Re: The reality of sets, on a scale of 1 to 10 [Was: The non-existence of "dark numbers"]

Am Thu, 03 Apr 2025 22:17:33 +0200 schrieb WM:

On 03.04.2025 16:40, Alan Mackenzie wrote:

WM <wolfgang.mueckenheim@tha.de> wrote:

On 28.03.2025 17:32, Alan Mackenzie wrote:

 

So tell us, O wise one, how many elements are there in {1, 3, 5, 7,
9, ...}?  And how many elements in {0, 4, 8, 12, 16, ...}?  Which of
these two numbers is bigger, and why?

|ℕ|/2 > |ℕ|/4.

Start out with a set of natural numbers.  Multiply each member by four,
giving a new set.  You'd have us believe that the new set contains
fewer elements than the original set.

Fact. Ifff the natural numbers are an actually infinite set, then its
elements are invariable and fixed.

How else could it be.

By multiplication no larger numbers
can be created. What you have in mind is a potentially infinite set.
 
Let me explain in detail:
Cantor created the sequence of the ordinal numbers by means of his first
and second generation principle:
0, 1, 2, 3, ..., ω, ω+1, ω+2, ω+3, ..., ω*2, ω*2+1, ω*2+2,
ω*2+3, ..., ω*3, ... .
This sequence, except its very first terms, has no relevance for
classical mathematics. But it is important for set theory that in actual
infinity nothing fits between ℕ and ω. Likewise before ω*2 and ω*3
there is no empty space.

What do „fits” and „space” mean?

According to Hilbert we can simply count beyond the
infinite by a quite natural and uniquely determined, consistent
continuation of the ordinary counting in the finite.  But we would
proceed even faster, when instead of counting, we doubled the numbers.
This leads to the central issue:  Multiply every element of the set ℕ by
2
{1, 2, 3, ...}*2 = {2, 4, 6, ...} .
 
The density of the natural numbers on the real axis is greater than the
density of the even natural numbers.

In which sense? There are infinitely many of either.

Therefore the doubled natural numbers cover twice as many space than
before.

Again, how? How much?

What is the result of
doubling? Either all doubled numbers are natural numbers, then not all
natural numbers have been doubled.

Why? Every even number has a natural half its value.

Natural numbers not available before
have been created. This is possible only based on potential infinity. Or
all natural numbers have been doubled, then the result stretches
farther, namely beyond all natural numbers.

No, there is no natural number whose double is larger than ω.
Your mistake is to think of infinity as a really big number of the same
kind as naturals, but it is rather more like a type or a special point.

It is more suggestive to double the set ℕ U {ω} = {1, 2, 3, ..., ω}
with the result
{1, 2, 3, ..., ω}*2 = {2, 4, 6, ..., ω*2} .
What elements fall between ω and ω*2?

Where is ω?

What size has the interval between 2ℕ and ω*2?

What is the interval between two sets?

The natural answer is (0, ω]*2 = (0, ω*2] with ω or ω+1
amidst.

No, that would be two consecutive infinities. You can do that, but not
with f(x)=2x and not without violating the order.

The number of doubled natural numbers is precisely |ℕ|. But half
of the doubled numbers are no longer natural numbers; they surpass ω. If
all natural numbers including all even numbers are doubled and if
doubling increases the value for all natural numbers because n < 2n,
then not all doubled even numbers fit below ω.

Yes they do. The product of two naturals is also a natural.

Natural numbers greater
than all even natural numbers however are not possible.
 
Every other result would violate symmetry and beauty of mathematics, for
instance the claim that the result would be ℕ U {ω, ω*2}. All numbers
between ω and ω*2, which are precisely as many as in ℕ between 0 and ω,
would not be in the result? Every structure must be doubled!

The „structure” (order type) of N is ω and not ω*2.

Like the
interval [1, 5] of lengths 4 by doubling gets [1, 5]*2 = [2, 10] of
length 8, the interval (0, ω]*2 gets (0, ω*2] with ω*2 = ω + ω =/= ω
where the whole interval between 0 and ω*2 is evenly filled with even
numbers like the whole interval between 0 and ω is evenly filled with
natural numbers before multiplication. On the ordinal axis the numbers
0, ω, ω*2, ω*3, ... have same distances because same number of ordinals
lie between them.

How do you define subtraction for ordinals?

This means that contrary to the collection of visible
natural numbers ℕ_def which only are relevant in classical mathematics
the whole set ℕ is not closed under multiplication. Some natural numbers
can become transfinite by multiplication.

No, the union of the set called N and those numbers you have added is not
closed. That’s your problem, not that of mathematics.

This resembles the displacement of the interval (0, 1] by one point to
the left-hand side such that the interval [0, 1) is covered.

What is „one point”? Can you give an explicit function?

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.