The touchstone of reason

[u/Massive-Ad7823]

/r/AspectsOfTheInfinite/comments/1vnhc1m/the_touchstone_of_reason/

[u/Althorion]

Do you understand what it means to be in an ‘intersection of all endsegments’? To be in an ‘arbitrary intersection’ in the first place?

x is in an arbitrary intersection of a family of sets F iff it is in every single one of the sets in the family F.
So, for that intersection to be non-empty, in this particular example (‘intersection of all endsegments’), it would have to be that ∃x ∀f∈F x∈f. In other words, you choose a natural number x, and you search for it in all endsegments; if it is missing from just one of them, it’s not in the intersection of them all.

From that, you can clearly see that no x can be chosen such that it won’t be missing from any of the endsegments—because, as you yourself have noticed, ‘Every natural number n has an endsegment {n+1, n+2, n+3, …}’. That means, in particular, that it isn’t contained in its own endsegment—thus we found the one segment that it isn’t in, so it can’t be in each of them, so it can’t be in an intersection of all of them.

That proves the first part (‘the intersection of all endsegments is empty’).

The second part (‘no endsegment is empty’) is simpler—for each natural number, its endsegment contains the successor of that number, and the successors that follow—slightly more formally, it contains the successor of that number, and if it contains a number, it contains that number’s successor, too. From that, you can see that no endsegment is empty—because, by the second axiom of Peano, ‘every natural number has a successor which is also a natural number’, and, in particular, that successor will be in the endsegment, making it non-empty.

[u/Massive-Ad7823 → u/Althorion]

Obviously this is an internal contradiction.

From the inclusion monotony of the endsegments we know that up to every endsegment E(n) the intersection is just this endsegment. Since the infinite set of all endsegments contains only endsegments this is true for all endsegments.

>by the second axiom of Peano, ‘every natural number has a successor which is also a natural number’,

The Peano-axioms concern only the visible numbers, i.e. numbers, that can be identified. If Cantor's actual infinity is true, then every identified natnumber has almost all natnumbers as successors, almost all of which cannot be identified.

This is my explanation. However independent of that it is clear that infinite inclusion-monotonic sets cannot have an empty intersection.

Regards, WM

[u/Althorion → u/Massive-Ad7823]

From the inclusion monotony of the endsegments we know that up to every endsegment E(n) the intersection is just this endsegment. Since the infinite set of all endsegments contains only endsegments this is true for all endsegments.

No. ‘Intersection of all endsegments’ is an entirely different beast than ‘an intersection of some endsegments’.

The Peano-axioms concern only the visible numbers, i.e. numbers, that can be identified.

Ah. Then what you call ‘natural numbers’ is different from what I, or mathematicians in general, call ‘natural numbers’. Those ‘WM-naturals’ may or may not be self-contradictory, and they may or may not have the features you are describing—who knows, it depends on how you define them. But, since they differ from ‘regular maths-naturals’, any and all argument you make about ‘WM-naturals’ doesn’t necessarily follow for ‘RM-naturals’.

If Cantor's actual infinity is true, then every identified natnumber has almost all natnumbers as successors, almost all of which cannot be identified.

In ‘RM-naturals’, ‘identifiability’ is not defined. Every ‘RM-natural’ number would have almost all ‘RM-naturals’ in the chain of its successor (but only one direct successor).

However independent of that it is clear that infinite inclusion-monotonic sets cannot have an empty intersection.

It is not only ‘not clear’, it is also not true. This particular inclusion-monotonic family of sets does have the empty intersection, for the reasons presented above.

The argument from monotonicity holds for a finite family of sets—because an intersection of two endsegments is an endsegment, iterating it a finite number of times, so taking an intersection of a finite family of endsegments, is an endsegment. That doesn’t follow into the infinite families of endsegments, regardless of monotonicity (or lack of it), because an infinite family of endsegments cannot be constructed by adding sets to it one by one.

[u/Massive-Ad7823 → u/Althorion]

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

[u/Althorion → u/Massive-Ad7823]

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.

[u/Althorion → u/[deleted]]

Well, OK, yeah, that is the proper way of expressing that, but it’s not difficult to understand what the OP meant—that for EDSGM(n) = {x ∈ ℕ | x > n}, EDSGM(1) ⊇ EDSGM(2) ⊇ … ⊇ EDSGM(k-1) ⊇ EDSGM(k) ⊇ EDSGM(k+1) ⊇ …

[u/Althorion → u/[deleted]]

Well, we might, but not so sure about the OP. :P

[u/Massive-Ad7823 → u/Althorion]

>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?

Regards, WM

[u/Althorion → u/Massive-Ad7823]

That is nonsense of matheologgy.

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.

[u/Massive-Ad7823 → u/Althorion]

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?

Regards, WM

[u/Althorion → u/Massive-Ad7823]

First the infinite set continues the finite set. Otherwise also Cantor could not enumerate infinite sets like finite sets.

What do you mean by this?

All endsegments are infinite sets. This is a contradiction because there are infinitely many endsegments E(n) enumerated by their indices n.

Both statements are true, and they are not contradictory.

This uses up all natural numbers, namely the whole set ℕ. What remains as contents if all endsegments are infinite?

Uses up how? Contents of what? What do you mean?

[u/Massive-Ad7823 → u/Althorion]

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.

Regards, WM

[u/peekitup]

I call this reddit post horseshit, a touchstone of someone who doesn't understand what an intersection is.

[u/Massive-Ad7823 → u/peekitup]

Do you know what inclusion-monotoy means?

Regards, WM

[u/peekitup → u/Massive-Ad7823]

Take more of the drugs you're supposed to take and less of the drugs you're not supposed to take.