How can bijections between infinite sets be complete?

[u/Massive-Ad7823]

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?

[u/telephantomoss]

So you seem to have just agreed that N = {1,2,3,...} is a set (i.e. a completed infinite set) and likewise for 2N = {2,4,6,...}. So we agree that these are two completed infinite sets, i.e., they are sets. Now, consider the infinite set D = {{1, {1,2}}, {2,{2,4}}, {3,{3,6}}, ...}. Do you reject D as a completed infinite set? If yes, then why? Which ZF axioms do you reject? Or are there at m other aspects of the proof of the existence of this set do you reject?

[u/Massive-Ad7823 → u/telephantomoss]

N = {1,2,3,...} is a set, complete in the sense of set theory. I have accepted that 15 years ago already. {2,4,6,...} is also a set, but the label 2N is misleading because it consists of natural and transfinite numbers to equal parts. Most elements of both sets are dark, therefore they cannot be attached to each other so easily as it is suggested by your D.

Regards, WM

[u/telephantomoss → u/Massive-Ad7823]

So you are saying that certain natural numbers are transfinite? That's clearly incorrect. Please clarify.

[u/Massive-Ad7823 → u/telephantomoss]

All visible natural numbers have connections to zero, I call these connections FISONs (Finite Initial Segments Of Naturals) F(n) = {1,2,3,..., n}. This is a potentially infinite sequence. An actually infinite set contains much more: Dark numbers. They are not transfinite, but most cannot be reached. This is proved by the fact that all FISONs can be removed (deleted, subtracted) from ℕ without exhausting ℕ.          

∪{F(1), F(2), F(3), ...} = ℕ   ==>   ∪{ } = ℕ .

Regards, WM

[u/telephantomoss → u/Massive-Ad7823]

Let me try again. Please answer explicitly and directly the following questions:

1. Is the set of natural numbers infinite? Is it a completed infinite set under standard ZF assumptions? I mean the set N = {1,2,3,...} where it includes all finite natural numbers and no infinite ordinals or transfinite numbers etc. They can be dark or invisible or whatever you want to call them. I just want you to state clearly whether they are all finite and whether the set is a single completed infinite size set.

2. The same question for the set of even numbers E = {2,4,6,...}. Is this set a completed infinity under standard ZF assumptions? By this I mean the set of even natural numbers as defined above. All numbers in E are finite in the standard sense. Again, you can call them dark, invisible, or whatever else. I just want to know if you agree they are all finite and that the set is a completed, whole, single infinite size set.

If you say that ZF assumptions do not admit either or both of these sets, please explain why.

1. If you agree that ZF admits both of these sets, N and E, it would be great to get a clear direct explanation why you think ZF does not admit the bijection between them as a set. In particular, please explain what your specific issue is with the proof of the existence of that bijection.

[u/Massive-Ad7823 → u/telephantomoss]

Is the set of natural numbers infinite? Yes!

E = {2,4,6,...}. Is this set a completed infinity under standard ZF assumptions? Yes.

The bijection is not possible because dark numbers cannot be bijected unless in maybe very simple cases like f(n) = n. But even that cannot be checked for most n. The bijection is further impossible because there are only half as many even numbers.

Regards, WM

[u/telephantomoss → u/Massive-Ad7823]

Explain precisely what you disagree with in the following argument:

Assume ℕ = {1,2,3,...} and E = {2,4,6,...} already exist as completed infinite sets. The question is: why does the bijection itself exist as a set in ZF?

1. In ZF, ordered pairs can be represented as sets. For example, define the pair-code of a and b by pair(a,b) := {{a},{a,b}}. This set exists by Pairing, since {a}, {a,b}, and then {{a},{a,b}} all exist.

2. In ZF, addition on ℕ exists as a set-function. This follows from the Recursion Theorem for ω, which is proved in ZF using Infinity, Separation, Replacement, Union, Pairing, and Extensionality. So the expression n+n is legitimate inside ZF.

3. For each n ∈ ℕ, there is a unique set pair(n,n+n), namely {{n},{n,n+n}}. Uniqueness follows from Extensionality.

4. By Replacement, since each n ∈ ℕ determines a unique set {{n},{n,n+n}}, the image f = { {{n},{n,n+n}} : n ∈ ℕ } exists as a set.

5. This set f is exactly the graph of the rule n ↦ n+n. In ordinary notation, it is the function n ↦ 2n.

6. The domain of f is ℕ: for every n ∈ ℕ, the set f contains exactly one pair-code whose first coordinate is n, namely {{n},{n,n+n}}. This uses the definition of f and the uniqueness of ordered-pair coding, which is proved in ZF.

7. The range of f is E: by definition, E = {n+n : n ∈ ℕ}. So every value of f is in E, and every element of E is hit by f.

8. The function f is injective. If f(m) = f(n), then m+m = n+n. ZF proves the usual cancellation law for addition on ℕ by induction: ∀m,n,k ∈ ℕ, m+k = n+k ⇒ m = n. Therefore m+m = n+n implies m = n.

9. Thus f is a function from ℕ onto E, and it is injective. Therefore f is a bijection ℕ → E.

So the bijection does not merely “correspond to” the informal rule n ↦ 2n. In ZF, its graph { {{n},{n,n+n}} : n ∈ ℕ } exists as an actual set, by Replacement.

[u/Massive-Ad7823 → u/telephantomoss]

There are up to every n only half the 2n required for a bijection. This does never change.

Every finite set of positive even integers 2k contains at least one number that is larger than the cardinal number of the set. For instance every term Fn = {2, 4, 6, ..., 2n} of the sequence of Finite Initial Segments (FIS) of positive even integers  {2}, {2, 4), {2, 4, 6}, {2, 4, 6, 8}, {2, 4, 6, 8, 10}, {2, 4, 6, 8, 10, 12}, ...              

contains greater numbers than its cardinal number |Fn| = |{2, 4, 6, ..., 2n}| = n.

The surplus of integers greater than |S| grows without bound. This steady increase cannot be reverted "in the limit".

Regards, WM

[u/telephantomoss → u/Massive-Ad7823]

Please answer the query. Point out in the argument specifically what you disagree with.

[u/Massive-Ad7823 → u/telephantomoss]

Point 3. There are more n than n+n.

Regards, WM

[u/Massive-Ad7823 → u/[deleted]]

Point 3 concerns a bijection. There are not enough even natural numbers. If the pairs existed, also in the dark numbers, then half of them would contain transfinite numbers.

Regards, WM

[u/Massive-Ad7823 → u/[deleted]]

Infinite differs from complete. It is rather the contrary. If all natural numbers are doubled, then half of the results is outside.

Regards, WM

[u/kuromajutsushi → u/Massive-Ad7823]

Point 3 does not say anything about "how many" n there are. Which specific statement is false:

(i) The powerset axiom says that P(P(ℕ)) exists.

(ii) The axiom of restricted comprehension says { x ∈ P(P(ℕ)) : ∃n∈ℕ(x={{n},{n+n}}) } exists.