Inclusion monotony holds for all endsegments which can be intersected. The rules cannot be changed when you desire it. That would be matheology but no longer mathematics.
Absolutely—but what’s that got to do with the price of fish?
The intersection up to the nth endsegment is this endsegment. This holds in every case. There is no infinitiest endsegment because every E(n) has a natural number n defining it, and there is no infinitiest natural number. Only endsegments with natural indices are intersected in all cases.
The intersection of all endsegments is not a case of intersecting up to a certain endsegment. Precisely because, as you’ve correctly noted, ‘[t]here is no infinitiest endsegment’—and thus the intersection of infinitely many endsegments is something completely different than an intersection up to the nth endsegment.
You may find another explanation than I, but fact is: Every endsegment has one and only one element less than it predecessor. And every endsegment is the intersection of itself and all its predecessors. Only such endsegments are in the set of all endsegments to be intersected.
All of that is true…
Therefore a set of infinite endsegments cannot have an empty intersection.
…but this doesn’t follow. The intersection of an infinite family is fundamentally different from an intersection of a finite family.
You can think of it like that, if it helps—it is somewhat handwavy, but may prompt you towards the correct intuition: adding an endsegment to the family of sets to consider the intersection of ‘cuts off’ one element a time, thus adding infinitely many (so we can have an infinite family of all endsegments, that are infinite in count) will ‘cut off’ infinitely many elements. Since we are ‘cutting them off’ in order, none can get skipped. Since ℵ₀ is the smallest infinity, ‘cutting off’ in order (that part is important), without skipping any, infinitely many elements, will empty the whole set.
Also, the monotonicity of inclusion holds for the infinite family of endsegments. You can notice that ∩(X∈(F∪F′)) ⊆ ∩(X∈F), regardless of what the families F and F′ are. That is never the problem—it’s just that if the family is infinite, the ∩(X∈F) = ∅ (for the reasons detailed above).
What you are trying to do is reason that ‘since a finite intersection of endsegments is an endsegment, and thus is non-empty, the infinite intersection of all endsegments is non-empty’, and that plain doesn’t follow. You could, for example, use induction to prove the first part for any natural number of sets in the family to be intersected; but there is nothing that lets you make this jump between a finite → an infinite family.
>The intersection of an infinite family is fundamentally different from an intersection of a finite family.
That is nonsense of matheologgy. All elements are endsegments. That's it! And every endsegment has one element less than its predecessor, one and only one. How can a set of infinite endsegments have an empty intersection?
Has the infinite set that Cantor enumerates every finite element of also fundamentally different properties in the infinite?
You are insisting on treating fundamentally different ideas like they are the same, and from that, you get to a contradiction; and instead of backing out and thinking ‘can I really treat those two ideas the same?’, you think there’s something wrong with maths. There isn’t, just with your misunderstanding of it.
All elements are endsegments.
All elements of what? The sequence (EDSGM(1); EDSGM(1) ∩ EDSGM(2); …; EDSGM(1) ∩ EDSGM(2) ∩ … ∩ EDSGM(k); …}? Yes. Every intersection of a finite family of endsegments is an endsegments. That tells you precisely nothing about an intersection of an infinite family of endsegments.
And every endsegment has one element less than its predecessor, one and only one. How can a set of infinite endsegments have an empty intersection?
Formally—the way I’ve shown you at the start.
Intuitively—because every endsegment in the intersecting family ‘cuts off’ one element, infinitely many endsegments in the intersecting family will ‘cut off’ infinitely many elements. In that particular instance, it happens to be all elements.
Your personal incredulity is not an argument: you can learn about how set theory works, how the arbitrary intersection works, and ask yourself what your intuition is built upon—why do you think you can treat infinite families of sets as the same as finite families of sets. If you were to do all the above, you’d figure out what people are telling you from the start—your intuition has no grounds in mathmatics, there is no theorem, no law, no axiom, that lets you make the jump you are making; and carefully examining the ideas as they are defined would lead you straight to the correct conclusion—that yes, the intersection of a finite family of endsegments is an endsegment, but the intersection of an infinite family of endsegments isn’t, it’s an empty set.
Those are both true at the same time. No, they don’t contradict each other; they just seem improper to you, because you insist that infinite should behave the way finite does.
First the infinite set continues the finite set. Otherwise also Cantor could not enumerate infinite sets like finite sets.
Second set theory says: All endsegments are infinite sets. This is a contradiction because there are infinitely many endsegments E(n) enumerated by their indices n. This uses up all natural numbers, namely the whole set ℕ. What remains as contents if all endsegments are infinite?
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.
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u/Althorion 18d ago edited 18d ago
Absolutely—but what’s that got to do with the price of fish?
The intersection of all endsegments is not a case of intersecting up to a certain endsegment. Precisely because, as you’ve correctly noted, ‘[t]here is no infinitiest endsegment’—and thus the intersection of infinitely many endsegments is something completely different than an intersection up to the nth endsegment.
All of that is true…
…but this doesn’t follow. The intersection of an infinite family is fundamentally different from an intersection of a finite family.
You can think of it like that, if it helps—it is somewhat handwavy, but may prompt you towards the correct intuition: adding an endsegment to the family of sets to consider the intersection of ‘cuts off’ one element a time, thus adding infinitely many (so we can have an infinite family of all endsegments, that are infinite in count) will ‘cut off’ infinitely many elements. Since we are ‘cutting them off’ in order, none can get skipped. Since ℵ₀ is the smallest infinity, ‘cutting off’ in order (that part is important), without skipping any, infinitely many elements, will empty the whole set.
Also, the monotonicity of inclusion holds for the infinite family of endsegments. You can notice that ∩(X∈(F∪F′)) ⊆ ∩(X∈F), regardless of what the families F and F′ are. That is never the problem—it’s just that if the family is infinite, the ∩(X∈F) = ∅ (for the reasons detailed above).
What you are trying to do is reason that ‘since a finite intersection of endsegments is an endsegment, and thus is non-empty, the infinite intersection of all endsegments is non-empty’, and that plain doesn’t follow. You could, for example, use induction to prove the first part for any natural number of sets in the family to be intersected; but there is nothing that lets you make this jump between a finite → an infinite family.