On 7/3/2022 5:35 PM, WM wrote:
> Jim Burns schrieb
> am Sonntag, 3. Juli 2022 um 19:43:14 UTC+2:
>> On 7/3/2022 10:23 AM, WM wrote:
>>> Jim Burns schrieb
>>> am Sonntag, 3. Juli 2022 um 00:10:53 UTC+2:
>>>> On 7/2/2022 12:42 PM, WM wrote:

>>>>> For all definable numbers we obtain
>>>>> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀
>>>>
>>>> No, you don't obtain that.>>>>
>>>> That definition is ill-formed.
>>>
>>> |∩{E(k) : k <10}| = ℵ₀
>>> |∩{E(k) : k <100}| = ℵ₀
>>> |∩{E(k) : k <10^10}| = ℵ₀
>>> |∩{E(k) : k <10^10^10}| = ℵ₀
>>> and so on for all definable numbers k.
>>> Can you find an exception?
>>
>> So, you want to say that ℕ_def is
>> the collection of all m such that
>> |∩{E(k) : k < m }| = ℵ₀
>> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀
>> is not how one says that.
>
> It is how one says that.

You intend to collect 10, 100, 10^10, 10^10^10,
and all numbers similar to them in a specified way.
Set to one side what you'll call the collection.

Suppose m refers to something in ℕ

If |∩{E(k) : k < m }| = ℵ₀
then m is in your intended collection.

If |∩{E(k) : k < m }| ≠ ℵ₀
then m is not in your intended collection.

What I think you intend to say by
|∩{E(k) : k ∈ ℕ_def}| = ℵ₀

is that ℕ_def refers to _one of many_ collections
such that |∩{E(k) : k ∈ ℕ_def}| = ℵ₀

that ℕ_def might
refer to {k ∈ ℕ : k < 10 }
because |∩{E(k) : k <10}| = ℵ₀
refer to {k ∈ ℕ : k < 100 }
because |∩{E(k) : k <100}| = ℵ₀
refer to {k ∈ ℕ : k < 10^10 }
because |∩{E(k) : k <10^10}| = ℵ₀
refer to {k ∈ ℕ : k < 10^10^10 }
because |∩{E(k) : k <10^10^10}| = ℵ₀

However,
that is saying something different.
None of those sets which ℕ_def might refer to
can be your intended collection.
It isn't {k ∈ ℕ : k < 10 }
because 11 is missing.
It isn't {k ∈ ℕ : k < 100 }
because 101 is missing.
It isn't {k ∈ ℕ : k < 10^10 }
because 10^10+1 is missing.
It isn't {k ∈ ℕ : k < 10^10^10 }
because 10^10^10+1 is missing.

In order for nothing to be missing from
your intended collection,
_everything_ which ℕ_def might refer to
should be included as a subset.

One can say that, too.
⋃{ ℕ_def ⊆ ℕ : |∩{E(k):k∈ℕ_def}| = ℵ₀ }

Here, I am using 'ℕ_def' as a variable name which
ranges over subsets of ℕ. It's a little unusual.
I may get called a crank by someone for doing that.

I'm doing that because I want to show where
it does make sense to write |∩{E(k):k∈ℕ_def}| = ℵ₀

|∩{E(k):k∈ℕ_def}| = ℵ₀
is a property of one of many sets ℕ_def
none of which are your intended set.

⋃{ ℕ_def ⊆ ℕ : |∩{E(k):k∈ℕ_def}| = ℵ₀ }
is one set, your intended set.

If you want to use 'ℕ_def' to refer to
your intended set, change the variable name.
ℕ_def = ⋃{ F ⊆ ℕ : |∩{E(k):k∈F}| = ℵ₀ }

----
Remember that _each of those sets_ misses something.
Different sets, different missing, but something.

ℕ_def does not miss any intended element.
It cannot be any of those sets F for which
|∩{E(k):k∈F}| = ℵ₀

However,
ℕ_def contains as subsets each of those sets F
for which |∩{E(k):k∈F}| = ℵ₀

On Monday, July 4, 2022 at 10:42:48 AM UTC-3, WM wrote:
> William schrieb am Montag, 4. Juli 2022 um 15:20:54 UTC+2:
> > On Monday, July 4, 2022 at 10:00:56 AM UTC-3, WM wrote:
> > > William schrieb am Sonntag, 3. Juli 2022 um 18:29:34 UTC+2:
> > >
> > > > I said that each of them [without exception] not "all of them". For *each* of them [without exception] ℵ₀ successors remain.
> > > For all of them they are gone.
> > The statement "For 'all of them', the colleciton, ℵ₀ successors remain." is not made.
> No,

Yes. The statement is not made.

> we know that for all of them [the successors] are gone.

So what? No statement about "all of them", the collection, is made. There is no implication that there are "gone numbers"

--
William Hughes

William schrieb am Montag, 4. Juli 2022 um 16:19:26 UTC+2:
> On Monday, July 4, 2022 at 10:42:48 AM UTC-3, WM wrote:

> > we know that for all of them [the successors] are gone.
>
> So what? No statement about "all of them",

All of them are the set. Or do you know more that is in the set?

> the collection, is made. There is no implication that there are "gone numbers"

For not all of the endsegments, ℵ₀ natural numbers are in the intersection:
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀ .
For all of the endsegments no natural numbers are in the intersection:
∩{E(k) : k ∈ ℕ} = { }
All natural numbers go only by the action of endsegments:
∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
What makes the last ℵ₀ numbers go, and how?

Regards, WM

Fritz Feldhase schrieb am Montag, 4. Juli 2022 um 15:56:13 UTC+2:
> On Monday, July 4, 2022 at 3:37:25 PM UTC+2, WM wrote:
>
> > I am only asking how the loss of ℵ₀ natural numbers between [any] finite intersection and [any] infinite intersection can be justified.
>
> Well, in the first case the intersection only concerns FINITELY many endsegments,

Since all finitely many endsegments with no exception are concerned, the infinite set is concerned.

> but in the other INFINITELY many endsegments.

Not more than above.
>
> You see: For all e IN: {1, 2, 3, ...,, n} is finite, while {1, 2, 3, ...} is infinite.

But you cannot name any natnumber that is in the infinite set but not in any finite set.That proves that either infinity is not larger than all finite sets or that there are dark numbers.

> Hint: There's no "step" which leads from finite to infinite or from infinite to finite ({ }).

Nevertheless ℵ₀ natnumbers are lost between all finite and the infinite intersection. How can that happen, if
∀k ∈ ℕ: E(k+1) = E(k) \ {k}
is valid for all lost numbers.

> Now,
>
> ∩{E(k) : k ∈ ℕ} = { }
> can easily be derived from
>
> Ak e IN: E(k+1) = E(k) \ {k} (*)
>
> Since (*) implies:
>
> ∀k ∈ ℕ: En ∈ ℕ: k !e E(n). (For each and every natural number there's an ensegement not containing it.)

But all definable endsegments keep ℵ₀ numbers, i.e., almost all.

Regards, WM

On Monday, July 4, 2022 at 12:10:16 PM UTC-3, WM wrote:
> William schrieb am Montag, 4. Juli 2022 um 16:19:26 UTC+2:
> > On Monday, July 4, 2022 at 10:42:48 AM UTC-3, WM wrote:
>
> > > we know that for all of them [the successors] are gone.
> >
> > So what? No statement about "all of them",
> All of them are the set.

Indeed "all of them" is the set. No statement is made about the set. Only a statement about the *elements* of the set.
Each *element* of set of natural numbers is followed by ℵ₀ successors does not mean the *set* , "all of them", is followed by ℵ₀ successors.

--
William Hughes

måndag 4 juli 2022 kl. 13:55:40 UTC+1 skrev WM:
> William schrieb am Sonntag, 3. Juli 2022 um 18:15:46 UTC+2:
> > On Sunday, July 3, 2022 at 12:37:50 PM UTC-3, WM wrote:
>
> > > > There is no contradiction with the statement
> > > > {∩E(k) : k ∈ ℕ} = { }
> > > > which is one statement about the set |N.
> > > No, it is about the elements of ℕ.
> > It is one statement about one collection.
> Above you can read: k ∈ ℕ. That means if all endsegments E(k) defined by an element k of ℕ are intersected, then the result is the empty set. There is no endsegment E(ℕ).
>
> ∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵ₀ is also a statement for all elements of ℕ. Accordingf to you it is about the set ℕ.
>
> But let us start from the fact that you accept
> |{∩E(k) : k =< mn ∈ ℕ}| = ℵ₀ ..
> This says that until *every* natural number n the intersection of endsegments is ℵ₀-infinite. {∩E(k) : k ∈ ℕ} = { }
> says that the intersection over *all* endsegments is empty:
>
> But no-one can say how the ℵ₀ elements are lost between every and all. It happens in an uncontrollable way, whether or not we accept the basic equation
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
> I call this state of ignorance dark. You may call it different, but you accept it.
>
> Regards, WM

We can explain it, BECAUSE IT IS NOT A FUCKING PROCESS!

The intersection is empty because for every element n, it is not in E(n+1)
Thus n is not in the intersection
This is arbitrary, meaning NO element is in it.

The end. That is the explination

On Sunday, July 3, 2022 at 11:35:54 PM UTC+2, WM wrote:
> Jim Burns schrieb am Sonntag, 3. Juli 2022 um 19:43:14 UTC+2:
> > On 7/3/2022 10:23 AM, WM wrote:
> > >
> > > For all definable numbers we obtain
> > > | ∩{E(k) : k ∈ ℕ_def} | = ℵ₀
> > >
> > No, you don't obtain that.
> >
> > That definition is ill-formed.

Of course, it's nonsense.

It's like the claim

"For all definable numbers we obtain
{ } = { }."

*** IN CONTRAST *** A SENSIBLE claim might be, say:

"For all natural numbers n we obtain
| ∩{E(k) : k e IN, k < n } | = ℵ₀ "

The following is NONSENSE again:

> > > |∩{E(k) : k <10}| = ℵ₀
> > > |∩{E(k) : k <100}| = ℵ₀
> > > |∩{E(k) : k <10^10}| = ℵ₀
> > > |∩{E(k) : k <10^10^10}| = ℵ₀
> > > and so on for all [natural] numbers k.

PURE CRANK STUFF.

Hint: "k" is already bound in the expression "{E(k) : k <10^10^10}". Hence "k" is not FREE in "{E(k) : k <10^10^10}".

This means that the universal quantor "for all natural numbers k" does NOT BIND the variable "k" in "{E(k) : k <10^10^10}".

Hence you claim doesn't make any sense.

Maybe you meant:

> > > |∩{E(k) : k e IN, k <10}| = ℵ₀
> > > |∩{E(k) : k e IN, k <100}| = ℵ₀
> > > |∩{E(k) : k e IN, k <10^10}| = ℵ₀
> > > |∩{E(k) : k e IN, k <10^10^10}| = ℵ₀

So what?

Hint:

For all natural numbers m: ∩{E(k) : k e IN, k < m } | = ℵ₀ .

> > So, you want to say that ℕ_def is
> > the collection of all m [in IN] such that
> > |∩{E(k) : [k e IN,] k < m}| = ℵ₀

Who knows? But seems to be a "reasonable" interpretation of WM's nonsense.

BUT:

> > |∩{E(k) : k ∈ ℕ_def}| = ℵ₀
> > is not how one says that.

Right.

> It is how one says that.

No. Learn some logic, learn some math, Mückenheim!

> > How one says that (in one way) is
> > ℕ_def = { m ∈ ℕ : |∩{E(k) : k < m }| = ℵ₀ }

Right.

@Dumbo/WM:

Look, dumbo, you want IN_def to be defined in a way SUCH THAT

m e IN_def <-> m e IN & | ∩{E(k) : k e IN, k < m} | = ℵ₀ (*)

holds (for all m).

So you may use (*) as a definition for "IN_def", or the "defining equation":

ℕ_def := {m ∈ ℕ : |∩{E(k) : k < m}| = ℵ₀} . (**)

> Say it that way if you like.

No, we say it in a CORRECT way.

> > If we correct your definition [...]
> > then [we] obtain |∩{E(k) : k ∈ ℕ_def}| = 0

Since in this case we get IN_def = IN.

> > The intersection of all the end segments
> > _which you intend to use_ is empty.

Right.

> You are a liar.

Nope.

> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀ is infinite by definition.

Nope.

Right, ℵ₀ is infinite, but |∩{E(k) : k ∈ ℕ_def}| IS NOT ℵ₀.

Actually, |∩{E(k) : k ∈ ℕ_def}| = 0 .

> > {E(k) : [k e IN,] k <10} does not have all the end segments.
> > {E(k) : [k e IN,] k <100} does not have all the end segments.
> > {E(k) : [k e IN,] k <10^10} does not have all the end segments.
> > {E(k) : [k e IN,] k <10^10^10} does not have all the end segments.

Right.
> > The intersection of _all_ the end segments is empty.

Right.

> Then ∀k ∈ ℕ : E(k+1)} = E(k) \ {k} would be violated.

Nope. It isn't "violated", dumbo.

> > Each element of ℕ has |ℕ|-many successors in ℕ.

Indeed!

An e IN: card({m e IN : m > n}) = |ℕ|.

> Each [...] element. That are those which I collect in ℕ_def.

If so then IN_def = IN.

Once more again:

If we DEFINE

IN_def := {n e IN : card({m e IN : m > n}) = |ℕ|}

then we get

IN_def = IN.

Too hard for you?

> But if ℕ_def = ℕ, then you do not cover each element. The successors are missing.

Huh?! ??????

Nothing is missing, dumbo.

You mean if ℕ_def = ℕ then there's an element in IN which is not in IN_def. Does this make ANY SENSE?

Seems that you are deluded, idiot!

> <nonsense deleted>

On Monday, July 4, 2022 at 5:19:16 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Montag, 4. Juli 2022 um 15:56:13 UTC+2:
> > On Monday, July 4, 2022 at 3:37:25 PM UTC+2, WM wrote:
> > >
> > > I am only asking how the loss of ℵ₀ natural numbers between [any] finite intersection and [any] infinite intersection can be justified.
> > >
> > Well, in the first case the intersection only concerns FINITELY many endsegments,
> >
> Since <bla>

Again, in the first case EACH AND EVERY intersection WE ARE CONSIDERING only contains FINITELY many endsegments,

Got that dumbo?!

> > but in the other case

EACH AND EVERY intersection WE ARE CONSIDERING contains INFINITELY many endsegments.

Too dumb or too psychotic to understand that FINITELY MANY =/= INFINITELY MANY?

> Not <bla>

It seems to me that in your case a psychiatrist might be more appropriate than a mathematician.

EOD

On Monday, July 4, 2022 at 5:19:16 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Montag, 4. Juli 2022 um 15:56:13 UTC+2:
> >
> > Now,
> >
> > ∩{E(k) : k ∈ ℕ} = { }
> >
> > can easily be derived from
> >
> > Ak e IN: E(k+1) = E(k) \ {k} (*)
> >
> > Since (*) implies:
> >
> > ∀k ∈ ℕ: En ∈ ℕ: k !e E(n). (For each and every natural number there's an ensegement not containing it.)
> >
> But

But??? What do you mean with "but"?

Did you get what I just explained to you or not? No "but", you silly crank!

A yes or no.

Hint:

__________ (*) implies: ∀k ∈ ℕ: En ∈ ℕ: k !e E(n). __________

In words: "For each and every natural number there's an ensegement not containing it."

In fact, from (*) we get for each and every k e IN: k is not an element in E(k+1).

Got that?

From this we get:

∀k ∈ ℕ: k !e ∩{E(k) : k ∈ IN}

by definition of ∩. [ Hint: x e ∩{E(k) : k ∈ IN} iff x e E(k) for ALL k in IN. In words: x is in ∩{E(k) : k ∈ IN} iff x is in each and every endsegment. ]

Hence

~Ek ∈ ℕ: k e ∩{E(k) : k ∈ IN}

Hence

∩{E(k) : k ∈ IN} = { }

(since ∩{E(k) : k ∈ IN} c ℕ).

qed

Why are you so bad in math, Mückenheim?

Is it the result of a psychosis or just a general lack of intelligence?

On 7/4/2022 7:57 AM, WM wrote:
> William schrieb am Sonntag, 3. Juli 2022 um 18:29:34 UTC+2:
>> On Sunday, July 3, 2022 at 12:44:45 PM UTC-3, WM wrote:
>
>>> But what is wrong here?
>>>
>>> 1, ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}
>>> 2, {∩E(k) : k ∈ ℕ} = { }
>>
>> Nothing. Each of the infinite number of statements in 1. is true. The statement 2. is true, Each of the statements of 1. is consistent with 2.
>
> According to (1) the empty set in (2) cannot be realized without a finite predecessor.

your conclusion is wrong.

your above equations simplify to these;

1 ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = E(k+1)

2 ∀k ∈ ℕ : ∩{E(1), E(2),..., E(k+1),...} = {}

which are correct.

>
> Regards, WM

On 7/4/2022 8:00 AM, WM wrote:
> William schrieb am Sonntag, 3. Juli 2022 um 18:29:34 UTC+2:
>
>> I said that each of them [without exception] not "all of them". For *each* of them [without exception] ℵ₀ successors remain.
>
> For all of them they are gone. This change cannot be investigated.
> The change from E(n) to E(n+1) can be investigated for definable n.

? all of that is wrong.

>
> Regards, WM

On 7/4/2022 10:58 AM, Fritz Feldhase wrote:
> On Monday, July 4, 2022 at 5:19:16 PM UTC+2, WM wrote:
>> Fritz Feldhase schrieb am Montag, 4. Juli 2022 um 15:56:13 UTC+2:
>>>
>>> Now,
>>>
>>> ∩{E(k) : k ∈ ℕ} = { }
>>>
>>> can easily be derived from
>>>
>>> Ak e IN: E(k+1) = E(k) \ {k} (*)
>>>
>>> Since (*) implies:
>>>
>>> ∀k ∈ ℕ: En ∈ ℕ: k !e E(n). (For each and every natural number there's an ensegement not containing it.)
>>>
>> But
>
> But??? What do you mean with "but"?
>
> Did you get what I just explained to you or not? No "but", you silly crank!
>
> A yes or no.
>
> Hint:
>
> __________ (*) implies: ∀k ∈ ℕ: En ∈ ℕ: k !e E(n). __________
>
> In words: "For each and every natural number there's an ensegement not containing it."
>
> In fact, from (*) we get for each and every k e IN: k is not an element in E(k+1).
>
> Got that?
>
> From this we get:
>
> ∀k ∈ ℕ: k !e ∩{E(k) : k ∈ IN}
>
> by definition of ∩. [ Hint: x e ∩{E(k) : k ∈ IN} iff x e E(k) for ALL k in IN. In words: x is in ∩{E(k) : k ∈ IN} iff x is in each and every endsegment. ]
>
> Hence
>
> ~Ek ∈ ℕ: k e ∩{E(k) : k ∈ IN}
>
> Hence
>
> ∩{E(k) : k ∈ IN} = { }
>
> (since ∩{E(k) : k ∈ IN} c ℕ).
>
> qed
>
> Why are you so bad in math, Mückenheim?
>
> Is it the result of a psychosis or just a general lack of intelligence?

perhaps a thinking disorder, where one can state #1, and then #2 but cannot put #1 and #2 together... happens to older folks.
only can process 2 or 3 thoughts at a time, or 5 or 6 words at a time... that kind of stuff. Which means Math is Out for WM.

William schrieb am Montag, 4. Juli 2022 um 17:20:43 UTC+2:
> On Monday, July 4, 2022 at 12:10:16 PM UTC-3, WM wrote:
> > William schrieb am Montag, 4. Juli 2022 um 16:19:26 UTC+2:
> > > On Monday, July 4, 2022 at 10:42:48 AM UTC-3, WM wrote:
> >
> > > > we know that for all of them [the successors] are gone.
> > >
> > > So what? No statement about "all of them",
> > All of them are the set.
> Indeed "all of them" is the set. No statement is made about the set. Only a statement about the *elements* of the set.
> Each *element* of set of natural numbers is followed by ℵ₀ successors does not mean the *set* , "all of them", is followed by ℵ₀ successors.

Correct. We have the elements 1, 2, 3, ... . Many of them are deleted by definable endsegments. Most of them are not deleted by definable endsegments.
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
But all remaining ℵo of them are deleted by the intersection of all endsegments.
∩{E(k) : k ∈ ℕ} = { }

How many endsegments are required to reduce ℵ₀ to 0 by
E(k+1) = E(k) \ {k} ?
None of them is visible.

Regards, WM

Fritz Feldhase schrieb am Montag, 4. Juli 2022 um 17:58:46 UTC+2:
> On Monday, July 4, 2022 at 5:19:16 PM UTC+2, WM wrote:

> But??? What do you mean with "but"?

We have the elements 1, 2, 3, ... . Many of them are deleted by definable endsegments. Most of them are not deleted by definable endsegments.
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
But all remaining ℵo of them are deleted by the intersection of all endsegments.
∩{E(k) : k ∈ ℕ} = { }

How many endsegments are required to reduce ℵ₀ to 0 by
E(k+1) = E(k) \ {k} ?
None of them is visible.

> In words: "For each and every natural number there's an ensegement not containing it."

Fine. For ℵ₀ natural number there is no definable endsegment not containg it. Every definable endsegment contains ℵ₀ natural numbers. How is the empty set realized?

Regards, WM

zelos...@gmail.com schrieb am Montag, 4. Juli 2022 um 17:22:57 UTC+2:
> måndag 4 juli 2022 kl. 13:55:40 UTC+1 skrev WM:

> > But no-one can say how the ℵ₀ elements are lost between every and all. It happens in an uncontrollable way, whether or not we accept the basic equation
> > ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
> > I call this state of ignorance dark. You may call it different, but you accept it.

> We can explain it, BECAUSE IT IS NOT A FUCKING PROCESS!

According to
∀k ∈ ℕ: E(k+1) = E(k) \ {k}
it is a step-by-step process.
>
> The intersection is empty because for every element n, it is not in E(n+1)

For almost all natural numbers every definable endsegment fails.

> Thus n is not in the intersection
> This is arbitrary, meaning NO element is in it.
>
For almost all natural numbers every definable endsegment fails.

∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo .

Regards, WM

On Monday, July 4, 2022 at 2:01:03 PM UTC-3, WM wrote:
> William schrieb am Montag, 4. Juli 2022 um 17:20:43 UTC+2:
> > On Monday, July 4, 2022 at 12:10:16 PM UTC-3, WM wrote:
> > > William schrieb am Montag, 4. Juli 2022 um 16:19:26 UTC+2:
> > > > On Monday, July 4, 2022 at 10:42:48 AM UTC-3, WM wrote:
> > >
> > > > > we know that for all of them [the successors] are gone.
> > > >
> > > > So what? No statement about "all of them",
> > > All of them are the set.
> > Indeed "all of them" is the set. No statement is made about the set. Only a statement about the *elements* of the set.
> > Each *element* of set of natural numbers is followed by ℵ₀ successors does not mean the *set* , "all of them", is followed by ℵ₀ successors.
> Correct. We have the elements 1, 2, 3, ...

Indeed, each element of the set is followed by ℵ₀ successors but the *set* not is followed by ℵ₀ successors.

--

William Hughes

On 7/4/2022 12:00 PM, WM wrote:
> William schrieb am Montag, 4. Juli 2022 um 17:20:43 UTC+2:
>> On Monday, July 4, 2022 at 12:10:16 PM UTC-3, WM wrote:
>>> William schrieb am Montag, 4. Juli 2022 um 16:19:26 UTC+2:
>>>> On Monday, July 4, 2022 at 10:42:48 AM UTC-3, WM wrote:
>>>
>>>>> we know that for all of them [the successors] are gone.
>>>>
>>>> So what? No statement about "all of them",
>>> All of them are the set.
>> Indeed "all of them" is the set. No statement is made about the set. Only a statement about the *elements* of the set.
>> Each *element* of set of natural numbers is followed by ℵ₀ successors does not mean the *set* , "all of them", is followed by ℵ₀ successors.
>
> Correct. We have the elements 1, 2, 3, ... . Many of them are deleted by definable endsegments.

Learn some Math!
"deleted", and "definable" are not Math. Endsegments are Set, all Sets remain fixed.

> Most of them are not deleted by definable endsegments.

"Most", "deleted", and "definable" are not Math.

> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀

> But all remaining ℵo of them are deleted by the intersection of all endsegments.

no, get out of your imagination, and use Math;

∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
∀k ∈ ℕ: ∩{E(1), E(2), ...)} = {}

> ∩{E(k) : k ∈ ℕ} = { }

you need to add the range for k on that one.

>
> How many endsegments are required to reduce ℵ₀ to 0 by
> E(k+1) = E(k) \ {k} ?

you cannot reduce ℵ₀ to 0 without using an Operator like Intersection ∩

>
> Regards, WM

The Deceiver and Miss Director.

On Monday, July 4, 2022 at 7:06:17 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Montag, 4. Juli 2022 um 17:58:46 UTC+2:
> > On Monday, July 4, 2022 at 5:19:16 PM UTC+2, WM wrote:
> >
> We have the elements 1, 2, 3, ... .

We have the elements in IN, dumbo, usually called the /natural numbers/.

> Many of them are deleted by definable endsegments. Most of them are not deleted by definable endsegments.

Who cares?

We aren't interested in natural numbesr which are or aren't "deleted by definable endsegments".

All we are interested in is the provable set-theoretic fact:

∩{E(k) : k ∈ ℕ} = { } .

> How many endsegments are required to

get an empty intersection?

This question has been answered MANY TIMES ALREADY, YOU SILLY ASSHOLE FULL OF SHIT!

The answer is: ___INFINITELY MANY ensegements.___

Now the important fact concerning

∩{E(k) : k ∈ ℕ} = { }

is the following:

∀k ∈ ℕ: En ∈ ℕ: k !e E(n). In words: "For each and every natural number there's an ensegement not containing it."

The latter is implyed by your favorite formula Ak e IN: E(k+1) = E(k) \ {k}.

> Fine.

Great.

From this we get:

∀k ∈ ℕ: k !e ∩{E(k) : k ∈ IN}

by definition of ∩. [ Hint: x e ∩{E(k) : k ∈ IN} iff x e E(k) for ALL k in IN. In words: x is in ∩{E(k) : k ∈ IN} iff x is in each and every endsegment. ]

Hence

~Ek ∈ ℕ: k e ∩{E(k) : k ∈ IN}

Hence

∩{E(k) : k ∈ IN} = { }

(since ∩{E(k) : k ∈ IN} c ℕ).

qed

You know, these steps form a so called PROOF.

Actually, they PROVE the THEOREM: ∩{E(k) : k ∈ IN} = { }

Hope this helps.

On Monday, July 4, 2022 at 7:12:06 PM UTC+2, sergi o wrote:

> > ∩{E(k) : k ∈ ℕ} = { }
>
> you need to add the range for k on that one.

Nope. It's ok. There are 3 main variants to write that "intersection":

1. ∩{E(k) : k ∈ ℕ} [ Her ∩ is a unary operator which works on the set {E(k) : k ∈ ℕ}. Note that "k" is already bounded in the expression "{E(k) : k ∈ ℕ}" which denotes the set {E(k) : k ∈ ℕ}. ]
2. ∩_(k e IN) E(k)
3. ∩_(k = 1)_^oo E(k) [In ASCII I'd rather write ∩_(k = 1...oo) E(k) in this case].

See: https://en.wikipedia.org/wiki/Union_(set_theory)#Notations
which mutatis mutandis holds for ∩ too.

On Monday, July 4, 2022 at 7:12:06 PM UTC+2, sergi o wrote:

> > How many endsegments are required to reduce ℵ₀ to 0 by
> > E(k+1) = E(k) \ {k} ?
> >
> you cannot reduce ℵ₀ to 0 without using an Operator like Intersection ∩

Good remark!

And even when adopting ∩ you have to "intersect" INFINITELY MANY endsegments to get ∩ ... = { }

_____________________________________

Maybe you have seen that I defined a "set-theoretic" operatator \\ which allows to "take infinitely many sets into account" _in the context of "set difference"_.

For example we'd have:

IN \\ {{1}, {2}, {3}, ...} = IN \ U{{1}, {2}, {3}, ...} = IN \ IN = { }.

In the same way:

IN \\ {{1}, {1, 2}, {1, 2, 3}, ...} = IN \ U{{1}, {1, 2}, {1, 2, 3}, ...} = IN \ IN = { }.

So we (!!!) might claim that "infinitely many endsegments are required to reduce IN to { }". (Knowing the definition of the machinery which allows for such a claim.)

FINITELY many endsegments won't do, Since for each and every n e IN:

IN \\ {{1}, {1, 2}, {1, 2, 3}, ..., {1, 2, 3, ..., n}} = IN \ U{{1}, {1, 2}, {1, 2, 3}, ..., {1, 2, 3, ..., n}} = IN \ {1, 2, 3, ..., n} = {n + 1, n + 2, n + 3, ...} =/= { }.

Using a rather "natural" definition (->notation) we may even write these things down in an "intuitively appealing" way:

IN \ {1} \ {1, 2} \ {1, 2, 3} \ ... = { } , but for all n e IN:

IN \ {1} \ {1, 2} \ {1, 2, 3} \ ... \ {1, 2, 3, ..., n} = {n + 1, n + 2, n + 3, ...} =/= { }.

It's facinating that WM does not even get it when formulated in this rather SIMPLE (or "simplified") form.

WM explained :
> zelos...@gmail.com schrieb am Montag, 4. Juli 2022 um 17:22:57 UTC+2:
>> måndag 4 juli 2022 kl. 13:55:40 UTC+1 skrev WM:
>
>>> But no-one can say how the ℵ₀ elements are lost between every and all. It
>>> happens in an uncontrollable way, whether or not we accept the basic
>>> equation ∀k ∈ ℕ: E(k+1) = E(k) \ {k} . I call this state of ignorance
>>> dark. You may call it different, but you accept it.
>
>> We can explain it, BECAUSE IT IS NOT A FUCKING PROCESS!
>
> According to
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> it is a step-by-step process.

No, it isn't.

On 7/4/2022 12:06 PM, WM wrote:
> Fritz Feldhase schrieb am Montag, 4. Juli 2022 um 17:58:46 UTC+2:
>> On Monday, July 4, 2022 at 5:19:16 PM UTC+2, WM wrote:
>
>> But??? What do you mean with "but"?
>
> We have the elements 1, 2, 3, ... . Many of them are deleted by definable endsegments. Most of them are not deleted by definable endsegments.
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
> But all remaining ℵo of them are deleted by the intersection of all endsegments.
> ∩{E(k) : k ∈ ℕ} = { }
>
> How many endsegments are required to reduce ℵ₀ to 0 by
> E(k+1) = E(k) \ {k} ?
> None of them is visible.
>
>> In words: "For each and every natural number there's an ensegement not containing it."
>
> Fine. For ℵ₀ natural number there is no definable endsegment not containg it. Every definable endsegment contains ℵ₀ natural numbers. How is the empty set realized?

Proof it is empty.

assume there is a natural number k in the intersection of all endsegments.

However, k is not in the endsegment(k+1)

therefore, k cannot be in the intersection of all endsegments.

since k ∈ ℕ, then there are no elements of ℕ in the intersection of all endsegments

Got That ?

WM, you need to restrain your math naivety.
We all went to Grade School here, Jr High, and took Algebra I

>
> Regards, WM

On Monday, July 4, 2022 at 7:52:07 PM UTC+2, Fritz Feldhase wrote:

> Using a rather "natural" definition (->notation) we may even write these things down in an "intuitively appealing" way:
>
> IN \ {1} \ {1, 2} \ {1, 2, 3} \ ... = { } , but for all n e IN:
>
> IN \ {1} \ {1, 2} \ {1, 2, 3} \ ... \ {1, 2, 3, ..., n} = {n + 1, n + 2, n + 3, ...} =/= { }.
>
> It's facinating that WM does not even get it when formulated in this rather SIMPLE (or "simplified") form.

@Mücke: You see:

For all n e IN:

IN \ {1} \ {1, 2} \ {1, 2, 3} \ ... \ {1, 2, 3, ..., n} = {n + 1, n + 2, n + 3, ...} =/= { } , but

IN \ {1} \ {1, 2} \ {1, 2, 3} \ ... = IN \ {1} \ {1, 2} \ {1, 2, 3} \ .... \ {1, 2, 3, ..., n} \ {1, 2, 3, ..., n + 1} \ {1, 2, 3, ..., n + 1, n+ 2} \ {1, 2, 3, ..., n + 1, n+ 2, n+3}, ...

Using Sergio's words: "We do not stop at n (for some n e IN)."

måndag 4 juli 2022 kl. 18:10:35 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Montag, 4. Juli 2022 um 17:22:57 UTC+2:
> > måndag 4 juli 2022 kl. 13:55:40 UTC+1 skrev WM:
>
> > > But no-one can say how the ℵ₀ elements are lost between every and all. It happens in an uncontrollable way, whether or not we accept the basic equation
> > > ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
> > > I call this state of ignorance dark. You may call it different, but you accept it.
> > We can explain it, BECAUSE IT IS NOT A FUCKING PROCESS!
> According to
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> it is a step-by-step process.

no it IS NOT!

That gives only a RELATION between things, it does not say there are STEPS!

> >
> > The intersection is empty because for every element n, it is not in E(n+1)
> For almost all natural numbers every definable endsegment fails.

"definable" is meaningless.

There needs no fucking specific endsegmetn, it is the COLLECTION AS A WHOLE that does it.

> > Thus n is not in the intersection
> > This is arbitrary, meaning NO element is in it.
> >
> For almost all natural numbers every definable endsegment fails.
>
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
>
> Regards, WM

None of which is relevant for the intersection to be empty you fucking retard

On 7/4/2022 12:10 PM, WM wrote:
> zelos...@gmail.com schrieb am Montag, 4. Juli 2022 um 17:22:57 UTC+2:
>> måndag 4 juli 2022 kl. 13:55:40 UTC+1 skrev WM:
>
>>> But no-one can say how the ℵ₀ elements are lost between every and all. It happens in an uncontrollable way, whether or not we accept the basic equation
>>> ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
>>> I call this state of ignorance dark. You may call it different, but you accept it.
>
>> We can explain it, BECAUSE IT IS NOT A FUCKING PROCESS!
>
> According to
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> it is a step-by-step process.

no. you are reading it wrong.

go find out your own self what it means

I am tired of spoon feeding you.

>>
>> The intersection is empty because for every element n, it is not in E(n+1)
>
>
>> Thus n is not in the intersection
>> This is arbitrary, meaning NO element is in it.
>>
>
> Regards, WM