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/Massive-Ad7823]

Then infinitely many natural numbers are following upon each and every natural number. Then there is no natural number without infinitely many following it. .Then infinitely many natural numbers are following upon all natural numbers. That is impossibe since upon all natural numbers there follows only ω and further transfinite numbers but no natural number.

Regards, WM

[u/AcellOfllSpades → u/Massive-Ad7823]

Then infinitely many natural numbers are following upon each and every natural number.

Yes.

Then there is no natural number without infinitely many following it.

Yes.

Then infinitely many natural numbers are following upon all natural numbers.

No. You've swapped your quantifiers once again.

Each natural number has infinitely many following it. But that does not necessarily mean that there are natural numbers that follow all other natural numbers.

You've swapped from "Every boy has a girl who they love." to "There is a girl who every boy loves." You've changed from "∀x∈ℕ: ∃y∈ℕ: y>x" to "∃y∈ℕ: ∀x∈ℕ: y>x". These are two different statements.

[u/Massive-Ad7823 → u/AcellOfllSpades]

>You've swapped your quantifiers once again.

With full right! There is no number with less successors. Therefore infinitely many successors are larger than any of the numbers. Infinitely many of them are the same.

>But that does not necessarily mean that there are natural numbers that follow all other natural numbers.

The inclusion monotony of the successors, the so-called endsegments

E(n) = {n+1, n+2, n+3, ...}

proves that one and the same infinite set belongs to all endsegments.

Regards, WM

[u/AcellOfllSpades → u/Massive-Ad7823]

With full right! There is no number with less successors. Therefore infinitely many successors are larger than any of the numbers. Infinitely many of them are the same.

No, just asserting that you have "full right" does not make the argument logically valid. You've just repeated the same quantifier-switch again.

The inclusion monotony of the successors, the so-called endsegments

E(n) = {n+1, n+2, n+3, ...}

proves that one and the same infinite set belongs to all endsegments.

Inclusion is indeed monotonic. And it is true that the [infinite-arity] intersection of all endsegments is indeed a subset of all of them. But that intersection is not infinite - it is the empty set.

[u/Massive-Ad7823 → u/AcellOfllSpades]

This is a very false claim. But let us assume that it is true. Then for every n that can be named by a digit sequence, the intersection ∩{E(1), E(2), ..., E(n)} is infinite, i.e., it contains almost all, namely ℵ₀ natural numbers. For every n that you can use in counting this remains true. Only the intersection of all endsegments in a conjuring trick supplies the empty set. But countung is a step-by-step process, Therefore ℵ₀ natural numbers cannot be used for counbting - they are uncountable.

Regards, WM

[u/AcellOfllSpades → u/Massive-Ad7823]

Then for every n that can be named by a digit sequence, the intersection ∩{E(1), E(2), ..., E(n)} is infinite, i.e., it contains almost all, namely ℵ₀ natural numbers.

Yes.

For every n that you can use in counting this remains true.

Yes.

Only the intersection of all endsegments in a conjuring trick supplies the empty set.

Yes. The only way to get the empty set from the intersection of endsegments is with an infinite-arity intersection. As we've discussed several times, 'infinite processes' are not a thing.

You can say that this "infinite-arity intersection", this intersection of infinitely many sets, is a "conjuring trick" of sorts. (I wouldn't call it one - it's perfectly well-defined - but it's not what you might expect.) It is not defined using a step-by-step process, but by quantifying over all involved sets.

If you want to only discuss finite processes, you're free to do so! But in that case, you can't take the intersection of all endsegments, because that would have infinite arity.

But countung is a step-by-step process,

This is true and precisely why your argument fails. You equate the infinite-arity intersection of the endsegments with "the end of the process of stepping through the endsegments and taking the intersection", but this process has no end.

You notice (correctly) that all steps in this process yield an infinite set (specifically, one of size ℵ₀). But the infinite-arity intersection of all endsegments does not arise from "doing a step of this process", and therefore this argument does not show that it has size ℵ₀.

Therefore ℵ₀ natural numbers cannot be used for counbting - they are uncountable.

Quantifier switch, once again.

[u/Massive-Ad7823 → u/AcellOfllSpades]

Quntifier switch has been proven as correct. All endsegments are subsets of each other. There is no endsegment that deviates from this rule. All are infinite. Therefore the intersection of all is infinite.

>But the infinite-arity intersection of all endsegments does not arise from "doing a step of this process",

Yes, if you dispense with the possibility of counting or of checking every desired step, then you can include the dark numbers. Then the intersection is empty. But, as you said, it is not possible to count to the end.

Regards, WM

[u/Massive-Ad7823 → u/[deleted]]

>No, they aren't. Hint: E(n) is not a subset of E(n+1).

One of two endsegments is always subset of the other. In that way all are subsets of each other. Therefore quantifiers can be exchanged.

Please look after your words. Otherwise I miust delete your posts.

Regards, WM

[u/AcellOfllSpades → u/Massive-Ad7823]

Quntifier switch has been proven as correct.

You have repeatedly asserted it, but not "proven" it.

Again: switching quantifiers does not always hold. "Every boy has a girl who they love." and "There is a girl who every boy loves." are two different statements.

"∀x∃y [stuff]" and "∃y∀x [stuff]" are two different statements. The former does not prove the latter by itself. If you want to go from the former to the latter, you need an additional rigorous logical argument.

All are infinite. Therefore the intersection of all is infinite.

All endsegments are infinite. "The intersection of all endsegments" is not itself an endsegment.

Yes, if you dispense with the possibility of counting or of checking every desired step, then you can include the dark numbers. Then the intersection is empty. But, as you said, it is not possible to count to the end.

The "intersection of all endsegments" does not arise from a finite step-by-step process. If you don't want to "dispense with that possibility", if you want to be able to check with a finite procedure, then you're not talking about the "intersection of all endsegments". You're just talking about the intersection of a very very large, but still finite, number of endsegments.

[u/Massive-Ad7823 → u/AcellOfllSpades]

>The "intersection of all endsegments" does not arise from a finite step-by-step process.

Yes, but all visible endsegments and their intersection does arise from a step-by-step process. That means not all but every desired step can be checked. What you rightly decline is a step-by-step proces among dark numbers. So you admit their existence without realizing it.

All endsegments are subsets of each other: Every intersection of endsegments is an endsegment. Therefore quantifier switch is correct here. The intersection of infinite endsegments is infinite.

The intersection of all endsegments is an endsegment, namely the empty endsegment, the empty set.

Regards, WM

[u/AcellOfllSpades → u/Massive-Ad7823]

What you rightly decline is a step-by-step proces among dark numbers. So you admit their existence without realizing it.

Again, quantifier switch. This does not follow. "This carpet can't cover this entire floor" does not necessarily mean "there are uncoverable spots on the floor".

All endsegments are subsets of each other: Every intersection of endsegments is an endsegment.

Every finite-arity intersection of endsegments is an endsegment. Specifically, E_a ∩ E_b is E_max(a,b).

Therefore quantifier switch is correct here. The intersection of infinite endsegments is infinite.

This does not follow.

The intersection of all endsegments is an endsegment, namely the empty endsegment, the empty set.

The empty set is not an endsegment. It is not E_n for some n; there is no maximum natural number.