May 10, 2021 at 4:20:58 PM UTC-4, WM wrote:
> > > Dan Christensen schrieb am Montag, 10. Mai 2021 um 21:22:02 UTC+2:
> > > > On Monday, May 10, 2021 at 1:53:58 PM UTC-4, WM wrote:
> > >
> > > > > The questions is: how can it be reconciled with the fact that only one element is lost per step, i.e., consecutive terms differ merely by one natural number.
> > > > There is nothing to "reconcile."
> > > You can't see it. The questions is: how can it be reconciled with the fact that only one element is lost per step, i.e., consecutive terms differ merely by one natural number.
> > >
> > If you are claiming that this leads to a contradiction in set theory, that would require a proof.
> It leads to a contradiction in mathematics.

Not in any axiomatic system that I am aware of. Unless you mean your worthless MuckeMath system were are all numbers are equal. (Can't even prove 1=/= 2. Hee, hee!)

> A descending sequence of sets with one element lost per term and an empty end but without finite terms is an inconsistency proof in mathematics.

You mean, for every natural number n, no matter how large, there exists infinitely many other natural numbers x such that x>n. Nothing wrong with that.. It leads to no known contradictions. You certainly have not demonstrated any here, Mucke. To do that, you would have to derive two contradictory theorems from whatever axiomatic system you may have in mind. But we know how you hate axioms and proofs. So, we won't hold our breath, Mucke.

Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

zelos...@gmail.com schrieb am Dienstag, 11. Mai 2021 um 14:52:00 UTC+2:
> >That is impossible by definition, because per term only one natnumber gets lost
> Yes but we aren't doing the process so it is irrelevant.
> >If all elements are removed but no finite endsegments are passed, then mathematics is violated.
> No it isn't because we aren't passing anything, it is not a process.
>
> >∀k ∈ ℕ: E(k+1) = E(k) \ {k} is a formula. No intuition
>
> I didn't say that, the fact we get the empty set is logical, but counterintuitive.

No it is not counter intuitive but obvious since ever number has an endsegment, where it appears last and an endsegmnet, where it is missing.
>
> Lim E_n = {x e N: Am e N (x e E_m)}
>
> But given that for any x, m=x+1, we have x /e E_m, then that means x is not part of Lim E_n, ergo, Lim E_n = {}
>
> There is no passing, no finites, nothing of the sort.

The sequence of endsegments is complete. None is missing. It connects |N and { }. Therefore

k ∈ ℕ: E(k+1) = E(k) \ {k}

enforces finite terms.

Regards, WM

On Monday, May 10, 2021 at 8:59:35 AM UTC-3, Gus Gassmann wrote:
> On Monday, 10 May 2021 at 07:58:54 UTC-3, zelos...@gmail.com wrote:
> > torsdag 6 maj 2021 kl. 13:00:42 UTC+2 skrev WM:

> > > 3) If a union of disjunkt intervals I(n) covers an interval I, then every point of I is covered by precisely one interval I(n). Set-theory contravenes this simple condition of logic and mathematics.
> This seems to be new. I am not sure what to make of the requirement that the intervals must be disjoint, or how set theory contradicts this.

Yes this does seem to be new. I think that WM is complaining that the set of disjoint intervals (1/n] for every n, is said to cover (0,1] even though no interval covers the (imaginary) real number "next" to 0.

--
William Hughes

On Tuesday, 11 May 2021 at 09:36:54 UTC-3, WM wrote:
> Gus Gassmann schrieb am Dienstag, 11. Mai 2021 um 12:20:24 UTC+2:
> > On Monday, 10 May 2021 at 17:24:56 UTC-3, WM wrote:
>
> > > Can a strictly decreasing sequence of sets where every set S(n+1) has precisely one element less than its predecessor S(n) lose every element and yield the empty set without passing through finite sets?
> > It is actually very funny to see you talk about mental capacities. It is quite clear that the deficiencies are on your side, and they are selective. You also have an inborn resistance to considering analogies and to learn from them.
> I see that there is no answer. Are you too much ashamed to give an answer?
> >
> > So you have on the one hand an insistence that 0.999... =/= 1, since the left-hand side represents a sequence of partial sums and the right-hand side is the *limit* of that sequence. You're not mainstream on this question, but it is a reasonable way to resolve the inherit ambiguity in the notation 0.999....
> Nice to hear.
> >
> > You must at some point have applied the same reasoning to the sequence {1, 1+1/2, 1+1/2+1/3,...}. Do you understand that the positive real numbers can be completed by adjoining to it the symbol oo, so that divergent sequences of partial sums such as {1, 1+1/2, 1+1/2+1/3,...} can be said to have a limit, namely oo?
> oo does not mean completion but represents the potentially infinite. When we say "it is the limit oo" that means only that the sum increases without end. Cantor said 1 + 1/2 + 1/3 + ... = ω. That is a fixed number.

You can think of oo or ω as a number, but sure as hell not a natural number, or a rational, or real, and especially not as that nowadays moronic "potentially infinite". Why don't you look up at some time the phrase "compactification of the real numbers"? It is almost certainly not going to lead to anything with your utterly frustrating resistance to *any kind* of learning, but --- hope springs eternal.

> > Have you ever thought what happens when you apply that same reasoning to the sequence { {1}, {1} U {2}, {1} U {2} U {3}, ...}?
> We get |N, that is all natural numbers, but we do not get ω, the first transfinite number. In the same way we get with 0.999... all finitely indexed nines, but that is not the analytical limit 1.

Clearly no comprehension of any kind from you. When you think of 0.999... as a sequence, it bloody well has limiting behaviour of *some* sort. Ditto for the sequences {1, 1+1/2, 1+1/2+1/3,...} and { {1}, {1} U {2}, {1} U {2} U {3}, ...}. How the *FUCK* do you think you can determine that limiting behavious? Sure as hell not by adding the next term to the previous ones until the end of time.

> The set theoretic identification of |N and omega is simply sloppy and worse.
Where, oh where, do you think there is an identification of |N and omega in anything that I wrote. The slop is entirely in your reading, comprehension, and ultimately, brain.

> > Have you ever thought *why* Halmos went to the trouble of defining a limit for sequences of sets such as the one above? You *know* what that limit is, but you refuse to accept its significance, and I have never understood why. I mean, the facts are staring you in the face.
> It is difficult to distinguish between unions and limits in set theory. The union of all natnumbers is |N. For that we don't need a limit, since we accept that all are there and can simply be unioned.

In other words, you know that Halmos went to some trouble, but you have no clue why he might have done so. Figures.

On Tuesday, 11 May 2021 at 12:31:30 UTC-3, William wrote:
> On Monday, May 10, 2021 at 8:59:35 AM UTC-3, Gus Gassmann wrote:
> > On Monday, 10 May 2021 at 07:58:54 UTC-3, zelos...@gmail.com wrote:
> > > torsdag 6 maj 2021 kl. 13:00:42 UTC+2 skrev WM:
> > > > 3) If a union of disjunkt intervals I(n) covers an interval I, then every point of I is covered by precisely one interval I(n). Set-theory contravenes this simple condition of logic and mathematics.
> > This seems to be new. I am not sure what to make of the requirement that the intervals must be disjoint, or how set theory contradicts this.
> Yes this does seem to be new. I think that WM is complaining that the set of disjoint intervals (1/n] for every n, is said to cover (0,1] even though no interval covers the (imaginary) real number "next" to 0.
Sancta simplicitas! Are you saying he means the disjoint intervals (1/(n+1),1/n]????

On 5/11/2021 7:45 AM, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 11. Mai 2021 um 14:08:28 UTC+2:
>
>>> A descending sequence of sets with one element lost per term and an empty end but without finite terms is an inconsistency proof in mathematics.
>> Nope, this is counterintuitive, but not a contradiction.
>
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k} is a formula. No intuition.
>
> Regards, WM
>

you failed to specify E(k),

is this what you mean ?

E(k) = {k,k+1,k+2,...} ?

and by

E(k+1) = E(k) \ {k}

do you mean {k,k+1,k+2,...} = {k+1,k+2,...} ∪ {k} right ?

what do you mean by "empty end" ? Both these sets are never empty, as
they are infinite sets, remember oo - k = oo

what do you mean "without finite terms" ?
do you mean the E(k) be finite ?

so your formula is simply

{k,k+1,k+2,...} = {k+1,k+2,...} ∪ {k}

how is that an inconsistency ??

William schrieb am Dienstag, 11. Mai 2021 um 17:31:30 UTC+2:

> Yes this does seem to be new. I think that WM is complaining that the set of disjoint intervals (1/n] for every n, is said to cover (0,1] even though no interval covers the (imaginary) real number "next" to 0.

If *only* intervals could be covered which leave infinitely many intervals between themselves and 0, then infinitely many intervals could not be covered.

Regards, WM

Gus Gassmann schrieb am Dienstag, 11. Mai 2021 um 18:33:40 UTC+2:
> On Tuesday, 11 May 2021 at 09:36:54 UTC-3, WM wrote:

> > oo does not mean completion but represents the potentially infinite. When we say "it is the limit oo" that means only that the sum increases without end. Cantor said 1 + 1/2 + 1/3 + ... = ω. That is a fixed number.
> You can think of oo or ω as a number, but sure as hell not a natural number

Of course not. It is an ordinal, a whole number.

>, or a rational

although Cantor defined transfinite fractions

>, or real, and especially not as that nowadays moronic "potentially infinite".

No, omega "contains nothing shaky, nothing indefinite, nothing varying, nothing potential, and the same is true for all other transfinite numbers."

> When you think of 0.999... as a sequence, it bloody well has limiting behaviour of *some* sort.

Of course. But the limit does not belong to the sequence as a term.
0.999... converges to 1.

> > The set theoretic identification of |N and omega is simply sloppy and worse.
> Where, oh where, do you think there is an identification of |N and omega in anything that I wrote.

It is usual in set theory to say |N or omega. Ask some people with enough technical knowledge.

> In other words, you know that Halmos went to some trouble,

I am not informed about the detailed fate of Halmos. I only know that he was bluffed by the diagonal procedure.

Regards, WM

Sergio schrieb am Dienstag, 11. Mai 2021 um 20:07:31 UTC+2:

> is this what you mean ?
>
> E(k) = {k,k+1,k+2,...} ?
>
>
>
> and by
>
> E(k+1) = E(k) \ {k}
>
> do you mean {k,k+1,k+2,...} = {k+1,k+2,...} ∪ {k} right ?

Yes.
>
>
>
>
> what do you mean by "empty end" ? Both these sets are never empty, as
> they are infinite sets, remember oo - k = oo

The intersection of all endsegments is empty.
>
> how is that an inconsistency ??

The sequence of intersections can only become empty if there are also finite terms:

∀k ∈ ℕ: E(k+1) = E(k) \ {k}.

Only one loss per term.

Regards, WM

On Tuesday, May 11, 2021 at 1:35:44 PM UTC-3, Gus Gassmann wrote:
> On Tuesday, 11 May 2021 at 12:31:30 UTC-3, William wrote:
> > On Monday, May 10, 2021 at 8:59:35 AM UTC-3, Gus Gassmann wrote:
> > > On Monday, 10 May 2021 at 07:58:54 UTC-3, zelos...@gmail.com wrote:
> > > > torsdag 6 maj 2021 kl. 13:00:42 UTC+2 skrev WM:
> > > > > 3) If a union of disjunkt intervals I(n) covers an interval I, then every point of I is covered by precisely one interval I(n). Set-theory contravenes this simple condition of logic and mathematics.
> > > This seems to be new. I am not sure what to make of the requirement that the intervals must be disjoint, or how set theory contradicts this.
> > Yes this does seem to be new. I think that WM is complaining that the set of disjoint intervals (1/n] for every n, is said to cover (0,1] even though no interval covers the (imaginary) real number "next" to 0.
> Sancta simplicitas! Are you saying he means the disjoint intervals (1/(n+1),1/n]????
Indeed, mea maxima culpa.

On Tuesday, 11 May 2021 at 15:43:58 UTC-3, WM wrote:
> Gus Gassmann schrieb am Dienstag, 11. Mai 2021 um 18:33:40 UTC+2:
> > On Tuesday, 11 May 2021 at 09:36:54 UTC-3, WM wrote:
>
> > > oo does not mean completion but represents the potentially infinite. When we say "it is the limit oo" that means only that the sum increases without end. Cantor said 1 + 1/2 + 1/3 + ... = ω. That is a fixed number.
> > You can think of oo or ω as a number, but sure as hell not a natural number
> Of course not. It is an ordinal, a whole number.
>
> >, or a rational
>
> although Cantor defined transfinite fractions
> >, or real, and especially not as that nowadays moronic "potentially infinite".
> No, omega "contains nothing shaky, nothing indefinite, nothing varying, nothing potential, and the same is true for all other transfinite numbers."
> > When you think of 0.999... as a sequence, it bloody well has limiting behaviour of *some* sort.
> Of course. But the limit does not belong to the sequence as a term.

Whoever said that? I didn't.

> 0.999... converges to 1.
> > > The set theoretic identification of |N and omega is simply sloppy and worse.
> > Where, oh where, do you think there is an identification of |N and omega in anything that I wrote.
> It is usual in set theory to say |N or omega. Ask some people with enough technical knowledge.
> > In other words, you know that Halmos went to some trouble,
> I am not informed about the detailed fate of Halmos. I only know that he was bluffed by the diagonal procedure.

Oh, fuck off. You know bloody well that Halmos defined limsup and liminf for sets, and hence a limit, and it has dick all to do with adding or subtracting one element at a time, in the same way that the limit of the implicit sequence 0.999... has nothing to do with adding one '9' at a time. So why pretend?

On Monday, 10 May 2021 at 15:00:47 UTC-3, WM wrote:
> Gus Gassmann schrieb am Montag, 10. Mai 2021 um 13:59:35 UTC+2:
[...]
> > > torsdag 6 maj 2021 kl. 13:00:42 UTC+2 skrev WM:
> > > > 3) If a union of disjunkt intervals I(n) covers an interval I, then every point of I is covered by precisely one interval I(n). Set-theory contravenes this simple condition of logic and mathematics.
> > This seems to be new. I am not sure what to make of the requirement that the intervals must be disjoint, or how set theory contradicts this.
> Disjoint intervals guarantee that the point is only covered by one interval.

What point? What I? What are the "disjunkt intervals" I(n)? And, once you write this down properly, which point of I do you think is not covered? Maybe you think it looks like this ı?

On 5/11/2021 1:48 PM, WM wrote:
> Sergio schrieb am Dienstag, 11. Mai 2021 um 20:07:31 UTC+2:
>
>
>> is this what you mean ?
>>
>> E(k) = {k,k+1,k+2,...} ?
>>
>>
>>
>> and by
>>
>> E(k+1) = E(k) \ {k}
>>
>> do you mean {k,k+1,k+2,...} = {k+1,k+2,...} ∪ {k} right ?
>
> Yes.
>>
>>
>>
>>
>> what do you mean by "empty end" ? Both these sets are never empty, as
>> they are infinite sets, remember oo - k = oo
>
> The intersection of all endsegments is empty.

no. Intersection of any three endsegments is empty, not "all".

{k,k+1,k+2,...} ∩ {k+1,k+2,k+3...} = {k}

{k+1,k+2,k+3...} ∩ {k+2,k+3,k+4...} = {k+1}

{k} ∩ {k+1} = {}

{k,k+1,k+2,...} ∩ {k+1,k+2,k+3...} ∩ {k+2,k+3,k+4...} = {}

>>
>> how is that an inconsistency ??
>
> The sequence of intersections can only become empty if there are also finite terms:
>
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}.
>
> Only one loss per term.

no. as shown above, only 3 intersections of endsegments is {}, using
infinite sets.

>
> Regards, WM
>

On 5/11/2021 7:11 PM, Sergio wrote:
> On 5/11/2021 1:48 PM, WM wrote:
>> Sergio schrieb am Dienstag, 11. Mai 2021 um 20:07:31 UTC+2:
>>
>>
>>> is this what you mean ?
>>>
>>> E(k) = {k,k+1,k+2,...} ?
>>>
>>>
>>>
>>> and by
>>>
>>> E(k+1) = E(k) \ {k}
>>>
>>> do you mean {k,k+1,k+2,...} = {k+1,k+2,...} ∪ {k} right ?
>>
>> Yes.
>>>
>>>
>>>
>>>
>>> what do you mean by "empty end" ? Both these sets are never empty, as
>>> they are infinite sets, remember oo - k = oo
>>
>> The intersection of all endsegments is empty.
>
>

>
> {k,k+1,k+2,...} ∩ {k+1,k+2,k+3...} = {k}
>
> {k+1,k+2,k+3...} ∩ {k+2,k+3,k+4...} = {k+1}
>
> {k} ∩ {k+1} = {}
>
>
> {k,k+1,k+2,...} ∩ {k+1,k+2,k+3...} ∩ {k+2,k+3,k+4...} = {}

no, that is not right Sergio! you mix ∪ with ∩ !, you try to trick me.

{k,k+1,k+2,...} ∩ {k+1,k+2,k+3...} = {k+1,k+2,k+3...}

and

{k,k+1,k+2,...} ∩{k+1,k+2,k+3...} ∩ {k+2,k+3,k+4...} = {k+2,k+3,k+4...}

so, one simply one elements out of a set each time.

BUT remember oo - k = oo for finite k

But for "all endsegements" or "infinite number of endsegments" the limit
is empty.

I see no contradiction at all. It is simply taking the limit.

>
>
>>>
>>> how is that an inconsistency ??
>>
>> The sequence of intersections can only become empty if there are also finite terms:
>>
>> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}.
>>
>> Only one loss per term.
>

you will never get empty that way, take the limit.

>
>>
>> Regards, WM
>>
>
>
>

tisdag 11 maj 2021 kl. 15:27:27 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 11. Mai 2021 um 14:52:00 UTC+2:
> > >That is impossible by definition, because per term only one natnumber gets lost
> > Yes but we aren't doing the process so it is irrelevant.
> > >If all elements are removed but no finite endsegments are passed, then mathematics is violated.
> > No it isn't because we aren't passing anything, it is not a process.
> >
> > >∀k ∈ ℕ: E(k+1) = E(k) \ {k} is a formula. No intuition
> >
> > I didn't say that, the fact we get the empty set is logical, but counterintuitive.
> No it is not counter intuitive but obvious since ever number has an endsegment, where it appears last and an endsegmnet, where it is missing.
> >
> > Lim E_n = {x e N: Am e N (x e E_m)}
> >
> > But given that for any x, m=x+1, we have x /e E_m, then that means x is not part of Lim E_n, ergo, Lim E_n = {}
> >
> > There is no passing, no finites, nothing of the sort.
> The sequence of endsegments is complete. None is missing. It connects |N and { }. Therefore
> k ∈ ℕ: E(k+1) = E(k) \ {k}
> enforces finite terms.
>
> Regards, WM

It doesn't, it does not say in any way it must have finite terms. This is a logical leap, a non-sequitor, of yours.

Gus Gassmann schrieb am Dienstag, 11. Mai 2021 um 23:33:06 UTC+2:
> On Tuesday, 11 May 2021 at 15:43:58 UTC-3, WM wrote:

> > > In other words, you know that Halmos went to some trouble,
> > I am not informed about the detailed fate of Halmos. I only know that he was bluffed by the diagonal procedure.
> You know bloody well that Halmos defined limsup and liminf for sets, and hence a limit,

I am not informed about his trouble. What kind of trouble was it?

Regards, WM

Gus Gassmann schrieb am Dienstag, 11. Mai 2021 um 23:39:25 UTC+2:
> On Monday, 10 May 2021 at 15:00:47 UTC-3, WM wrote:
> > Gus Gassmann schrieb am Montag, 10. Mai 2021 um 13:59:35 UTC+2:
> [...]
> > > > torsdag 6 maj 2021 kl. 13:00:42 UTC+2 skrev WM:
> > > > > 3) If a union of disjunkt intervals I(n) covers an interval I, then every point of I is covered by precisely one interval I(n). Set-theory contravenes this simple condition of logic and mathematics.
> > > This seems to be new. I am not sure what to make of the requirement that the intervals must be disjoint, or how set theory contradicts this.
> > Disjoint intervals guarantee that the point is only covered by one interval.
>
> What point?

Every point of the interval I = (0, 1] is covered by precisely one interval (1/(n+1). 1/n]

> What I? What are the "disjunkt intervals" I(n)? And, once you write this down properly, which point of I do you think is not covered?

The point next to 0 is covered by the union of the I(n) but is not covered by any I(n).

Regards, WM

zelos...@gmail.com schrieb am Mittwoch, 12. Mai 2021 um 06:51:42 UTC+2:
> tisdag 11 maj 2021 kl. 15:27:27 UTC+2 skrev WM:

> > The sequence of endsegments is complete. None is missing. It connects |N and { }. Therefore
> > k ∈ ℕ: E(k+1) = E(k) \ {k}
> > enforces finite terms.
> >
> It doesn't, it does not say in any way it must have finite terms.

Try again. The sequence of endsegments connects |N and { } and has maximum stepwidth 1 element.

Regards, WM