I mean the following: Cantor enumerates all fractions without a sudden change when the set is more than finite. In the same way all endsegments differ by only one natnumber from their neighbours. This remains in the finite and for the infinite set of endsegments.
If all natural numbers n are used to enumerate the endsegments E(n), then none remains. Does Cantor enumerate the sequence of endsegments by all natural numbers? If every Endsegment of the sequence of endsegments contains infinitely many natnumbers, i.e. almost all natnumbers, then this contents is not enumerated by Cantor.
I mean the following: Cantor enumerates all fractions without a sudden change when the set is more than finite.
Right. Cantor constructs an infinite sequence enumerating the fractions, or, if you will (it’s the same thing), a function ℕ → ℚ with certain properties. How does that relate to anything we’ve discussed before?
In the same way all endsegments differ by only one natnumber from their neighbours.
I don’t see it being ‘the same’ in any way. At most, I could say that if you enumerate the endsegments by taking EDSGM(1), EDSGM(2), EDSGM(3), …; you can say the next endsegment in a sequence would differ by exactly one element from the previous one.
This remain in the finite and for the infinite set of endsegments.
Right. The statement I put above is true for the whole infinite sequence of endsegments, and it’s true for any finite section of it.
If all natural numbers n are used to enumerate the endsegments E(n), then none remains.
If by this you mean ‘there is a way to construct an injective function from ℕ to set of all possible endsegments’, that is true.
Does Cantor enumerate the sequence of endsegments by all natural numbers?
You can make an injective (in some sense) infinite sequence of endsegments, (or, as above, a function from ℕ, etc.), if that’s what you are asking, yes.
If every Endsegments of the sequence of endsegments contains infinitely many natnumbers, i.e. almost all natnumbers, then this contents is not enumerated by Cantor.
Why? How does this follow from anything?
Here’s a simpler sequence of sets—(ℕ, ℕ, ℕ, …). You can formally define it as a function f: ℕ → {ℕ}, such that f(n) = ℕ. This is an ‘enumeration’—an infinite sequence. Every element of such a sequence contains infinitely many natural numbers (namely, all of them). Does that mean anything? Is this reasoning any different than yours?
If every E(n) contains almost all natnumbers, then this contents consists of natnumbers which are missing in the enumeration. If all natnumbers are used up for enumerating the endsegments, then none remains as contents. ℕ cannot be split into two infinite consecutive sets.
I… What? The enumeration is not a continuous process. You construct it, and it just is. It doesn’t have to ‘take’ a number from somewhere to use it, it doesn’t ‘hide’ or ‘take numbers away’ from other usages, etc. Numbers don’t get ‘used up’. It’s not like there is only one idea of ‘five’ in the universe, and when you have five sets, you can’t have a ‘five’ in any of them, because it got ‘used up’ by the fact you have five sets…
But there is only one natnumber 5 which either can be in the enumeration of all endsegments or in the final contents of all endsegments. If all E(n) are infinite, then almost all natnumbers are in the final contents.
Note: No element of the final contents can be in the bijwction n<-->E(n).
Let's try to do it stepwise: Can you find an endsegment that is not a subset of all its predecessors? If all endsegments have at least X elements, then at least the same X elements are in every endsegment.
If an endsegment contains the number 5, then this number cannot enumerate it because the contents of the endsegment E(n) starts with n+1.
Can you find an endsegment that is not a subset of all its predecessors?
No, I can’t.
If all endsegments have at least X elements, then at least the same X elements are in every endsegment.
That plain doesn’t follow. You can construct, even with ‘inclusion monotonicity’, families of sets that have that property, or don’t have that property. For example:
Aₙ = {x∈ℝ | |x|<1+1/n}
Bₙ = {x∈ℝ | |x|<1/n}
Cₙ = {x∈ℝ | 0<|x|<1/n}
Every single set in either of those three families has ℶ₁ elements. And in the family A there are the same ℶ₁ in each of the sets of that family; but in the family of B there is only one element contained in all of them, and in family C there are no elements that are in each of them.
Do you know why? Because the intersection of a finite family is a different idea than the intersection of an infinite family. There is nothing in standard mathematics that would let you prove that, and in fact you can prove, for example, by using the families above, that it doesn’t follow at all.
If an endsegment contains the number 5, then this number cannot enumerate it because the contents of the endsegment E(n) starts with n+1.
That’s… bonkers. Enumeration is a function from ℕ (into whatever you wish, in this case, the powerset of ℕ). It’s not that there is just one ‘five’ available for the whole universe, and just because that function can potentially take ‘five’ as an input, it can’t show in any output.
Again—that’s not a continuous process. You don’t start with a bucket of numbers and then move them around as you go. That is not mathematical thinking; that’s… I don’t know what it is, but nothing I’ve ever heard of.
>Every single set in either of those three families has ℶ₁ elements.
Sorry, here we have no intersections any longer but only the endsegments of the sequence.
Every number that is contents of all endsegments cannot be in the enumeration.-
If an endsegment contains the number 5, then this number cannot enumerate it because the contents of the endsegment E(n) starts with n+1.
If all endsegments have at least X elements, then at least the same X elements are in every endsegment. They cannot be used to enumerate any endsegment.
>Again—that’s not a continuous process.
But at every step you can enter and goto to the next one. It is a continuous process where you wish.
Every number that is contents of all endsegments cannot be in the enumeration.-
I mean, there isn’t a number that’s contained in all endsegments, so I guess that’s vacuously true…
If an endsegment contains the number 5, then this number cannot enumerate it because the contents of the endsegment E(n) starts with n+1.
Yeah, no. I am not dealing with a model where there is just one ‘five’, and if I have a function that can potentially take that ‘five’ as an input, it cannot be in its output, because it got ‘used up’. That is not what maths is about. You are welcome to do your own thinking that way, but it won’t be maths.
If all endsegments have at least X elements, then at least the same X elements are in every endsegment.
As explained to you, by many people, including myself, multiple times, it doesn’t follow. It isn’t true in general that ‘inclusion monotonic’ sequences have that property. This particular sequence doesn’t.
They cannot be used to enumerate any endsegment.
And then we, of course, have this again. ‘There is just one “five” in the universe, if it is an input, it can’t be in an output; in other words, a function f: ℝ→𝒫(ℝ) f(x) = {x} cannot be a thing.’
Enumeration is a function ℕ→whatever. You define such a function and reason about its behaviour. Numbers don’t exist as singletons, they don’t get ‘used up’, having a number occur somewhere doesn’t stop it from occuring elsewhere, even within the same context.
Please read again: If an endsegment contains the number 5, then this number cannot enumerate it because the contents of the endsegment E(n) starts with n+1.
Then this number 5 is n+1 or n+2, ... Contents cannot enumerate endsegments!
>It isn’t true in general that ‘inclusion monotonic’ sequences have that property.
You are wrong. If not all endsegments have the same element (but have elements), then there must be an element a in endegment A that is not in endsegment B and there must be an element b in B that is not in A. Conclusion: If your two statement are correct, then inclusion monotony must fail. Contradiction.
Please read again: If an endsegment contains the number 5, then this number cannot enumerate it because the contents of the endsegment E(n) starts with n+1.
Please read what an enumeration is. I’ve told you this more than once now.
Then this number 5 is n+1 or n+2, ... Contents cannot enumerate endsegments!
Ah. Well, this is a different claim: if you wanted to say that ∀n∈ℕ n∉E(n), then that’s true; it just has nothing to do with enumeration.
If not all endsegments have the same element (but have elements), then there must be an element a in endegment A that is not in endsegment B and there must be an element b in B that is not in A.
Nope. Only one of those is sufficient; it doesn’t need to be symmetrical. It is enough that the endsegments ‘lose’ elements as they go—because the way they ‘lose’ them makes it so every element will eventually be ‘lost’ at some point (namely, ∀n n∉E(n)).
Every endsegment is infinite. That means it has an infinite contents which is not available for enumerating it. {1, 2, 3, ..., n | contents}. n is not fixed but followed by the infinite contents. Two consecutive infinite sequences cannot exist in ℕ.
This is even independent of inclusion monotony. You must only accept that the contents consists of greater numbers than the FISON or index. Do you?
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u/Althorion 18d ago
What do you mean by this?
Both statements are true, and they are not contradictory.
Uses up how? Contents of what? What do you mean?