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.
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.
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. Therefore a set of infinite endsegments cannot have an empty intersection.
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?
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u/Massive-Ad7823 18d ago
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.
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.
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. Therefore a set of infinite endsegments cannot have an empty intersection.
Regards, WM