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

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. 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, ..., 2ω, 2ω+1, 2ω+2, 2ω+3, ..., 3ω, ... . In 1884 he exchanged the positions of multiplier and the number to be multiplied with the result
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. 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. Therefore the doubled natural numbers cover twice as many space than before. What is the result of doubling? Either all doubled numbers are natural numbers, then not all natural numbers have been doubled. 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.
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? What size has the interval between 2ℕ and ω*2? The natural answer is (0, ω]*2 = (0, ω*2] with ω or ω+1 amidst. 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. 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! 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.  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.
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. Of course these natural numbers are dark like every result of ω/k with k ∈ ℕ_def, for instance ω/2 or ω/10^10^100. It is with certainty excluded to reduce positive real numbers by division to negative numbers. This impossibility might be taken as an argument that it is also impossible to produce transfinite numbers by multiplying dark natural numbers. This argument would be tantamount to denying actual infinity at all.
Regards, WM