Readers may be aware of one of the principal claims of the moderator of this sub: that the natural numbers contain "dark" natural numbers: that is, not all natural numbers are "visible".

In this post I will demonstrate an inconsistency in the theory of dark natural numbers with respect to the moderator's own definitions.

TL;DR. The moderator holds that there is a last (dark) finite natural number ω − 1. The moderator acknowledges that adding 1 to a finite natural number is still a finite natural number. But (ω − 1) + 1 = ω, which the moderator holds is infinite. Thus the theory of dark natural numbers is inconsistent.

If you're still reading, here's the details. First, a caveat: the moderator views ZFC (and possibly ZF) as inconsistent (p. 120), as well as saying that

ZFC is inconsistent with mathematics

so any argument invoking set theory isn't going to be convincing. (In particular, it should be noted that the von Neumann definition of the ordinals is a model of the Peano axioms for the natural numbers, and shouldn't be confused with those axioms—appealing to their set-theoretic definition doesn't typically hold any weight for the moderator.) Also, for the moderator,

modern mathematics is nonsense

so, instead, I'll use only definitions and statements that the moderator either acknowledges or has outright stated.

(1) Natural number. For the moderator, a natural number is defined by three of the Peano axioms:

1 ∈ M (4.1)
n ∈ M ⇒ (n+1) ∈ M (4.2)
If a set M satisfies (4.1) and (4.2), then ℕ ⊆ M. Of course ℕ has also to satisfy these axioms. 

where we'll take + 1 to indicate the Peano successor operation S(n). (In this formulation 1 is the initial natural number, not 0, but that's not important for the current discussion, and we can accept 1 for that role.) So, 1 is a natural number, and the natural numbers are closed under the operation + 1.

Subsequent queries to the moderator indicate that he accepts that the operation + 1 is injective and that 1 is initial, so that gives us all of the Peano axioms we need.

It should go without saying (since definitions are "if and only if" statements) that if an object doesn't follow the properties, then it's not a natural number.

(2) Visible number. A natural number is visible if

. . . it can be communicated such that sender and receiver understand the same and can link it by a finite initial segment to the origin 0. All other natural numbers are called dark natural numbers.

 Communication can occur

- by direct description in the unary system like ||||||| or as many beeps, flashes, or raps,
- by a finite initial segment of natural numbers (1, 2, 3, 4, 5, 6, 7) called a FISON,
- as n-ary representation, for instance binary 111 or decimal 7,
- by indirect description like "the number of colours of the rainbow",
- by other words known to sender and receiver like "seven"

which is a quote (p. 212). FISON is an acronym for a finite initial segment of natural numbers.

(3) Ordinals and the first infinite ordinal. The moderator defines the ordinals to be

1, 2, 3, ..., ω, ω+1, ω+2, ..., ω2, ω2+1, ... 

where we can take ω2 to mean ω · 2. The moderator says that ω is

the first infinite ordinal

as well as saying that

ω is the limit of the sequence (n)

and that

upon all natural numbers there follows only ω and further transfinite numbers but no natural number

so that ω is not finite.

(4) Last (dark) natural number. The moderator has stated that there is a last (dark) natural number ω – 1:

These dark natural numbers end at ω–1.

as well as saying that

In fact every natural number is finite, even ω-1.

To flesh out the picture slightly, the moderator goes on to say that these dark numbers descend, and that the least element of the set {. . . , ω – 3, ω – 2, ω – 1} is:

It is 1. The sequence is 1 2, 3, ..., n, ...,ω − 3, ω − 2, ω − 1

when I asked him to confirm that the natural numbers were indeed well-ordered.

These definitions are enough to show the inconsistency of the moderator's own reasoning (which he hasn't yet answered in that thread), which I'll summarize:

The moderator acknowledges above that ω − 1 is a (dark) finite natural number, and also that

(ω − 1) + 1 = ω

If ω − 1 is a natural number, since the natural numbers are closed under the + 1 operation by the moderator's definition, then (ω − 1) + 1 = ω is also a natural number, which by definition and the moderator's acknowledgement must be finite. But this contradicts the moderator's assertion that ω is infinite. Because the moderator holds both that ω − 1 is a natural number and that ω is infinite, the theory of dark numbers is inconsistent.

Possible criticism of this argument by the moderator. The moderator may claim that the Peano axioms (which he acknowledges) apply only to visible natural numbers and not to dark natural numbers. But the Peano axioms are what it means to be a natural number: the classification of some natural numbers as dark means that they are still natural numbers, and therefore bound by their definition. If the Peano axioms don't work for dark natural numbers, then they can't be natural numbers.

Cantor's measure of infinite sets has been disproved. See https://www.reddit.com/r/AspectsOfTheInfinite/comments/1tc6v1l/proof_of_the_existence_of_dark_numbers/ and https://www.reddit.com/r/AspectsOfTheInfinite/comments/1tdz9dm/classical_mathematics_contradicts_set_theory/

Not all infinite sets can be compared by size, but we can establish some useful rules.

- The rule of subset proves that every proper subset has fewer elements than its superset. So there are more natural numbers than prime numbers,  and more complex numbers than real numbers. Even finitely many exceptions from the subset-relation are admitted for infinite subsets. Therefore there are more odd numbers than prime numbers.

- The rule of construction yields the number of integers |Z| = 2|N| + 1 and the number of fractions |Q| = 2|N|² + 1 (there are fewer rational numbers). Since all products of rational numbers with an irrational number are irrational, there are many more irrational numbers than rational numbers.

- The rule of symmetry yields precisely the same number of real geometric points in every interval (n, n+1] and with at most a small error same number of odd numbers and of even numbers in every finite interval and in the whole real line.

This theory makes the number of natural numbers (and of course of other sets too) depending on the numerical representation. The set {1, 11, 111, ...} of natural numbers has only comparatively few elements. Therefore the set of natural numbers in unary or binary notation has fewer, in hexadecimal notation more than |N| elements. The set {10, 20, 30, ...} has |N|/10 elements, but if the zeros are only applied as decoration, this set, like {1', 2', 3', ...}, has |N| elements.

If every rational number were equal to |N| fractions, then only |N| rational numbers would exist. This is clearly wrong. The solution of this paradox lies in the fact that small rationals are equal to more fractions than large rationals. 1 = 1/1 = 2/2 = 3/3 = ... has |N| equal fractions,  100 = 100/1 = 200/2 = 300/3 = ... has only |N|/100 equal fractions. All definable rational numbers have ℵ₀ equal fractions, but almost all numbers are undefinable.

There are fewer real numbers of the form 100.1415... than of the form 0.1415... (see above).

It will be a matter of future research to investigate the effect of different numerical systems in detail.

[W. Mückenheim: "Evidence for Dark Numbers", ELIVA Press, Chisinau (2024) pp. 1-36.] https://www.elivapress.com/en/book/book-5647011244/]

Regards, WM

Let X(n) = {1, 2, 3, ..., n} be a finite initial segement of ℕ. For every natural number n: ℕ \ X(n) is nonempty. That means it is impossible to insert all n into the template X(n). Almost all remain outside. How can it be explained that all n can completely be inserted into the template (m, n) of a bijection f(n) = m between the sets M and ℕ?

edderiofer asserted in https://www.reddit.com/r/badmathematics/comments/17luuxl/retired_physics_professor_and_ultrafinitist/?sort=old: However, it seems like OP is […] saying that there is some nonempty set of "natural numbers that can never be defined with any finite amount of time/space". This is of course nonsense by the very definition of natural numbers in PA or ZFC or most other sensible definitions of the naturals.

This shows that edderiofer's claim and the definition in ZFC are bad math.

(The potential infinity of the collection of Peanos's natural numbers however is correct.)

Proof: Between every x ∈ (0, 1] that can be defined and 0 there are infinitely many real numbers, infinitely many of which can never be defined. Among them there are infinitely many unit fractions that can never be defined. The natural numbers belonging to them can also never be defined (otherwise the unit fractions were defined too).

Regards, WM

Contra the claim here, there exist linear orderings where not every nonempty set has a minimal element. The reversed regular ordering on the naturals is such a one; no infinite set of naturals has a minimal element under this ordering and in particular N doesn't. (The regular ordering on the reals is also a linear order that's not a well-order but the naturals are simpler to work with in Isabelle.)

Here is the machine-checked proof. If you paste this into Isabelle (it sadly doesn't have an online client that I know of, but you can download it here), it'll verify it in less than a second.

theory Linorder
  imports Main
begin

(* Isabelle's nats are defined in the standard Peano way: nat ::= 0 | Suc nat. *)

(* We define our own "backwards" ordering of natural numbers. (The proof also works for
   reals under the standard ordering, but the proof with a backwards order on ℕ is 
   much, much simpler and I'm pasting this into reddit, so I'm going with that.) *)

fun bnat_order :: "nat ⇒ nat ⇒ bool" (infix "≽" 60) where
  "x ≽ 0 = True"
| "0 ≽ Suc y = False"
| "Suc x ≽ Suc y = x ≽ y"

(* Some straightforward helper lemmas for later: *)

lemma zero_is_maximal_under_the_bnat_order [dest]: "0 ≽ x ⟹ x = 0"
  by (induction x) simp_all

lemma bnat_order_is_reversed [simp]: "¬ x ≽ Suc x"
  by (induction x) simp_all

(* We now show that this is indeed a linear order: *)

theorem bnat_order_is_reflexive: "x ≽ x"
  by (induction x) simp_all

theorem bnat_order_is_antisymmetric: "x ≽ y ⟹ y ≽ x ⟹ x = y"
  by (induction x y rule: bnat_order.induct) auto

theorem bnat_order_is_transitive: "x ≽ y ⟹ y ≽ z ⟹ x ≽ z"
proof (induction x y arbitrary: z rule: bnat_order.induct)
  case (3 x y)
  thus ?case by (induction z) simp_all
qed auto

theorem bnat_order_is_total: "x ≽ y ∨ y ≽ x"
  by (induction x y rule: bnat_order.induct) simp_all

(* However, despite being a linear order, it is *not* a well-order, and we can construct
   an explicit counterexample (the set ℕ itself, in fact; although Isabelle writes the
   universal set of a given type as "UNIV"). *)

definition minimal_element_prop :: "'a set ⇒ ('a ⇒ 'a ⇒ bool) ⇒ bool" 
    (infix "hasMinimalElementUnder" 60) where 
  "A hasMinimalElementUnder R ≡ ∃x ∈ A. ∀y ∈ A. R x y"

definition well_ordering :: "('a ⇒ 'a ⇒ bool) ⇒ bool" ("_ isAWellOrder") where
  "R isAWellOrder ≡ ∀A. A ≠ {} ⟶ A hasMinimalElementUnder R"

theorem bnat_order_is_not_well_ordered: "¬ ((≽) isAWellOrder)"
(* by (constructive) contradiction *)
proof 
  assume "(≽) isAWellOrder"
  (* unfold the definition of isAWellOrder *)
  hence "∀A. A ≠ {} ⟶ A hasMinimalElementUnder (≽)" by (unfold well_ordering_def)
  (* instantiate the ∀ with ℕ *)
  hence "(UNIV :: nat set) hasMinimalElementUnder (≽)" by blast 
  (* unfold the definition of hasMinimalElementUnder *)
  hence "∃x ∈ UNIV :: nat set. ∀y ∈ UNIV. x ≽ y" by (unfold minimal_element_prop_def) 
  (* Isabelle is strongly typed, so ∃x ∈ UNIV :: 'a set is equivalent to just ∃x :: 'a *)
  hence "∃x :: nat. ∀y. x ≽ y" by blast 
  (* obtain a concrete witness x from the ∃ *)
  then obtain x :: nat where "∀y. x ≽ y" by blast
  (* instantiate the ∀ with Suc x *)
  hence "x ≽ Suc x" by blast
  (* this contradicts the bnat_order_is_reversed lemma from the top *)
  thus False using bnat_order_is_reversed by blast
qed

end

I know two alternatives:

In potential infinity there is nothing next to 1. We can come as close as we like, but we can never close the gap. A gap remains.

In actual infinity, there is a point next to 1. Of course this point cannot be known. It is dark.

Is there a third alternative?

Choose two fractions as close together as you can. Between them there remain infinitely many fractions. Even if you divide the interval by 2 or by 10^10000000 this will never change.

The basic element or atom of the Binary Tree is a node. Each node has two child nodes:

o

/ \

Every sheaf of paths (containing not yet distinguished paths) splits at a node into two distinct sheaves of paths. One sheaf is coming in from above, two sheaves are going out downwards. That results in 2 - 1 = 1 more sheaf per node. The set of nodes in the whole Binary Tree is countable. Therefore the set of distinct sheaves of paths is countable too. By what could uncountably many paths be distinguished? The width of the Binary Tree is too narrow a tunnel to contain and distinguish uncountably many paths.

Here is a proof of a contradiction in set theory:

(1) Cantor's diagonal argument finds for every countable set of reals a real number not in that set.

(2) According to Cantor's definition of countable set the set of nodes of the Binary Tree is countable.

(3) If we map every node onto a path, then the mapped set of paths is countable. More paths cannot exist because paths consist of nodes, but after having mapped them all, there are no further nodes to define further paths. The mapping is a surjection since every path as an individual that cannot be covered by other paths gets a node.*)

(4) For every n ∈ ℕ: map the nth node onto a path containing this node. Choose what path you like. In case of too restricted imagination take the path going always left below the node.

(5) Every node belongs to paths of this set.

(6) From the root to every level L(k) the Binary Tree is completely covered by this set of paths, for every k ∈ ℕ.

(7) These paths represent the real numbers between 0 and 1.

(8) It is impossible to find a further real number between 0 and 1. So these real numbers are countable.

*) It is often claimed that the set of paths S = {RLLL..., RRLLL..., RRRLLL..., ...} can completely cover all nodes of the path RRR... . But that is wrong. All paths having a tail of LLL... cannot cover the always devitating path RRR... . Otherwise also an infinite sum of even numbers could be odd. Further all paths of S can be omitted by induction without changing the result.

The complete infinite Binary Tree (see left-hand figure) has, according to set theory, countably many nodes and uncountably many infinite paths.

But classical mathematics gives different results.

(1) If we look at the upper levels only, then between root node and level n we can distinguish 2ⁿ paths and 2ⁿ⁺¹ - 1 nodes. In the limit there are twice as many nodes as paths.

(2) If we delete the paths (see right-hand figure) but fix three infinite ribbons to every node instead, then every level n is reached by R(n) = 3(2ⁿ - 1) ribbons and P(n) = 2ⁿ paths, i.e., by more ribbons than paths. By the majorant-criterion (as well as by the simple continuity criterion) there cannot be more paths than ribbons in the limit. However R(n) is countable in the limit.

You start with one cent. For a cent you can buy an infinite path of your choice in the Binary Tree. For every node covered by this path you will get a cent. For every cent you can buy another path of your choice. For every node covered by this path (and not yet covered by previously chosen paths) you will get a cent. For every cent you can buy another path. And so on. Since there are only countably many nodes yielding as many cents but uncountably many paths requiring as many cents, the player will get bankrupt before all paths are conquered. If no player gets bankrupt, the number of paths cannot surpass the number of nodes.

If all positive fractions m/n are existing, then they all are contained in the matrix

1/1, 1/2, 1/3, 1/4, ...

2/1, 2/2, 2/3, 2/4, ...

3/1, 3/2, 3/3, 3/4, ...

4/1, 4/2, 4/3, 4/4, ...

5/1, 5/2, 5/3, 5/4, ...

...   .

If all natural numbers k are existing, then they can be used as indices to index the integer fractions m/1 of the first column. Denoting indexed fractions by X and not indexed fractions by O, we obtain the matrix

XOOO...

XOOO...

XOOO...

XOOO...

XOOO...

...

Cantor claimed that all natural numbers k are existing and can be applied to index all positive fractions m/n. They are distributed according to

k = (m + n - 1)(m + n - 2)/2 + m .

The result is a sequence of fractions

1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ... .

This sequence is modelled here in the language of matrices. The indices are taken from their initial positions in the first column and are distributed in the given order.

Index 1 remains at fraction 1/1, the first term of the sequence. The next term, 1/2, is indexed with 2 which is taken from its initial position 2/1 

XXOO...

OOOO...

XOOO...

XOOO...

XOOO...

...

Then index 3 is taken from its initial position 3/1 and is attached to 2/1

XXOO...

XOOO...

OOOO...

XOOO...

XOOO...

...

Then index 4 is taken from its initial position 4/1 and is attached to 1/3

XXXO...

XOOO...

OOOO...

OOOO...

XOOO...

...

Then index 5 is taken from its initial position 5/1 and is attached to 2/2

XXXO...

XXOO...

OOOO...

OOOO...

OOOO...

...

And so on. When finally all exchanges of X and O have been carried out and, according to Cantor, all indices have been issued, it turns out that no fraction without index is visible any longer

XXXX...

XXXX...

XXXX...

XXXX...

XXXX...

...

but by the process of lossless exchange of X and O no O can have left the matrix as long as finite natural numbers are issued as indices. Therefore there are not less fractions without index than at the beginning.

We know that all O and as many fractions without index are remaining, but we cannot find any one. Where are they? The only possible explanation is that they are attached to dark positions.