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).

If all sets S(n) have ℵo elements, then all sets contain the same infinite subset. Set theory contravenes this simple condition of logic and mathematics.

2) A strictly decreasing sequence of sets where every set S(n+1) has precisely one element less than its predecessor S(n) cannot lose every element and yield the empty set without passing through finite sets. Set theory contravenes this simple condition of logic and mathematics.

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.

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).

Regards, WM

On Thursday, May 6, 2021 at 7:00:42 AM UTC-4, WM wrote:
> 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).
>
> If all sets S(n) have ℵo elements, then all sets contain the same infinite subset. Set theory contravenes this simple condition of logic and mathematics.
>

Can you write this in the notation of set theory? One never quite knows what you might mean by some phrase, Mucke. Likewise, your other alleged "mistakes" of set theory here.

More absurd quotes from Wolfgang Muckenheim (WM):

“In my system, two different numbers can have the same value.”
-- sci.math, 2014/10/16

“1+2 and 2+1 are different numbers.”
-- sci.math, 2014/10/20

“1/9 has no decimal representation.”
-- sci.math, 2015/09/22

"0.999... is not 1."
-- sci.logic 2015/11/25

“Axioms are rubbish!”
-- sci.math, 2014/11/19

“Formal definitions have lead to worthless crap like undefinable numbers.”
-- sci.math 2017/02/05

“No set is countable, not even |N.”
-- sci.logic, 2015/08/05

“Countable is an inconsistent notion.”
-- sci.math, 2015/12/05

Slipping ever more deeply into madness...

“There is no actually infinite set |N.”
-- sci.math, 2015/10/26

“|N is not covered by the set of natural numbers.”
-- sci.math, 2015/10/26

“The set of all rationals can be shown not to exist.”
--sci.math, 2015/11/28

“Everything is in the list of everything and therefore everything belongs to a not uncountable set.”
-- sci.math, 2015/11/30

"'Not equal' and 'equal can mean the same.”
-- sci.math, 2016/06/09

“The set of numbers will get empty after all have numbers been used..”
-- sci.math, 2016/08/24

“I need no set theory.”
-- sci.math, 2016/09/01

A special word of caution to students: Do not attempt to use WM's “system” (MuckeMath) in any course work in any high school, college or university on the planet. You will fail miserably. MuckeMath is certainly no shortcut to success in mathematics.

Using WM's “axioms” for the natural numbers, he cannot even prove that 1=/=2. His goofy system is truly a dead-end.

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

Dan Christensen schrieb am Donnerstag, 6. Mai 2021 um 18:20:46 UTC+2:
> On Thursday, May 6, 2021 at 7:00:42 AM UTC-4, WM wrote:
> > 1) An inclusion-monotonic sequence of sets is a sequence where every set S(n+1) contains only elements which are in the preceding set S(n). All elements of all sets are in the first set S(1).
> >
> > If all sets S(n) have ℵo elements, then all sets contain the same infinite subset. Set theory contravenes this simple condition of logic and mathematics.
> >
> Can you write this in the notation of set theory?

Why? You confessed yourself that you never could understand a formal proof written by others. And all other readers will grasp this simple sentence as I wrote it.

Regards, WM

On 5/6/2021 7:00 AM, WM wrote:

> 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).

Consider an endless inclusion-monotonic decreasing
_sequence_ Q(1),Q(2),Q(3),... of sets such that
adjacent sets differ by one element.
Q(n)\Q(n+1) = { b{n) }
and all sets Q(n) have aleph_0 elements.

( Note that we must assume such a set sequence exists in order
( for there to be an infinite subset in all of them.

Q(n) is a _sequence_
By "sequence", I mean that, for each Q(n) in the sequence,
a completable linearly-ordered crowd Q(1),...,Q(n) exists
in the sequence such that, for adjacent sets Q(j),Q(j+1) in
Q(1),...,Q(n), Q(j) and Q(j+1) differ by one element.

Define
R(m) = Q(1)\Q(m)

R(1) = { b(1) }
R(2) = { b(1),b(2) }
R(3) = { b(1),b(2),b(3) }
....

For all b(j) in Q(1), either b(j) is in R(m), for some m,
or there isn't any such R(m).
I'll say any x with that property is _visible_

Visible(x) <->
x in Q(1) & exists m, x in R(m)

Define the visible-only subset S(n) of Q(n)
x in S(n) <->
Visible(x) & x in Q(n)

S(1),S(2),S(3),...
is an endless inclusion-monotonic decreasing
sequence S(1),S(2),S(3),... of sets such that
adjacent sets differ by one element.

Let S(n) be a set in that sequence.
For each set S(m+1) following S(n), there is a unique
element b(m) such that b(m) is not in S(m+1) and
is in all the sets S(n),...,S(m).

There are aleph_0 sets S(m) following S(n).
Therefore,
all sets S(n) have aleph_0 elements or more.

> If all sets S(n) have ℵo elements, then all sets contain
> the same infinite subset.

Lemma.
If exists S(n), x in S(n),
then exists S(m), x not in S(m).

Assume x in S(n).
x is visible
x in R(m), for some m.
R(m) = Q(1)\Q(m)
x not in Q(m)
x not in S(m)

Therefore, there is no element which is in all S(n).

> Set theory contravenes this simple condition of logic and
> mathematics.

On one side, we have the counter-example above.

On the other side, we have your simple condition of logic
and math, asserted but not argued for, unless you count
using cute names like "matheologian" as argument.

On Thursday, May 6, 2021 at 1:41:53 PM UTC-4, WM wrote:
> Dan Christensen schrieb am Donnerstag, 6. Mai 2021 um 18:20:46 UTC+2:
> > On Thursday, May 6, 2021 at 7:00:42 AM UTC-4, WM wrote:
> > > 1) An inclusion-monotonic sequence of sets is a sequence where every set S(n+1) contains only elements which are in the preceding set S(n). All elements of all sets are in the first set S(1).
> > >
> > > If all sets S(n) have ℵo elements, then all sets contain the same infinite subset. Set theory contravenes this simple condition of logic and mathematics.
> > >
> > Can you write this in the notation of set theory?
> Why? You confessed yourself that you never could understand a formal proof written by others. And all other readers will grasp this simple sentence as I wrote it.
>

If you can't express it in the abstract, maybe you could consider a concrete example: Let S_n = {x in Q: 0 <= x <= 1 + 1/n}. Let S = {x in Q: 0 <= x <= 1}. The elements of S are common to each of the S_n. S and each of the S_n also have ℵo elements. Somehow, you believe that this setup leads to a contradiction. It is NOT enough to say simply that it is intuitively obvious or whatever. This really does require a proof, Mucke. Given it a try. Maybe the experts here will understand. (Hee, hee!)

You haven't been doing your homework, have you, Mucke?

Dan

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

Jim Burns schrieb am Donnerstag, 6. Mai 2021 um 23:07:22 UTC+2:
> On 5/6/2021 7:00 AM, WM wrote:
>
> > 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).
> Consider an endless inclusion-monotonic decreasing
> _sequence_ Q(1),Q(2),Q(3),... of sets such that
> adjacent sets differ by one element.
> Q(n)\Q(n+1) = { b{n) }
> and all sets Q(n) have aleph_0 elements.
>
> ( Note that we must assume such a set sequence exists in order
> ( for there to be an infinite subset in all of them.
>
> Q(n) is a _sequence_
> By "sequence", I mean that, for each Q(n) in the sequence,
> a completable linearly-ordered crowd Q(1),...,Q(n) exists
> in the sequence such that, for adjacent sets Q(j),Q(j+1) in
> Q(1),...,Q(n), Q(j) and Q(j+1) differ by one element.

An example is the sequence (E(n)) of endsegments E(n). Adjacent terms are connected by

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

We can define the bijection

subtract(n) = E(n+1) with subtract(0) = E(1) = |N.

>
> Define
> R(m) = Q(1)\Q(m)

R(1) is then empty. R(2) = {1} when using endsegments.
>
> R(1) = { b(1) }
> R(2) = { b(1),b(2) }
> R(3) = { b(1),b(2),b(3) }
> ...
>
> For all b(j) in Q(1), either b(j) is in R(m), for some m,
> or there isn't any such R(m).
> I'll say any x with that property is _visible_
>
> Visible(x) <->
> x in Q(1) & exists m, x in R(m)
>
> Define the visible-only subset S(n) of Q(n)
> x in S(n) <->
> Visible(x) & x in Q(n)

Visible elements do not form a set. Potential infinity!
>
> S(1),S(2),S(3),...
> is an endless inclusion-monotonic decreasing
> sequence S(1),S(2),S(3),... of sets such that
> adjacent sets differ by one element.

S(n) are similar to endsegments E(n), but not all natural numbers of an endsegment are visible.
>
> Let S(n) be a set in that sequence.
> For each set S(m+1) following S(n), there is a unique
> element b(m) such that b(m) is not in S(m+1) and
> is in all the sets S(n),...,S(m).

That is same for all visible numbers in endsegments.
>
> There are aleph_0 sets S(m) following S(n).

There are not aleph_0 definable elements!

> Therefore,
> all sets S(n) have aleph_0 elements or more.

No, you said that they have only definable elements: "visible-only subset S(n) of Q(n)"

> > If all sets S(n) have ℵo elements

Impossible. So we can stop here.
All visible E(n) have aleph_0 elements, most of them dark.

>> , then all sets contain
> > the same infinite subset.

Not your S(n). They are potentially infinite, i.e., finite.

> Lemma.
> If exists S(n), x in S(n),
> then exists S(m), x not in S(m).
>
> Assume x in S(n).
> x is visible
> x in R(m), for some m.
> R(m) = Q(1)\Q(m)
> x not in Q(m)
> x not in S(m)
>
> Therefore, there is no element which is in all S(n).

Small wonder.

> > Set theory contravenes this simple condition of logic and
> > mathematics.
> On one side, we have the counter-example above.

No.

Repeat this with the potentially infinite collection of definable endsegments (defined by their first member n). Find ∀k ∈ ℕ_def: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo.
All last endegments can be put into one collection and being intersected. The result is not empty.

Regards, WM

On Friday, May 7, 2021 at 10:04:17 AM UTC-4, WM wrote:
> Jim Burns schrieb am Donnerstag, 6. Mai 2021 um 23:07:22 UTC+2:
> > On 5/6/2021 7:00 AM, WM wrote:
> >
> > > 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).
> > Consider an endless inclusion-monotonic decreasing
> > _sequence_ Q(1),Q(2),Q(3),... of sets such that
> > adjacent sets differ by one element.
> > Q(n)\Q(n+1) = { b{n) }
> > and all sets Q(n) have aleph_0 elements.
> >
> > ( Note that we must assume such a set sequence exists in order
> > ( for there to be an infinite subset in all of them.
> >
> > Q(n) is a _sequence_
> > By "sequence", I mean that, for each Q(n) in the sequence,
> > a completable linearly-ordered crowd Q(1),...,Q(n) exists
> > in the sequence such that, for adjacent sets Q(j),Q(j+1) in
> > Q(1),...,Q(n), Q(j) and Q(j+1) differ by one element.
> An example is the sequence (E(n)) of endsegments E(n). Adjacent terms are connected by
>
> E(k+1) = E(k) \ {k}
>

A poor example. The intersection U {E(k): k in N} is empty, i.e. there is no element of N that common to all of the E(k). Have a look at my example, if you dare. There, the intersection of all the S(k) is {x in Q: 0 <= x <= 1}

Dan

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

torsdag 6 maj 2021 kl. 13:00:42 UTC+2 skrev WM:
> 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).
>
> If all sets S(n) have ℵo elements, then all sets contain the same infinite subset. Set theory contravenes this simple condition of logic and mathematics.
>
> 2) A strictly decreasing sequence of sets where every set S(n+1) has precisely one element less than its predecessor S(n) cannot lose every element and yield the empty set without passing through finite sets. Set theory contravenes this simple condition of logic and mathematics.
>
> 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.
>
> 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).
>
> Regards, WM

No, Card Lim and Lim Card are different because we can show cases where htey are different.

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:
> > 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).
> >
> > If all sets S(n) have ℵo elements, then all sets contain the same infinite subset. [...]
Muckenheim's first "axiom". Other than wishful thinking, there is of course no truth to his assertion, and he knows it.
> >
> > 2) A strictly decreasing sequence of sets where every set S(n+1) has precisely one element less than its predecessor S(n) cannot lose every element and yield the empty set without passing through finite sets. Set theory contravenes this simple condition of logic and mathematics.
Muckenheim's second "axiom". Again unprovable nonsense.
> > 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.
> > 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).
> >
> > Regards, WM
> No, Card Lim and Lim Card are different because we can show cases where htey are different.

On 5/6/2021 6:00 AM, WM wrote:
> 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 the sets have identically the same elements*, no more no
less, so they are all equal.

> If all sets S(n) have ℵo elements, then all sets contain the same infinite subset.

and...

> Set theory contravenes this simple condition of logic and mathematics.

how ?

>
> 2) A strictly decreasing sequence of sets where every set S(n+1) has precisely one element less than its predecessor S(n) cannot lose every element and yield the empty set without passing through finite sets.

are you assuming S(n) is an infinite set, or not ?

what do you mean by "lose" ?

> Set theory contravenes this simple condition of logic and mathematics.

how ?

>
> 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).

and...

>Set-theory contravenes this simple condition of logic and mathematics.

how ?

>
> Set theorists claim that LimCardS(n) can differ from CardLimS(n).

provide references or URL, Define
LimCardS(n)
CardLimS(n)
LimS(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).

imagination.

>
> Regards, 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.

Regards, WM

Dan Christensen schrieb am Samstag, 8. Mai 2021 um 05:39:48 UTC+2:
> On Friday, May 7, 2021 at 10:04:17 AM UTC-4, WM wrote:
> > Jim Burns schrieb am Donnerstag, 6. Mai 2021 um 23:07:22 UTC+2:
> > > On 5/6/2021 7:00 AM, WM wrote:
> > >
> > > > 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).
> > > Consider an endless inclusion-monotonic decreasing
> > > _sequence_ Q(1),Q(2),Q(3),... of sets such that
> > > adjacent sets differ by one element.
> > > Q(n)\Q(n+1) = { b{n) }
> > > and all sets Q(n) have aleph_0 elements.
> > >
> > > ( Note that we must assume such a set sequence exists in order
> > > ( for there to be an infinite subset in all of them.
> > >
> > > Q(n) is a _sequence_
> > > By "sequence", I mean that, for each Q(n) in the sequence,
> > > a completable linearly-ordered crowd Q(1),...,Q(n) exists
> > > in the sequence such that, for adjacent sets Q(j),Q(j+1) in
> > > Q(1),...,Q(n), Q(j) and Q(j+1) differ by one element.
> > An example is the sequence (E(n)) of endsegments E(n). Adjacent terms are connected by
> >
> > E(k+1) = E(k) \ {k}
> >
> The intersection U {E(k): k in N} is empty,

Intersection is usually denoted by ∩.

> i.e. there is no element of N that common to all of the E(k).

Why do you mention that? It is not doubted. 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.

Regards, WM

Gus Gassmann schrieb am Montag, 10. Mai 2021 um 13:59:35 UTC+2:
> 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:
> > > 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).
> > >
> > > If all sets S(n) have ℵo elements, then all sets contain the same infinite subset. [...]
>
> Muckenheim's first "axiom". Other than wishful thinking, there is of course no truth to his assertion, and he knows it.

No axiom, no theorem. An inclusion-monotonic sequence cannot but satisfy this triviality.
> > >
> > > 2) A strictly decreasing sequence of sets where every set S(n+1) has precisely one element less than its predecessor S(n) cannot lose every element and yield the empty set without passing through finite sets. Set theory contravenes this simple condition of logic and mathematics.
> Muckenheim's second "axiom". Again unprovable nonsense.

No axiom, no theorem. No reason to prove this triviality. Everybody not completely stupid recognizes it.

> > > 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..

Regards, WM

On Monday, May 10, 2021 at 1:53:58 PM UTC-4, WM wrote:
> Dan Christensen schrieb am Samstag, 8. Mai 2021 um 05:39:48 UTC+2:
> > On Friday, May 7, 2021 at 10:04:17 AM UTC-4, WM wrote:
> > > Jim Burns schrieb am Donnerstag, 6. Mai 2021 um 23:07:22 UTC+2:
> > > > On 5/6/2021 7:00 AM, WM wrote:
> > > >
> > > > > 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).
> > > > Consider an endless inclusion-monotonic decreasing
> > > > _sequence_ Q(1),Q(2),Q(3),... of sets such that
> > > > adjacent sets differ by one element.
> > > > Q(n)\Q(n+1) = { b{n) }
> > > > and all sets Q(n) have aleph_0 elements.
> > > >
> > > > ( Note that we must assume such a set sequence exists in order
> > > > ( for there to be an infinite subset in all of them.
> > > >
> > > > Q(n) is a _sequence_
> > > > By "sequence", I mean that, for each Q(n) in the sequence,
> > > > a completable linearly-ordered crowd Q(1),...,Q(n) exists
> > > > in the sequence such that, for adjacent sets Q(j),Q(j+1) in
> > > > Q(1),...,Q(n), Q(j) and Q(j+1) differ by one element.
> > > An example is the sequence (E(n)) of endsegments E(n). Adjacent terms are connected by
> > >
> > > E(k+1) = E(k) \ {k}
> > >
> > The intersection U {E(k): k in N} is empty,

> Intersection is usually denoted by ∩.

Yes. That should be, the intersection ∩ {E(k): k in N} is empty,

> > i.e. there is no element of N that common to all of the E(k).

> Why do you mention that? It is not doubted.

One can never tell with you, Mucke.

> 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 wrote:

"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).

"If all sets S(n) have ℵo elements, then all sets contain the same infinite subset."

In your example here, there are no elements common to all sets, as you have just admitted. My example has a sequence of intervals in which each set DOES contain the same infinite subset:

Let S_n = {x in Q: 0 <= x <= 1 + 1/n}.

Let S = {x in Q: 0 <= x <= 1}.

The elements of S are common to each of the S_n.

S and each of the S_n have ℵo elements.

Dan

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

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:
> > 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:
> > > > 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).
> > > >
> > > > If all sets S(n) have ℵo elements, then all sets contain the same infinite subset. [...]
> >
> > Muckenheim's first "axiom". Other than wishful thinking, there is of course no truth to his assertion, and he knows it.
> No axiom, no theorem. An inclusion-monotonic sequence cannot but satisfy this triviality.

BULLSHIT. Your stating this nonsense a second time (or a thousandths) doesn't make it any less false than the first time.
> > > > 2) A strictly decreasing sequence of sets where every set S(n+1) has precisely one element less than its predecessor S(n) cannot lose every element and yield the empty set without passing through finite sets. Set theory contravenes this simple condition of logic and mathematics.
> > Muckenheim's second "axiom". Again unprovable nonsense.
> No axiom, no theorem. No reason to prove this triviality. Everybody not completely stupid recognizes it.
See my previous statement.
> > > > 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.
Sure. Didn't expect you to explain how set theory contradicts your single coverage requirements, but perhaps you can actually write this one down coherently. (At least you haven't yet repeated it thousands of times, so there might be hope yet.)

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.

Try to focus on one topic.

Regards, WM

Gus Gassmann schrieb am Montag, 10. Mai 2021 um 21:26:45 UTC+2:
> On Monday, 10 May 2021 at 15:00:47 UTC-3, WM wrote:

> > > > > 2) A strictly decreasing sequence of sets where every set S(n+1) has precisely one element less than its predecessor S(n) cannot lose every element and yield the empty set without passing through finite sets. Set theory contravenes this simple condition of logic and mathematics.
> > > Muckenheim's second "axiom". Again unprovable nonsense.
> > No axiom, no theorem. No reason to prove this triviality. Everybody not completely stupid recognizes it.
> See my previous statement.

Let's focus on one topic. More would exceed your ressources:

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?

Regards, WM

On Monday, May 10, 2021 at 10:48:00 AM UTC-7, WM wrote:
> 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..

One wonders how the phenomenon is 'explained' in the 'Best Selling Textbook'

But perhaps there it is taught that

0 = lim(m -> oo)lim(n -> oo)2^(m-n) = lim(n -> oo)lim(m -> oo)2^(m-n) = oo

and that

0 = lim(m -> oo)lim(n -> oo)(1/n)^(1/m) = lim(n -> oo)lim(m -> oo)(1/n)^(1/m) = 1

https://math.stackexchange.com/questions/15240/when-can-you-switch-the-order-of-limits
https://en.wikipedia.org/wiki/Interchange_of_limiting_operations

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. Good luck with that.

Don't hold your breath, folks. Mucke, was not even able to prove that 1=/=2 his OWN goofy little system.

What Mucke snipped:

<Quote>

You wrote:

"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).

"If all sets S(n) have ℵo elements, then all sets contain the same infinite subset."

In your example here, there are no elements common to all sets, as you have just admitted. My example has a sequence of intervals in which each set DOES contain the same infinite subset:

Let S_n = {x in Q: 0 <= x <= 1 + 1/n}.

Let S = {x in Q: 0 <= x <= 1}.

The elements of S are common to each of the S_n.

S and each of the S_n have ℵo elements.

<End Quote>

Mucke messed up! And STILL no proof of a contradiction in sight. Oh, well.....

Dan

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

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..
>
> Regards, WM

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.

On Monday, 10 May 2021 at 17:24:56 UTC-3, WM wrote:
> Gus Gassmann schrieb am Montag, 10. Mai 2021 um 21:26:45 UTC+2:
> > On Monday, 10 May 2021 at 15:00:47 UTC-3, WM wrote:
>
> > > > > > 2) A strictly decreasing sequence of sets where every set S(n+1) has precisely one element less than its predecessor S(n) cannot lose every element and yield the empty set without passing through finite sets. Set theory contravenes this simple condition of logic and mathematics.
> > > > Muckenheim's second "axiom". Again unprovable nonsense.
> > > No axiom, no theorem. No reason to prove this triviality. Everybody not completely stupid recognizes it.
> > See my previous statement.
> Let's focus on one topic. More would exceed your ressources:
>
> 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.

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....

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?

Have you ever thought what happens when you apply that same reasoning to the sequence { {1}, {1} U {2}, {1} U {2} U {3}, ...}?

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.

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

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

On Thursday, May 6, 2021 at 7:00:42 AM UTC-4, WM wrote:
> 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).
>
> If all sets S(n) have ℵo elements, then all sets contain the same infinite subset. Set theory contravenes this simple condition of logic and mathematics.
>
> 2) A strictly decreasing sequence of sets where every set S(n+1) has precisely one element less than its predecessor S(n) cannot lose every element and yield the empty set without passing through finite sets. Set theory contravenes this simple condition of logic and mathematics.
>
> 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.
>
> 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).
>
> Regards, WM

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

Nextly, possibly the subset deserves scrutiny as well as being a blurred version of the set. When the elemental form is taken seriously the notion that we might arrive at a complete set of elements would suggest that we have arrived at a rather serious set, as for instance we see in:
{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 }
which is well in use as a standard that does not generally get stated because we operate with it as an assumption. If you were to use another form it would be rather important that you state that form explicitly for fear of miscommunication; misinformation; something that we are learning to live with at a rather large level in our society. As mathematics attempts to maintain its integrity, that an authentic set such as the digits does in fact exist and can in fact be stated explicitly and seriously, is a great relief given the dubious works that have accumulated, many of which do care to dodge the very usage of these elementary forms in their work.

The idea that a reference standard set of elements could exist rather suggests that such a set deserves a stamp which is markedly different than the light form that mathematics toys with ad nauseum. Indeed attempts to pare down such a set to its minimal and thus simplest form seems to yield:
{ 1, 0 }
or even just
{ 1 }
which is arguably a form of unification. That we are free then to repeat such an element; well redundancy is entirely present within physical reality, so this is not problematic, yet to claim that
11111
is elemental does seem a misnomer, no? Nicely enough this brings us to my own third criticism of set theory on order. Using the unital set there is a lack of need of regard for order, whereas in all the higher reference sets the order of the word has to matter. Of course set theorists claim that their primitive set form lacks any order, all the while carrying on orderly operations such as unions and intersections like birds from under a hat.

>It leads to a contradiction in mathematics.

Prove it, you have yet to do it.

>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.

>Many set theorists are too blind to understand this.

You are too stupid to understand definitions.

>But that is no reason to change mathematical arguments. Either you understand mathematics, or you stay outside.

Then stay the fuck outside!