Wikipedia:Reference desk/Archives/Mathematics/2022 December 11

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December 11

What did Einstein think about Srinivasa Ramanujan?

Googling Einstein and Ramanujan didn't return anything. In any event, 1 year before Einstein died, he was asked who he thought was the greatest thinkers. He gave 2 answers. He said Josiah Gibbs, but also said he never met Gibbs before. For people that Einstein has met, then he said it was Lorentz. But I'm trying to find if Einstein said anything about Ramanujan. Also, there's this famous quote that said Einstein's difficulties in math were great. What math was his difficulties, I don't imagine it was Calc I-III? And differential equations? I also asked a similar math scenario question in the Humanities desk. 67.165.185.178 (talk) 14:30, 11 December 2022 (UTC)

There is no reason (that I can think of) to expect that Einstein was particularly interested in Ramanujan's work or even aware of it, other than by reading about Ramanujan in newspaper articles. Ramanujan's results had no applicability to any of Einstein's work. I think one should not read more in Einstein's quip than that he was trying to be kind to an admirer who confessed to be struggling with mathematics – one would assume in high school. Einstein's development of the theory of General relativity, expressed in the Einstein field equations, required the mastery of the topic of pseudo-Riemannian manifolds, which is some serious stuff. The four papers that propelled him to fame in his annus mirabilis also show considerable mathematical prowess.  --Lambiam 15:18, 11 December 2022 (UTC)

Einstein was not himself a mathematician; much of his work actually required people other than himself to invent the mathematics he later used, and the relationship between Einstein and various mathematicians and physicists is complex. I'm saying this not to diminish Einstein, but rather to poke holes in the Great Man Theory of scientific progress; the notion that Einstein worked by himself inventing all of modern physics is just foolish. Einstein had better PR than other people, but he's clearly indebted to people he worked alongside. People like Hendrik Lorentz, Bernhard Riemann, David Hilbert, Hermann Minkowski, Henri Poincaré, Emmy Noether, Élie Cartan, etc. etc. all did work alongside, or ahead of, Einstein in order to develop things like the two theories of relativity. Einstein's work was mainly in taking these mathematical tools and using them to create a consistent theoretical framework to make physics more accurate at high energy/mass situations. Marcel Grossmann in particular was highly important in teaching Einstein the math he would need to develop General relativity. Which is not to say that the mathematics was beyond Einstein; he clearly was able to learn it and apply it to solving the problems of relativity. He just didn't invent the mathematics, others did so and he learned it as he needed it. As noted, Einstein wasn't particularly interested in the kinds of mathematics that Ramanujan was doing. Most of Einstein's work required non-Euclidian geometry and the tools to work in non-Euclidian spacetimes, Ramanujan worked mainly in number theory, which held little use for Einstein. Einstein's admiration of Lorentz was in developing the Lorentz transform, an invaluable mathematical tool for being able to describe motion in a world that has a fixed speed limit like light speed. His admiration of Gibbs is undoubtedly the fact that Gibbs was of being one of the first to develop rigorous and complex mathematical theories to explain physical phenomena, while in a different field (thermodynamics rather than Einstein's field of kinematics), was still of the same sort of work that Einstein was involved in. --Jayron32 14:51, 12 December 2022 (UTC)

Don't forget Tullio Levi-Civita and Elwin Christoffel, probably more important to the early development of GR than Hilbert, Poincaré, Noether, or Cartan. The last two in particular seem to have made the formulation more elegant after the theory already existed, rather than contributing to its initial version. --Trovatore (talk) 08:01, 14 December 2022 (UTC)

Indeed. My list was meant to be representative and not exhaustive. Given the literal hundreds of people who have contributed to the development of GR, in terms of laying the groundwork, in terms of helping develop the original theory, and in terms of extending and refining the theory in the years since 1915, your two certainly atleast as as important as any of them. --Jayron32 14:35, 16 December 2022 (UTC)

I am not sure if this anecdote is true or apocryphal, but here it goes. Some random visitor challenges Einstein to calculate the square root of some large number. Einstein replies, "that is why they publish square root tables". Cullen328 (talk) 09:41, 13 December 2022 (UTC)

As he should; arithmetic is not equal to mathematics, and the ability to do quick arithmetic has nothing to do with mathematical skill. --Jayron32 18:33, 13 December 2022 (UTC)

Arithmetic is not equal to mathematics, but surely it is one of the basic building blocks required for learning other mathematical concepts, advancing from there to geometry, trigonometry, algebra and calculus. Cullen328 (talk) 07:35, 14 December 2022 (UTC)

To an extent. It's important to have enough facility with basic calculation that you have a "feel" for numbers. But there's not a huge correlation between mathematical achievement and the ability to do prodigious feats of mental arithmetic. --Trovatore (talk) 08:05, 14 December 2022 (UTC)

Per Trovatore, sort of. I mean, presumably anyone at any level of mathematics is familiar with the basic operations of arithmetic, including radicals, but it's one of those things that people unfamiliar with mathematics think mathematics is which is of course a misconception; that someone really skilled in math means they do really complex arithmetic. I know very few mathematicians that do anything of the sort; as a modern Einstein would say, I have a calculator for that. It is no statement on someone's skill as a mathematician that they can't be bothered to do trivial arithmetic problems. --Jayron32 14:33, 16 December 2022 (UTC)

Is the following sequence of points finite?

Let there be a sequence of geometric points (different from each other), on a line. For the time being, let's avoid any pre-assumption about whether the sequence is finite, or infinite, or even transfinite. We may, though, change our mind later.

A. The sequence contains a first point, whereas every other point in the sequence is a (unique) successor of a (unique) point in the sequence.

B. The sequence contains a last point, whereas every other point in the sequence is a (unique) predecessor of a (unique) point in the sequence.

C. There is a ball, that can reach a given point other than the first one, if and only if the ball jumps to the given point from its predecessor the ball has already reached, and only if it has. In other words: From every point that the ball has reached, the ball jumps to the successor, and only this way does the ball manage to reach this successor, whereas the ball is not allowed to skip any points on its way from a given point to another.

D. We let the ball reach the first point (e.g. by placing the ball on the first point). The ball starts jumping according to #C, and reaches the last point.

Is there any infinite model, of such a sequence of points, satisfying A-D?

Please notice, that if the sequence is the infinite ordered set $\omega +{\overline {\omega }},$ then it does not satisfy #D. 147.236.144.145 (talk) 15:54, 11 December 2022 (UTC)

To talk about a successor relation, we first need an order on the geometric line. So I assume that we treat the line as an affine image of the real number line.

A and B together already imply finiteness here: Assume not and define a map $n\rightarrow a_{n}$ from the natural numbers to the sequence by induction: Map 0 to the first point and map n + 1 to the successor of point n.

If the sequence is infinite, this is well-defined. Note that no $a_{n}$ is the last element. Therefore there is an element $b$ of the sequence which is greater than all $a_{n}$.

So let $a$ be the smallest upper bound of $\{a_{n}\}$ among the sequence elements.

By A, $a$ has a predecessor $a'$ in the sequence.

By definition, $a'$ is smaller than, say, $a_{N}$. But then $a_{N+1}>a_{N}\geq a$, contradicting the fact that $a$ is a smallest upper bound of $\{a_{n}\}$ among the sequence elements. Felix QW (talk) 16:24, 11 December 2022 (UTC)

#A and #B do not imply finiteness. Check the infinite ordered set $\omega +{\overline {\omega }}.$ It satisfies #A and #B. However, it does not satisfy #D. 147.236.144.145 (talk) 17:08, 11 December 2022 (UTC)

Could you remind us what ${\bar {\omega }}$ stands for? Felix QW (talk) 17:14, 11 December 2022 (UTC)

Take the elements of $\omega$ but reverse the order. Tagging the elements with an overline to avoid confusion with those of $\omega$, we have

$\cdots <{\overline {2}}<{\overline {1}}<{\overline {0}}.$

For $\omega +{\overline {\omega }}$ we can then write

$0<1<2<\cdots <{\overline {2}}<{\overline {1}}<{\overline {0}}.$

As to A, this sequence has a first element $0$. Every other element is either of the form $n{+}1,$ with predecessor $n,$ or of the form ${\overline {n}},$ with predecessor ${\overline {n{+}1}}.$ For B we have the mirror situation.  --Lambiam 17:40, 11 December 2022 (UTC)

Thank you very much for clearing this up! So my argument is nonsense, since the subset of sequence elements is not necessarily complete. Never mind me... Felix QW (talk) 17:46, 11 December 2022 (UTC)

It means something like "minus omega". But let's forget it, and have a look at the following infinite model:

1. The distance from the first point to the last one is four meters.
2. The distance from the first point to the second one is one meter, and the distance from the second point to the third one is half a meter, and the distance from the third point to the fourth one is a quarter of a meter, and so forth. That is, for every finite natural n, the distance from the n-th point to its succesoor is $2^{1-n}$ meter.
3. The distance from the last point to the penultimate one is one meter, and the distance from the penultimate point to the antepenultimate one is half a meter, and the distance from the antepenultimate point to its predecessor is a quarter of a meter, and so forth, analogously to what I've described in the previous section.

In this infinite model, when the ball starts jumping from the first point, it will never (even after an infinite period of time) manage to pass even just two meters (but rather necessarily less), and hence it will never manage to reach the last point, which is (as mentioned in section 1) four meters away from the first point, contrary to #D, that states that the ball does manage to reach the last point.

To sum up: this infinite model satisfies #A and #B, yet not #D. 147.236.144.145 (talk) 17:38, 11 December 2022 (UTC)

Confessions of Zeno.  --Lambiam 02:13, 12 December 2022 (UTC)

I didn't read it. How is it related to our topic? 147.236.144.145 (talk) 12:09, 12 December 2022 (UTC)

Only through its title; see Zeno's paradox. (I can recommend reading it; it is among my most favoured books.)  --Lambiam 13:47, 12 December 2022 (UTC)

(edit conflict) I'm not sure that the question is well-posed. Consider the set {0, 1/2, 2/3, 3/4, ... , 1} where 0 is the first point, 1 is the last point, and the successor of (n-1)/n is n/(n+1). You don't say that every point must have a predecessor, and in fact 1 does not have one. But you do say that the only way to reach the last point is with successive jumps. Are you saying that there must be a way to reach the last point with successive jumps or that if there is a way to reach the last point then it can only be with successive jumps? The example satisfies the second interpretation but not the first. The example is infinite so that shows in the second interpretation that there are infinite examples. In the first interpretation there is a sequence of jumps with a first and last element, and such a sequence must be finite, which would imply the that set must be finite. It seems like your setup is basically an equivalent version of the properties of well-ordered sets, or ordinal numbers if you prefer. Any ordinal < ω is finite by definition, but there are ordinals > ω with a terminal element, in particular any ordinal of the form u+1 where u is an infinite ordinal. --RDBury (talk) 16:40, 11 December 2022 (UTC)

Doesn't A say precisely that every point must have a predecessor? Felix QW (talk) 16:46, 11 December 2022 (UTC)

The sequence of natural numbers contains a first number, whereas every other number in the sequence is a successor of some number in the sequence. It does not follow that every natural number has a predecessor.  --Lambiam 17:21, 11 December 2022 (UTC)

Of course, I meant every element but the first. It was a response to RDBury (talk · contribs)'s comment that this axiomatises exactly the well-orders, as those do not have this property. Felix QW (talk) 17:25, 11 December 2022 (UTC)

@Felix QW: Yes, I missed that, so my example, essentially ω+1, doesn't work. I think the $\omega +{\overline {\omega }},$ raises the same issue though. If by 'only' you mean exactly one then the set must be finite, but if you mean at most one then there is an infinite counterexample. --RDBury (talk) 19:22, 11 December 2022 (UTC)

Yes, "only" means exactly one. However, you didn't prove your claim about the finiteness. My original question was about whether the sequence must be finite. I, too, thought it must, but I asked because I couldn't prove what I thought. 147.236.144.145 (talk) 19:34, 11 December 2022 (UTC)

As Felix pointed out, #A states that every point other than the first one must have a predecessor. As for the correct interpretation you've asked about: Well, the first one is the correct one, and I thought it was clear in #C, but I've just added a few words to #C to make it even clearer. 147.236.144.145 (talk) 17:08, 11 December 2022 (UTC)

The point that's still not quite precisely stated is the clause about "only this way does the ball manage to reach the last point". If "only this way" means "in a finite sequence of steps of the sort given", then the question is circular.

As a test case, suppose that the ball positions do have order type $\omega +{\overline {\omega }}$. You said in the question that this is ruled out by clause D. How is it ruled out, exactly?

One possible meaning of "only this way" would be to take some sort of minimal closure under the operations given. That I think will work, and give you the answer that the number of points must be finite. But you haven't stated the question quite explicitly enough to make that conclusion yet. --Trovatore (talk) 19:49, 11 December 2022 (UTC)

I didn't want to use the concept of minimal closure, because I found it needless. By "only this way" I simply mean "only by jumping to a given point (other than the first one) from its (unique) predecessor". I thought #C was pretty explicit in meaning that. Anyway, let's put my question the following way (as I've just phrased it above): Is there any infinite model, of such a sequence of points, satisfying A-D (including C), under the meaning of "only this way" I've just clarified? 147.236.144.145 (talk) 22:49, 11 December 2022 (UTC)

@147.236.144.145: I'm afraid the notion of "only this way" has not in fact been clarified. The ball can arrive at a given point only by jumping to a given point (other than the first one) from its unique predecessor. OK. But how did it get there?

Lambiam, below, alludes to needing infinitely many jumps if there are infinitely many points. That is correct. However you have not yet succeeded in excluding infinitely many jumps, in any precise way. I assume you don't want to exclude it by fiat, because that would make the whole question trivial. But your current formulation of D is not well enough specified to exclude it. --Trovatore (talk) 06:19, 12 December 2022 (UTC)

It seems you ignore the way I formultaed my question in bold letters, so let me quote myself: "Is there any infinite model, of such a sequence of points, satisfying A-D?". Without you pointing at a specific inifinite model, just "infinitely many jumps" alone cannot explain constructively how jumping from every point to its successor can take the ball - from a point whose index is finite - to another point whose index is not finite. Just as Cantor, without him pointing at the specific set of real numbers, couldn't explain constructively how an infinite set doesn't have any one-to-one correspondence to the set of natural numbers.

So, if you think you can present such an infinite model that satisfies A-D, please present it. Once you do that, this is not me who will have to explain why your infinite model satisfies A-D or doesn't satisfy A-D (I'm only asking whether it does), but this is you who will have to explain constructively how your infinite model lets the ball both satisfy A-C and get (per D) to the last point or at least to any point (you choose) whose index is not finite. 147.236.144.145 (talk) 12:09, 12 December 2022 (UTC)

Suppose I say the model is just $\omega +{\overline {\omega }}$, and the ball gets to the end "only" by starting at the first point and going to immediate successors, in $\omega +{\overline {\omega }}$ steps. Each individual step certainly conforms to the given criteria, so presumably you intend D to rule this out. But exactly how does it rule it out? I don't think you have explained that yet. You've added a new requirement, that the process has to happen "constructively", but you haven't said what that means. --Trovatore (talk) 17:25, 12 December 2022 (UTC)

According to your attitude, Cantor didn't have to use his well-known method in order to prove there's a one-to-one-correspondence between the set of natural numbers and the set of rational numbers. Analogously to what you've just claimed, he could claim something like: "Suppose I say I get to every rational number 'only' by drawing a line from a given natural number to a unique rational number, in $\omega$ steps. Each individual line certainly conforms to the given criteria, so presumably you intend something to rule this out. But exactly how does it rule it out? I don't think any mathematician has explained that yet. Mathematicians added a new requirement, that the correspondence has to be shown 'constructively', but they haven't said what that means". 147.236.144.145 (talk) 19:36, 12 December 2022 (UTC)

To formulate a precise version of the question: does every element of the sequence other than the last have a unique successor? The wording "the ball jumps to the successor" hints strongly in this direction. If so, the ball needs to make an infinitude of jumps to reach the end when the sequence is infinite.  --Lambiam 22:08, 11 December 2022 (UTC)

Thanks to your comment (excluding your last sentence), I've just added the word "unique" to the conditions A-B. However, I couldn't understand your last sentence, probably because I didn't understand whether you claim that such a sequence can be infinite and still can satisfy A-D. So let's be clear now: We assume that every point (other than the first one) has a unique predecessor, and that every point (other than the last one) has a unique successor; Assuming that, can you suggest any infinite model, of such a sequence of points, satisfying A-D? 147.236.144.145 (talk) 22:49, 11 December 2022 (UTC)

The question was whether A—D imply finiteness of any model. As shown by the model $\omega +{\overline {\omega }},$ A+B do not imply finiteness. The formulation of C is not entirely unambiguous, but taken what I assume to be its intention, this clause seems to be subsumed by D. If every next jump takes half the time of the previous one, the ball can make an infinite number of jumps in a finite amount of time. Presumably that is not intended. If the ball can get to its destination in $n$ jumps, where $n$ is a finite, natural number, the number of points visited on its journey equals $1+n,$ which is also finite. I think the conditions do not allow the ball to skip points on its way to the final destination, so this then excludes infinite models.  --Lambiam 00:56, 12 December 2022 (UTC)

As for your first comment, you are right: If every next jump takes half the time of the previous one, the ball can make an infinite number of jumps in a finite amount of time. Further, as opposed to what you thought ("Presumably that is not intended"), I don't exclude this option. However, please note that those infinitely many jumps can only explain how the ball can get to all of the infinitely many points whose indices/indexes are finite. However, my original question was about another matter: Is there an infinite sequence of points, satisfying A-D, i.e. so that infinitely many jumps explain how the ball can get to all of the infinitely many points whose indices/indexes are not finite, i.e. until it gets to the last point of the sequence? Please distinguish between a finite index and an infinite one.

As for your second comment, your conjecture is right: The conditions do not allow the ball to skip points on its way to the final destination.

As for your last comment: "this then excludes infinite models". Actually this was my intended question: Does this exclude infinite models? You claim this does, but I want to be sure (i.e. by a proof). Actually, that's why I asked my original question. 147.236.144.145 (talk) 12:09, 12 December 2022 (UTC)

As others have noted, you need to formalize C and D more precisely, and then the answer to your question will be trivial. You're saying there is a path from the first element to the last element, where successive elements of the path are successive elements of the sequence. For this purpose, does a path need to have a finite domain? Then the sequence will need to be finite. Do you instead allow a path with domain $\omega +{\overline {\omega }}$? Then you can have an infinite sequence.--2600:4041:5640:8400:CCED:C8D5:34D:7FD9 (talk) 01:00, 12 December 2022 (UTC)

This is exactly what I'm asking: Is there a path with an infinite domain, satisfying A-D? If you think there is one, please present it, and show constructively how it really satisfies A-D. Just as I've shown the ordered set $\omega +{\overline {\omega }},$ as an infinite model satisfying #A and #B. If you, too, show $\omega +{\overline {\omega }},$ as a model of the infinite domain, please show the exact [directed] path that satisfies C-D. Please note that the path is expected to be directed because it shows the way the ball walks/jumps, which is directed as well. 147.236.144.145 (talk) 12:09, 12 December 2022 (UTC)

We define an order-preserving embedding $f:\omega +{\overline {\omega }}\to [0,1]$ as follows. Each element of the source domain is either an untagged element $n$ or a tagged element ${\overline {n}}.$ Then

$f(n)={\tfrac {1}{2}}(1-2^{{-}n}),$

$f({\overline {n}})={\tfrac {1}{2}}(1+2^{{-}n}).$

The ball travels at a uniform speed in one unit of time along the line from $0$ to $1.$ Clearly, it visits all points in order, from the first to the last. From every point that the ball has reached, it goes to the successor, not skipping any points on the way – there are no intermediate points it could skip.

More specifically, when the ball reaches the image $f(n)$ of an untagged element, the ball goes next to its successor's image $f(n{+}1).$ When the ball reaches the image $f({\overline {n}})$ of a tagged element, either $n=0,$ so the ball has reached the final destination, or $n\geq 1,$ and the ball goes next to its successor's image $f({\overline {n{-}1}}).$ In all cases, the ball goes from each embedded point – other than the last – to its successor.

Conversely, when the ball reaches the image $f(n)$ of an untagged element, either $n=0,$ the point of departure, or $n\geq 1,$ and the ball came from its predecessor's image $f(n{-}1),$ which it reached earlier. When the ball reaches the image $f({\overline {n}})$ of a tagged element, the ball came from its predecessor's image $f({\overline {n{+}1}}).$ In all cases, the ball goes to each embedded point – other than the first – from its predecessor.  --Lambiam 13:31, 12 December 2022 (UTC)

If you've presented a physical model in which the ball travels along a physical route, then you must notice that the route is a continuum which contains, not only the rational numbers of the form defined by your two functions, but also every other rational number in the segment [0,1], as well as every irrational number in that segment, so when the ball travels from 0 to its successor 1/4 it physically passes through infinitely many rational points (as well as irrational points) lying between 0 and its successor 1/4, as opposed to #C.

However, if you've presented a theoretical model in which the ball only passes through the points of the form defined by your two functions, then for every point smaller than 1/2, you can really show constructively how it's reached by the ball, but no point greater than 1/2 has this constructive property. That's why I wanted an infinite model that explains how the ball, reaches the last point, while satisfying A-D. You have presented an infinite theoretical model, but you didn't show constructively how the ball reaches any point greater than 1/2 in this infinite theoretical model. 147.236.144.145 (talk) 19:36, 12 December 2022 (UTC)

Can you give a mathematical characterization of "constructive"? Can you "show constructively" that the ball can make discrete jumps? Otherwise, how can it even reach the second point in the sequence? You keep adding new but undefined requirements. The definition of the embedding and of the successor relation between the points is constructive in the sense of constructive mathematics, as is the proof that all points are reached and that every next point reached is the successor of the previous point.  --Lambiam 20:08, 12 December 2022 (UTC)

As for your question about whether I can show constructively that the ball can make discrete jumps: I didn't determine it can: I've only asked: Let's assume it can make discrete jumps, then can we infer so and so (e.g. that the sequence is finite), as a necessary conclusion following from the assumption?

As for your main question: By "constructive" I mean something that can be built by ZF (also by physics is OK). For example, the existence of a set of successors is constructive because it can be built by the axioms of ZF. The same is true for the existence of $\omega +{\overline {\omega }}.$ However, you didn't show by ZF (or by physics) whether/how the ball can reach the last point of any infinite model while satisfying A-D. 147.236.144.145 (talk) 20:58, 12 December 2022 (UTC)

It is (sometimes) possible to show that a given wff $P$ is, or is not, a theorem of ZF. I thought my construction above, which remains well within the limited expressiveness of ZF and which I thought satisfies A–D, was constructive enough. Apparently not. I have no idea how to formulate a formula of ZF that can be understood to represent the statement, "there is a constructive way in which the ball can reach the last point of any infinite model while satisfying A–D". I don't know what that means, and so I wouldn't even know where to start. Lacking the smarts to construct such a formula, I give up.  --Lambiam 23:05, 12 December 2022 (UTC)

Actually, I'm facing an analogous trouble. I'm trying to prove that the steps the ball can carry out have an upper bound other than the last point in any infinite model (e.g. the upper bound $\omega$ in the model $\omega +{\overline {\omega }}$), but I suspect it's not provable. 147.236.144.145 (talk) 07:30, 13 December 2022 (UTC)

@147.236.144.145: I have a sort of a guess at what you may ultimately getting at here. Of course my guess may be completely wrong. But if it's right, let me point out a fundamental limitation that may mean you can't quite get what you want.

Are you familiar with nonstandard models of arithmetic? Essentially they have all the genuine natural numbers, plus some "fake" ones that are actually infinite (in the sense that there are infinitely many elements of the model less than one of them), but the model can't tell that they are infinite. The model "thinks" they're finite.

Now, you could rule out all the nonstandard models, if you could just add an axiom that says "every natural number is finite".

But it turns out you can't do that, not in first-order logic with the given non-logical concepts. First-order logic isn't strong enough.

You can do it in second-order logic, quantifying over sets. But you seem to want to avoid that (you claimed it was "needless" to talk about minimal closures).

Or you could do it in infinitary logic in a number of ways, but I suspect that defeats the point for you. That's the "guess" I was alluding to; that you're looking for a way to distinguish between finite and infinite in the first place. --Trovatore (talk) 18:41, 12 December 2022 (UTC)

Yes, I'm familiar with nonstandard models of arithmetic, but I'm not looking for a way to distinguish between finite and infinite, because I know it's impossible in first order logic and possible in ZF and likewise. I've asked my original question, beacsue I really don't know its answer. 147.236.144.145 (talk) 19:36, 12 December 2022 (UTC)

Like Lambiam I've given up on following your question so I can't help you there. Small correction though: ZF is a theory of first-order logic, and as such is just as incapable of distinguishing between finite and infinite. Oh, you can define a predicate "is a finite set", but it's not guaranteed that that predicate, interpreted in a given model of ZF, will be correct about which sets are finite. It will be correct provided that the model is wellfounded, but first-order logic is not strong enough to capture that. --Trovatore (talk) 08:03, 13 December 2022 (UTC)

It seems it's just a misunderstaning. By "infinite" I mean any model having a (first-order) function from the model on the set of natural numbers. This set is defined in ZF, and I call it "infinite", as a starting point (while this set trivially satisfies the definition I've just given for what I mean by "infinite"). However, when I claimed it's impossible to define finiteness in first order logic, I meant, in any first order language, even with Peano Axioms, but without ZF. Probably I had to be more careful in my previous response, but now I hope my current clarification is sufficient to remove the misunderstanding. 147.236.144.145 (talk) 09:52, 13 December 2022 (UTC)

If C means every movement has to be in the same direction and there's an infinite number of point then there will be an accumulation point and a least upper bound - which doesn't have a predecessor before it for any point at or after as for any predecessor there must be another between it and the bound. If the jumps can be all over the place so we drop C we're into the well ordered theorem territory and it is possible - just no-one can describe how :-) NadVolum (talk) 10:46, 13 December 2022 (UTC)

The accumulation point doesn't have to be contained in the infinite sequence. Check the following model:

1. The distance from the first point to the last one is four meters.
2. The distance from the first point to the second one is one meter, and the distance from the second point to the third one is half a meter, and the distance from the third point to the fourth one is a quarter of a meter, and so forth. That is, for every finite natural n, the distance from the n-th point to its succesoor is $2^{1-n}$ meter.
3. The distance from the last point to the penultimate one is one meter, and the distance from the penultimate point to the antepenultimate one is half a meter, and the distance from the antepenultimate point to its predecessor is a quarter of a meter, and so forth, analogously to what I've described in the previous section. 147.236.144.145 (talk) 15:01, 13 December 2022 (UTC)

I didn't say the accumlation point had to be contained in the sequence. You just have to form an interval from the first point before the accumulation to the first after (or the end point if it is an end point). Those points can be determined according to C. That interval can't contain any points except possibly the accumulation point. But if it only has one point it isn't an accumulation point. NadVolum (talk) 18:15, 13 December 2022 (UTC)