How many unnatural numbers are created?

[WM]

How many unnatural numbers are created by multiplying all terms of the
sequence

1, 2, 3, 4, 5, ... ω by 2 with the result
2, 4, 6, ... 2ω?

Regards, WM

[FromTheRafters]

Post by WM
How many unnatural numbers are created by multiplying all terms of the
sequence
1, 2, 3, 4, 5, ... ω by 2 with the result
2, 4, 6, ... 2ω?

Could you explain how your 'unnatural number' system works?

[WM]

Post by FromTheRafters

Post by WM
How many unnatural numbers are created by multiplying all terms of the
sequence
1, 2, 3, 4, 5, ... ω by 2 with the result
2, 4, 6, ... 2ω?

Could you explain how your 'unnatural number' system works?

If all doubled natural numbers remain smaller than ω, then the infinite
space between ω and 2ω remains empty. If the doubling creates these
numbers ω+2, ω+4, ω+6, ..., then they are no longer natural numbers.
What is going on?

Regards, WM

[FromTheRafters]

Post by FromTheRafters

Post by WM
How many unnatural numbers are created by multiplying all terms of the
sequence
1, 2, 3, 4, 5, ... ω by 2 with the result
2, 4, 6, ... 2ω?

Could you explain how your 'unnatural number' system works?

If all doubled natural numbers remain smaller than ω, then the infinite space
between ω and 2ω remains empty. If the doubling creates these numbers ω+2,
ω+4, ω+6, ..., then they are no longer natural numbers.
What is going on?

They are the second, fourth, and sixth transfinite ordinals.

[Moebius]

Post by FromTheRafters

these numbers ω+2, ω+4, ω+6, ..., are no natural numbers. What is going on?

They are the second, fourth, and sixth transfinite ordinals.

Not quite. I mean, usually we would say

| ω is the first, ω+1 is the second, ω+2 is the third ... transfinite
ordinal

Of course, using the ordinals themselves you might state

| ω is the 0-th, ω+1 is the 1-th, ω+2 is the 2-th ... transfinite ordinal

On the other hand: "Did he win the 0-th prize?" Who would say that?

[Phil Carmody]

Post by Moebius

Post by FromTheRafters

these numbers ω+2, ω+4, ω+6, ..., are no natural numbers. What is going on?

They are the second, fourth, and sixth transfinite ordinals.

Not quite. I mean, usually we would say
| ω is the first, ω+1 is the second, ω+2 is the third ... transfinite
ordinal
Of course, using the ordinals themselves you might state
| ω is the 0-th, ω+1 is the 1-th, ω+2 is the 2-th ... transfinite ordinal
On the other hand: "Did he win the 0-th prize?" Who would say that?

WM always wins 0-th prize.

Phil

[Moebius]

Post by WM

Post by FromTheRafters

Post by WM
How many unnatural numbers are created by multiplying all terms of
the sequence
1, 2, 3, 4, 5, ... ω by 2 with the result
2, 4, 6, ... 2ω?

Could you explain how your 'unnatural number' system works?

If all doubled natural numbers remain smaller than ω,

which they do (hint: An e IN: n < ω and hence An e IN: 2n < ω since An e
IN: 2n e IN)

Post by WM
then the infinite space between ω and 2ω

There is no "infinite space between ω and 2ω" since ω = 2ω.

It seems that you are mixing up 2ω with ω2.

Post by WM
If the doubling creates these numbers ω+2, ω+4, ω+6, ...,

It doesn't.

Post by WM
then <whatever>
What is going on?

Ex falso quod libet, that's going on, Mückenheim.

[WM]

If all are doubled, then many are new.

Post by Moebius
There is no "infinite space between ω and 2ω" since ω = 2ω.
It seems that you are mixing up 2ω with ω2.

So did Cantor until 1884.

Post by Moebius

Post by WM
If the doubling creates these numbers ω+2, ω+4, ω+6, ...,

It doesn't.

Einfach über hinwegzählen.

Regards, WM

[Moebius]

Post by WM

Post by Moebius
It seems that you are mixing up 2ω with ω2.

So did Cantor until 1884.

So what?! Using erroneous notions is a hobby of yours?

Post by WM

Post by Moebius

Post by WM
If the doubling creates these numbers ω+2, ω+4, ω+6, ...,

It doesn't.

Einfach über hinwegzählen.

Hint: You aren't Cantor nor Hilbert. You are just a silly crank.

1. "doubling" is not counting

2. doubling the (infinitely many) natural numbers 1, 2, 3, ... results
in the (infinitely many) natural numbers 2, 4, 6, ...

3. multipliying the (single) ordinal number ω with 2 from left results
in the (single) ordinal sumber ω, multipliying the (single) ordinal
number ω with 2 from right results in the (single) ordinal sumber ω2.
(Hint: 2ω =/= ω2.)

4. Hence {2x : x e {1, 2, 3, 4, ... ω}} = {2x : x e {1, 2, 3, 4, ...} u
{ω}} = {2, 4, 6, 8, ...} u {2ω} = {2, 4, 6, 8, ...} u {ω} = {2, 4, 6, 8,
... w} and {x2 : x e {1, 2, 3, 4, ... ω}} = {x2 : x e {1, 2, 3, 4, ...}
u {ω}} = {2, 4, 6, 8, ...} u {ω2} = {2, 4, 6, 8, ... w2}.

5. ω+2, ω+4, ω+6, ... !e {2, 4, 6, 8, ... ω} and ω+2, ω+4, ω+6, ... !e
{2, 4, 6, 8, ... ω2} since (a) for all n e {1, 2, 3, 4, ...}: n < ω < ω2
and hence for all n e {2, 4, 6, 8, ...}: n < ω < ω2 and (b) ω, ω2 !e {
ω+2, ω+4, ω+6, ...}. (Hint: ω < ω+1, ω+2, ω+3, ω+4, ... < ω + ω = ω2.)

6. Geht's noch dümmer, Mückenheim?

Hint: {2x : x e {1, 2, 3, ω}} = {2, 4, 6, ω} and {x2 : x e {1, 2, 3, ω}}
= {2, 4, 6, ω2}.

[WM]

Post by Moebius

Post by WM

Post by Moebius
It seems that you are mixing up 2ω with ω2.

So did Cantor until 1884.

So what?! Using erroneous notions is a hobby of yours?

It is not erroneous but simply a matter of another definition.

Post by Moebius

Post by WM

Post by Moebius

Post by WM
If the doubling creates these numbers ω+2, ω+4, ω+6, ...,

It doesn't.

Einfach über ω hinwegzählen.

1. "doubling" is not counting

Doubling is faster.

Post by Moebius
2. doubling the (infinitely many) natural numbers 1, 2, 3, ... results
in the (infinitely many) natural numbers 2, 4, 6, ...

Then you are not accepting that all natnumbers had been there already and
were doubled?
Or you believe that doubling creates the same numbers which have been
doubled?

Regards, WM

[Moebius]

Post by WM

Post by Moebius
 >>

Post by WM

Post by Moebius
It seems that you are mixing up 2ω with ω2.

So did Cantor until 1884.

So what?! Using erroneous notions is a hobby of yours?

It is not erroneous but simply a matter of another definition.

Using non-standard definitions does not help your case, Mückenheim.

So let's stick to the usual definitions when using set theoretic notions
/ discussing _set theory_, idiot!

Hint: In set theory: 2ω = ω and w < ω + ω = ω2.

Post by WM

Post by Moebius

Post by WM

Post by Moebius

Post by WM
If the doubling creates these numbers ω+2, ω+4, ω+6, ...,

It doesn't.

Doubling the (infinitely many) natural numbers 1, 2, 3, ... results
in the (infinitely many) natural numbers 2, 4, 6, ...

Then you are not accepting that all natnumbers had been there already
and were doubled?

Sure, I do.

Post by WM
Or you believe that doubling [results in] the same numbers which have been
doubled?

Some of them, yes.

Hint: 1, 2, 3, 4 -- * 2 --> 2, 4, 6, 8

Got it?! 2 and 4 were "already there" before doubling.

Now 2IN = {2n : n e IN} c IN.

EOD

[WM]

Post by Moebius

Post by WM
Or you believe that doubling [results in] the same numbers which have been
doubled?

Some of them, yes.
Hint: 1, 2, 3, 4 -- * 2 --> 2, 4, 6, 8
Got it?! 2 and 4 were "already there" before doubling.

Of course, but many were not. For instance ω+ω.

Regards, WM

[Moebius]

Post by WM

Post by Moebius

Post by WM
Or you believe that doubling [results in] the same numbers which have
been doubled?

Some of them, yes.
Hint: 1, 2, 3, 4 -- * 2 --> 2, 4, 6, 8
Got it?! 2 and 4 were "already there" before doubling.

Of course, but many were not. For instance ω+ω.

Nein, nicht "many", Mückenheim, sondern genau _eine_ Zahl. nämlich ω2 = ω+ω.

Nochmal:

{1 2, 3, ... ω} = {1, 2, 3, ...} u {w} = IN u {w} mit ω !e IN.

Und daher:

{x2 : x e {1, 2, 3, ... ω}} = {x2 : x e {1, 2, 3, ...} u {ω}} = {2, 4,
6, ...} u {ω2}} = {2x : x e IN} u {ω2} = G u {ω2} mit G c IN und ω2 !e
IN. (Wo G := {2x : x e IN} die Menge der geraden natürlichen Zahlen ist.)

Wie dumm kann man eigentlich sein, Mückenheim?

[WM]

Post by Moebius

Post by WM

Post by Moebius
Got it?! 2 and 4 were "already there" before doubling.

Of course, but many were not. For instance ω+ω.

Nein, nicht "many", Mückenheim, sondern genau _eine_ Zahl. nämlich ω2 = ω+ω.

You do not accept that doubling of 1, 2, 3, ... ω doubles all structures?

The interval [3, 7] of length 4 becomes the interval [6, 14] of length 8.

Post by Moebius
{1 2, 3, ... ω} = {1, 2, 3, ...} u {w} = IN u {w} mit ω !e IN.
{x2 : x e {1, 2, 3, ... ω}} = {x2 : x e {1, 2, 3, ...} u {ω}} = {2, 4,
6, ...} u {ω2}} = {2x : x e IN} u {ω2} = G u {ω2} mit G c IN und ω2 !e
IN. (Wo G := {2x : x e IN} die Menge der geraden natürlichen Zahlen ist.)

I do not accept your claim but accept mathematics: The interval between
ℕ and ω is not longer than a natural number k. Hence the doubled
interval is not longer than 2k. Hence almost all of the infinitely many
ordinal places ω+1, ω+2, ω+3 between ω and ω+ω must be occupied by
numbers n + n, where n ∈ ℕ.

Regards, WM

[Moebius]

Na, wunderbar!

[...] almost all of the infinitely many ordinal places ω+1, ω+2, ω+3 between ω and ω+ω must be occupied by numbers n + n, where n ∈ ℕ.

Für alle n in IN ist n + n in IN.

Für alle n in IN: n < ω < ω+1 < ω+2 < ω+3 < ...

Also ist kein ω+1, ω+2, ω+3, ... in IN.

Also ist für kein n in IN: n + n in {ω+1, ω+2, ω+3, ...}.

Einfacher formuliert, dass es vielleicht sogar ein geistesgestörter
Spinner versteht: IN enthält NUR (und alle) ENDLICHEN ORDINALZAHLEN. Das
DOPPELTE einer natürlichen Zahl ist wieder eine natürliche Zahl. Die
Zahlen ω+1, ω+2, ω+3, ... sind aber UNENDLICHE ORDINALZAHLEN. Was
unendlich ist, ist nicht endlich. Also ist KEINE der Zahlen ω+1, ω+2,
ω+3, ... gleich n + n (wo n eine natürliche Zahl ist).

[Alan Mackenzie]

Post by Moebius
Na, wunderbar!

[...] almost all of the infinitely many ordinal places ω+1, ω+2, ω+3
between ω and ω+ω must be occupied by numbers n + n, where n ∈ ℕ.

Für alle n in IN ist n + n in IN.
Für alle n in IN: n < ω < ω+1 < ω+2 < ω+3 < ...
Also ist kein ω+1, ω+2, ω+3, ... in IN.
Also ist für kein n in IN: n + n in {ω+1, ω+2, ω+3, ...}.
Einfacher formuliert, dass es vielleicht sogar ein geistesgestörter
Spinner versteht: IN enthält NUR (und alle) ENDLICHEN ORDINALZAHLEN. Das
DOPPELTE einer natürlichen Zahl ist wieder eine natürliche Zahl. Die
Zahlen ω+1, ω+2, ω+3, ... sind aber UNENDLICHE ORDINALZAHLEN. Was
unendlich ist, ist nicht endlich. Also ist KEINE der Zahlen ω+1, ω+2,
ω+3, ... gleich n + n (wo n eine natürliche Zahl ist).

Was du sagst ist nicht falsch, aber ... s.m. sollte eine
englischsprachige Gruppe sein.

What you're saying isn't wrong, but ... s.m. is supposed to be an
English language group.

[Moebius]

Post by Alan Mackenzie
What you're saying isn't wrong, but ... s.m. is supposed to be an
English language group.

Kommt sonst die Sprachpolizei?

[Tom Bola]

Post by Moebius
Na, wunderbar!

[...] almost all of the infinitely many ordinal places ω+1, ω+2, ω+3 between ω and ω+ω must be occupied by numbers n + n, where n ∈ ℕ.

Für alle n in IN ist n + n in IN.
Für alle n in IN: n < ω < ω+1 < ω+2 < ω+3 < ...
Also ist kein ω+1, ω+2, ω+3, ... in IN.
Also ist für kein n in IN: n + n in {ω+1, ω+2, ω+3, ...}.
Einfacher formuliert, dass es vielleicht sogar ein geistesgestörter
Spinner versteht: IN enthält NUR (und alle) ENDLICHEN ORDINALZAHLEN. Das
DOPPELTE einer natürlichen Zahl ist wieder eine natürliche Zahl. Die
Zahlen ω+1, ω+2, ω+3, ... sind aber UNENDLICHE ORDINALZAHLEN. Was
unendlich ist, ist nicht endlich. Also ist KEINE der Zahlen ω+1, ω+2,
ω+3, ... gleich n + n (wo n eine natürliche Zahl ist).

WM does not get that natural numbers and limit ordinals are totally
different kind of numbers and sees all of them on a "natural line" whose
points (i.e. numbers) have distances that are measured in the unit of two
adjacent natural numbers...

[Moebius]

Post by Tom Bola

Post by Moebius
Na, wunderbar!

[...] almost all of the infinitely many ordinal places ω+1, ω+2, ω+3 between ω and ω+ω must be occupied by numbers n + n, where n ∈ ℕ.

Für alle n in IN ist n + n in IN.
Für alle n in IN: n < ω < ω+1 < ω+2 < ω+3 < ...
Also ist kein ω+1, ω+2, ω+3, ... in IN.
Also ist für kein n in IN: n + n in {ω+1, ω+2, ω+3, ...}.
Einfacher formuliert, dass es vielleicht sogar ein geistesgestörter
Spinner versteht: IN enthält NUR (und alle) ENDLICHEN ORDINALZAHLEN. Das
DOPPELTE einer natürlichen Zahl ist wieder eine natürliche Zahl. Die
Zahlen ω+1, ω+2, ω+3, ... sind aber UNENDLICHE ORDINALZAHLEN. Was
unendlich ist, ist nicht endlich. Also ist KEINE der Zahlen ω+1, ω+2,
ω+3, ... gleich n + n (wo n eine natürliche Zahl ist).

WM does not get that natural numbers and limit ordinals are totally
different kind of numbers and sees all of them on a "natural line" whose
points (i.e. numbers) have distances that are measured in the unit of two
adjacent natural numbers...

Right. You you nailed it (I guess).

[Tom Bola]

Post by Moebius

Post by Tom Bola

Post by Moebius
Na, wunderbar!

[...] almost all of the infinitely many ordinal places ω+1, ω+2, ω+3 between ω and ω+ω must be occupied by numbers n + n, where n ∈ ℕ.

Für alle n in IN ist n + n in IN.
Für alle n in IN: n < ω < ω+1 < ω+2 < ω+3 < ...
Also ist kein ω+1, ω+2, ω+3, ... in IN.
Also ist für kein n in IN: n + n in {ω+1, ω+2, ω+3, ...}.
Einfacher formuliert, dass es vielleicht sogar ein geistesgestörter
Spinner versteht: IN enthält NUR (und alle) ENDLICHEN ORDINALZAHLEN. Das
DOPPELTE einer natürlichen Zahl ist wieder eine natürliche Zahl. Die
Zahlen ω+1, ω+2, ω+3, ... sind aber UNENDLICHE ORDINALZAHLEN. Was
unendlich ist, ist nicht endlich. Also ist KEINE der Zahlen ω+1, ω+2,
ω+3, ... gleich n + n (wo n eine natürliche Zahl ist).

WM does not get that natural numbers and limit ordinals are totally
different kind of numbers and sees all of them on a "natural line" whose
points (i.e. numbers) have distances that are measured in the unit of two
adjacent natural numbers...

Right. You you nailed it (I guess).

Sure, but WM has "stated" this one of his mythical conceptions
already a few weeks ago (as far as I remind that).

[FromTheRafters]

Post by Tom Bola

Post by Moebius

Post by Tom Bola

Post by Moebius
Na, wunderbar!

[...] almost all of the infinitely many ordinal places ω+1, ω+2, ω+3
between ω and ω+ω must be occupied by numbers n + n, where n ∈ ℕ.

Für alle n in IN ist n + n in IN.
Für alle n in IN: n < ω < ω+1 < ω+2 < ω+3 < ...
Also ist kein ω+1, ω+2, ω+3, ... in IN.
Also ist für kein n in IN: n + n in {ω+1, ω+2, ω+3, ...}.
Einfacher formuliert, dass es vielleicht sogar ein geistesgestörter
Spinner versteht: IN enthält NUR (und alle) ENDLICHEN ORDINALZAHLEN. Das
DOPPELTE einer natürlichen Zahl ist wieder eine natürliche Zahl. Die
Zahlen ω+1, ω+2, ω+3, ... sind aber UNENDLICHE ORDINALZAHLEN. Was
unendlich ist, ist nicht endlich. Also ist KEINE der Zahlen ω+1, ω+2,
ω+3, ... gleich n + n (wo n eine natürliche Zahl ist).

WM does not get that natural numbers and limit ordinals are totally
different kind of numbers and sees all of them on a "natural line" whose
points (i.e. numbers) have distances that are measured in the unit of two
adjacent natural numbers...

Right. You you nailed it (I guess).

Sure, but WM has "stated" this one of his mythical conceptions
already a few weeks ago (as far as I remind that).

WM is a potentially infinite collection of misinterpretations. That is
to say, as many as he needs.

[Tom Bola]

The interval between IN and ω is not longer than a natural number k.

Bullshit - because ω is not a natural number.

[Moebius]

Post by Tom Bola

The interval between IN and ω is not longer than a natural number k.

Bullshit - because ω is not a natural number.

Die Frage ist: Was ist ein "interval between IN and ω". Wenn man d a s
weiß, d a n n kann man sich Gedanken dazu machen, ob die Behauptung "is
not longer than a natural number k" darauf zutrifft oder nicht.

Vermutlich meint er hier aber eigentlich: "The length of the interval
between IN and ω is not larger than a natural number k."

In jedem Fall haben wir es hier wieder einmal mit einer Behauptung der
Kategorie "not even wrong" zu tun. Typisch Mückenheim halt.

[WM]

Post by Moebius

Post by Tom Bola

The interval between IN and ω is not longer than a natural number k.

ω is not a natural number.

ω is a point on the ordinal axis.

Post by Moebius
Die Frage ist: Was ist ein "interval between IN and ω". Wenn man d a s
weiß, d a n n kann man sich Gedanken dazu machen, ob die Behauptung "is
not longer than a natural number k" darauf zutrifft oder nicht.

We know that no ordinal fits on the ordinal axis between ℕ and ω. That
is enough!

Post by Moebius
Vermutlich meint er hier aber eigentlich: "The length of the interval
between IN and ω is not larger than a natural number k."

We could also use less than 1. But the estimation k ∈ ℕ is sufficient.

Regards, WM

[Moebius]

How many [new] numbers are created by multiplying all terms of the sequence
1, 2, 3, 4, 5, ... ω by 2 with the result 2, 4, 6, ... 2ω?

Actually, none.

Why? Because no number is "created" this way which wasn't already there.

Hint: {2n : n e IN} c IN and 2ω = ω.

Hence {2, 4, 6, ... 2ω} = {2n : n e {1, 2, 3, 4, 5, ... ω} c {1, 2, 3,
4, 5, ... ω}, where {1, 2, 3, 4, 5, ... ω} = {1, 2, 3, 4, 5, ...} u {ω}
= IN u {ω} and {2, 4, 6, ... 2ω} = {2, 4, 6, ...} u {2ω} = {2n : n e IN}
u {2ω}.

[Tom Bola]

Post by Moebius
...
--
Diese E-Mail wurde von Avast-Antivirussoftware auf Viren geprüft.
www.avast.com

Warum teilst du der Welt immer wieder mit dass deine "E-Mail" geprüft wurde?
;)

[Moebius]

Post by Tom Bola

Post by Moebius
...
--
Diese E-Mail wurde von Avast-Antivirussoftware auf Viren geprüft.
www.avast.com

Warum teilst du der Welt immer wieder mit dass deine "E-Mail" geprüft wurde?
;)

Keine Ahnung. Schadet aber auch nicht, oder? :-)

Gleich kommt's wieder!
vvvvvvvvvvvvvvvvvvvvv

[Tom Bola]

Post by Moebius

Post by Tom Bola

Post by Moebius
...
--
Diese E-Mail wurde von Avast-Antivirussoftware auf Viren geprüft.
www.avast.com

Warum teilst du der Welt immer wieder mit dass deine "E-Mail" geprüft wurde?
;)

Keine Ahnung. Schadet aber auch nicht, oder? :-)

Nein, das nicht, aber man hat ja eine heutzutage natürliche Aversion
gegen permanent (und auch noch die gleiche) Werbung...

Post by Moebius
Gleich kommt's wieder!
vvvvvvvvvvvvvvvvvvvvv

;)

[WM]

Post by Moebius

How many [new] numbers are created by multiplying all terms of the sequence
1, 2, 3, 4, 5, ... ω by 2 with the result 2, 4, 6, ... 2ω?

Actually, none.
Why? Because no number is "created" this way which wasn't already there.
Hint: {2n : n e IN} c IN and 2ω = ω.

Nonsense. Cantor said 2ω is ω + ω =/= ω.

Post by Moebius
Hence

Wrong argument.

Regards, WM

[Moebius]

Post by WM

Post by Moebius

How many [new] numbers are created by multiplying all terms of the sequence
1, 2, 3, 4, 5, ... ω by 2 with the result 2, 4, 6, ... 2ω?

Actually, none.
Why? Because no number is "created" this way which wasn't already there.
Hint: {2n : n e IN} c IN and 2ω = ω.

Nonsense.

No, not nonsense, Mückenheim.

Post by WM
Cantor said 2ω is ω + ω =/= ω.

He only said that in the early years of set theory, later he corrected

Post by WM

Post by Moebius
Anmerkung von Zermelo [3] Zu S. 195. Hier und im folgenden stellt
Cantor den Multiplikator voran und schreibt 2ω für ω + ω; in der
späteren systematischen Darstellung III 9 stellt er umgekehrt den
Multiplikandus voran und schreibt ω2, was aus Gründen der Analogie
entschieden vorzuziehen ist, weil auch bei der Addition nur der zweite
Summand (der Addendus), wenn er endlich ist, die transfinite Summe
modifiziert, vergrößert. Vgl. S. 302, 322.
Cantor 1884: Ich habe in den "Grundlagen" den Multiplikator links, den
Multiplikandus rechts geschrieben; es hat sich mir aber gezeigt, daß
der entgegengesetzte Gebrauch, den Multipikandus links zuerst und dann
rechts den Multiplikator zu schreiben, für die weitere Entwicklung der
transfiniten Ordnungszahlenlehre der zweckmäßigere, ja fast
unentbehrliche ist; aus diesem Grunde kehre ich also die betreffende
Schreibweise der "Grundlagen", soweit sie sich auf Produkte bezieht,
von jetzt ab immer um.

Wie dumm kann man eigentlich sein, Mückenheim?

Also nochmal, vielleicht verstehst Du es ja jetzt besser:

| No number is "created" this way which wasn't already there.
|
| Hint: {2n : n e IN} c IN and 2ω = ω.
|
| Hence {2, 4, 6, ... 2ω} = {2n : n e {1, 2, 3, 4, 5, ... ω} c {1, 2, 3,
4, 5, ... ω}.