Yes ok, so the coach can turn into a pumpkin and the lamp can turn into a plate of spaghetti. Are you agreeing with me on that point? — fishfry

Yes. But I have an obstinate feeling that that fact is a reductio of the process that generated it. So I'm not questioning what you say, but rather what we make of it.

I was making my point about mathematical convergent sequences. Don't know whether it strictly applies to walking. — fishfry

It may be a bad habit to think of applications of a mathematical process. But that's what's going on with the infinite staircase. So it might be relevant to that.

Yes. But I have an obstinate feeling that that fact is a reductio of the process that generated it. So I'm not questioning what you say, but rather what we make of it. — Ludwig V

Right. So why is a lamp circuit that can switch states in arbitrarily small slices of time reasonable, and spaghetti isn't? That's one of the cognitive traps of the lamp problem. IMO the final state is simply not defined by the premises of the problem, AND there is no solution that makes the sequence continuous, therefore spaghetti is as sensible as anything else. And I've convinced myself that this is the solution to the problem.

It may be a bad habit to think of applications of a mathematical process. But that's what's going on with the infinite staircase. So it might be relevant to that.
3 minutes ago — Ludwig V

The staircase is different from the lamp. The walker is on step one, the walker is on step two, etc. So if we define the final state to be that the walker is at the bottom of the stairs, that definition has the virtue of making the walker's sequence continuous. So it's to be preferred over all other possible solutions.

That's one of the cognitive traps of the lamp problem. IMO the final state is simply not defined by the premises of the problem, AND there is no solution that makes the sequence continuous, therefore spaghetti is as sensible as anything else. And I've convinced myself that this is the solution to the problem. — fishfry

I agree with you. It suits my approach well, in that the existence of the problem is a result of the way it is defined, or not defined.

The walker is on step one, the walker is on step two, etc. So if we define the final state to be that the walker is at the bottom of the stairs, that definition has the virtue of making the walker's sequence continuous. — fishfry

That's the way ω is defined, isn't it? Although I'm not sure what you mean by "continuous" there.
I still feel uncomfortable, because it does get to the bottom of the stairs by placing a foot on each of the stairs, in sequence. But that's exactly the hypnotism of the way the problem is defined. And if an infinite physical staircase is the scenario, then anything goes..

Well between the two of you I have no idea what a supertask is anymore. — fishfry

A supertask is "a countably infinite sequence of operations that occur sequentially within a finite interval of time."

You believe in limits, you said so. And if you believe even in the very basics of set theory, in the principle that I can always union two sets, then I can adjoin 1 to {1/2, 1/3, 1/4, 1/5, ...} to create the set {1/2, 1/3, 1/4, 1/5, ..., 1}.

It's such a commonplace example, yet you claim to not believe it? — fishfry

I said I had no problem with any of that.
Is it a belief thing, like it is some kind of religious proposition or something? "Hey, I'm going rogue here and will suspend belief that 7 is a factor of 35".

Or what is your objection, exactly?

Treating infinity as a number, something you didn't do in your unionized set above

It's an infinite sequence. I stuck the number 1 on the end.

Yea, when it normally is depicted at the beginning. From what I know, a set is a set regardless of the ordering. There must be a different term (ordered set?) that distinguishes two identical sets ordered differently, sort of like {1, 3, 5, 7 --- --- 8, 6, 4, 2}

The entire set is ordered by the usual order on the rational numbers. So why is it troubling you that I called 1 the "infinitieth" member of the ordered set?

It violates thebijunction. You can't say what number comes just before it, which you can for any other element except of course the first. You can do that with any other element.

It's a perfect description of what's going on. And it's a revealing and insightful way to conceptualize the final state of a supertask. Which is why I'm mentioning it so often in this thread.

OK, but what problem does it solve? It doesn't solve Zeno's thing because there's no problem with it. It doesn't solve the lamp thing since it still provides no answer to it.

In terms of known physics as of this writing, we can not sensibly discuss what might be going on below the Planck length.

Nobody's asking the particle to meaningfully discuss (mathematically or not) the step. It only has to get from one side to the other, and it does. Your argument is similar to Michael wanting a person to recite the number of each step, a form of meaningful discussion.

Maybe we live in a discrete grid of points -- which would actually resolve Zeno's paradoxes.

It would falsify the first premise. Continuous space falsifies the second premise. Zeno posits two mutually contradictory premises, which isn't a paradox, only a par of mutually contradictory premises,.

But you can't say "you can traverse the space of that step, even when well below the Planck length" because there is no evidence, no theory of physics that supports that claim.

But I can say "for all we know, ....", and then there's no claim. I'm not making the claim you state. I'm simply saying we don't know it's not true. I even put out my opinion that I don't think it's true, but the chessboard thing isn't the alternative. That's even worse. It is a direct violation of all the premises of relativity theory (none of which has been proved).

IMO the final state is simply not defined by the premises of the problem, — fishfry

Spot on, yes.

A supertask is "a countably infinite sequence of operations that occur sequentially within a finite interval of time." — Michael

Yea, I don't know how that could have been lost. I don't think anybody attempted to redefine it anywhere.

I agree with you. — Ludwig V

I love when people agree with me. It happens so seldom around here :-)

It suits my approach well, in that the existence of the problem is a result of the way it is defined, or not defined. — Ludwig V

I agree with you too!

The walker is on step one, the walker is on step two, etc. So if we define the final state to be that the walker is at the bottom of the stairs, that definition has the virtue of making the walker's sequence continuous.
— fishfry
That's the way ω is defined, isn't it? Although I'm not sure what you mean by "continuous" there.
I still feel uncomfortable, because it does get to the bottom of the stairs by placing a foot on each of the stairs, in sequence. But that's exactly the hypnotism of the way the problem is defined. And if an infinite physical staircase is the scenario, then anything goes.. — Ludwig V

Let me see if I can clarify my point.

In the lamp problem, we have the sequence 0, 1, 0, 1, 0, 1, ... We can "complete" the sequence by defining the state at $\omega$ as 0, or 1, or a plate of spaghetti. In no conceivable completion can the sequence be made continuous, because 0, 1, 0, 1, ... simply does not have a limit.

But in the staircase problem, if 1 is "walker is on the step" and 0 otherwise, then we have the sequence 1, 1, 1, 1, ... which has the limit 1. So 1, the walker is on the step, is the natural state at the end of the sequence.

Does that make sense? The staircase has a natural answer; but the lamp has no natural answer. Any completion whatsoever is as equally bad as any other.

Well between the two of you I have no idea what a supertask is anymore.
— fishfry

A supertask is "a countably infinite sequence of operations that occur sequentially within a finite interval of time." — Michael

Well ok, then why don't I complete a supertask when I walk across the room, first going halfway, etc.? Can you distinguish this case from your definition?

What I think about supertasks is:

* Either they are already possible in the sense of Zeno, when I walk across the room; or

* They are physically impossible in currently known physics (because of Planck) but may be possible in future physics, by analogy with previous scientific revolutions; and

* You have not convinced me or even made me understand your reasoning that supertasks are "metaphysically impossible" or that they entail a logical contradiction.

I said I had no problem with any of that.
Is it a belief thing, like it is some kind of religious proposition or something? "Hey, I'm going rogue here and will suspend belief that 7 is a factor of 35". — noAxioms

I'm making the point that you are perfectly willing to accept {1/2, 3/4, 7/8, ..., 1} as a valid set that contains an infinite sequence and its limit; but you are having trouble accepting {1, 2, 3, ..., $\omega$}, simply because it's far less familiar. But in terms of their order, they are exactly the same set. They have very different metric properties; but strictly with respect to order, they are two different representations of the same ordered set.

Treating infinity as a number, something you didn't do in your unionized set above — noAxioms

Transfinite ordinal numbers are numbers. It's just a matter of expanding one's concept of a number. $\omega$ is a number. It's the first transfinite ordinal number. I am casually calling it a "point at infinity," but if that bothers you, just think of it as 1 in the set {1/2, 3/4, 7/8, ..., 1}. It's exactly the same thing wearing a different suit of clothes.

It's an infinite sequence. I stuck the number 1 on the end.
Yea, when it normally is depicted at the beginning. From what I know, a set is a set regardless of the ordering. There must be a different term (ordered set?) that distinguishes two identical sets ordered differently, sort of like {1, 3, 5, 7 --- --- 8, 6, 4, 2} — noAxioms

Yes, ordered set. I have been casually using the curly braces, but you are absolutely correct. {1/2, 3/4, 7/8, ..., 1} has no order, I could stick the 1 in the middle or at the beginning and it would be the same set, but I'd lose the order that I consider important.

Perhaps a notation like <1/2, 3/4, 7/8, ..., 1> would be better, to indicate an ordered set. You are absolutely right. I did not want to add any more complications earlier, but the curly braces are inaccurate in the way I'm using them. I'm speaking of ordered sets. So I'll use angle brackets from now on.

The entire set is ordered by the usual order on the rational numbers. So why is it troubling you that I called 1 the "infinitieth" member of the ordered set?
It violates thebijunction. You can't say what number comes just before it, which you can for any other element except of course the first. You can do that with any other element. — noAxioms

Well then here yet another representation of the same idea. Suppose I reorder the natural numbers

<1, 2, 3, 4, ...>

by putting 1 at the end, so that I have:

<2, 3, 4, ..., 1>

You can see that I still have a bijection. As you noted, sets don't have order, so it's still the same set.

Note that I no longer have an order-preserving bijection. I merely have a set bijection. I can still correspond 1 to 1, 2 to 2, and so forth. But I can't do it in an order-preserving manner.

But now I have another representation of an ordered set that consists of an infinite sequence followed by a "point at infinity," or a largest element. That largest element does not have a predecessor, you are right about that.

And in fact we have a name for that. In ordinal theory, an ordinal with a predecessor is a successor ordinal. And an ordinal without a predecessor is a limit ordinal. So your intuitions are spot on.

OK, but what problem does it solve? It doesn't solve Zeno's thing because there's no problem with it. It doesn't solve the lamp thing since it still provides no answer to it. — noAxioms

Ah yes, why am I doing all this?

It solves the lamp problem. The lamp state is a function on <1/2, 3/4, 7/8, ..., 1> defined as "on" at 1/2, "off" at 3/4, "on" at 7/8, and so forth.

But now we see (more clearly, IMO) that the state at 1 is simply undefined. The statement of the problem defines the lamp state at each element of the sequence; but does NOT define the state at the limit.

We also note that there is no way to make the sequence 0, 1, 0, 1, ... continuous.

And since you didn't tell me what is the state at 1, and there is no natural way to define the state at 1, I am free to define the state at 1 any way I like. And inspired by Cinderella, I define the state of the lamp at 1 to be a plate of spaghetti. That's the solution to the problem. The final state is anything you like. It doesn't even have to be on or off since it's not a real lamp, just as Cinderella's coach is not a real coach. The lamp problem is every bit as much a fairy tale as Cinderella.

So for many of these supertask problems, the ordered set <1/2, 3/4, 7/8, ..., 1> is the natural setting for the problem.

Note that the staircase is different. The walker is on step 1, on step 2, etc. So the natural, continuous way of completing the sequence is to say that the walker is at the bottom of the stairs. This is totally different than the lamp, which can not be made continuous or sensible in any way at all.

So my entire point is that <1/2, 3/4, 7/8, ..., 1> is the natural way to think about these problems. The question is always: how did we define the state at the elements of the infinite sequence; and then, how are we free to define the final state at the limit.

Nobody's asking the particle to meaningfully discuss (mathematically or not) the step. It only has to get from one side to the other, and it does. Your argument is similar to Michael wanting a person to recite the number of each step, a form of meaningful discussion. — noAxioms

I'm not sure what you mean by referring to the subjective state of the particles. When Newton wrote down his great law of gravitation, he did not care how the masses feel about it. I'm not following your analogy.

Maybe we live in a discrete grid of points -- which would actually resolve Zeno's paradoxes.
It would falsify the first premise. Continuous space falsifies the second premise. Zeno posits two mutually contradictory premises, which isn't a paradox, only a par of mutually contradictory premises,. — noAxioms

I confess to not knowing the answer to Zeno. It's a clever argument. Unless the answer is that we satisfy Zeno and execute a supertask every time we walk across the room. But @Michael objects to that, for reasons I don't yet understand.

But I can say "for all we know, ....", and then there's no claim. I'm not making the claim you state. I'm simply saying we don't know it's not true. I even put out my opinion that I don't think it's true, but the chessboard thing isn't the alternative. That's even worse. It is a direct violation of all the premises of relativity theory (none of which has been proved). — noAxioms

Some speculative physicists (at least one, I believe) think the world is a large finite grid. It's not out of the realm of possibility as I understand it. I think I read that in Penrose's The Road to Reality. And if Sir Roger thinks it's good enough to put in a book, it must be of interest.

In other words the chessboard universe is not ruled out by any known theory or experiment. And we know that quantum and relativity have yet to be integrated, and perhaps that's a clue.

A supertask is "a countably infinite sequence of operations that occur sequentially within a finite interval of time."
— Michael
Yea, I don't know how that could have been lost. I don't think anybody attempted to redefine it anywhere. — noAxioms

Yes ok but then ... how is walking across the room by first traversing 1/2, then half of the remaining half, etc., not a supertask? I don't understand this point.

Well ok, then why don't I complete a supertask when I walk across the room, first going halfway, etc.? Can you distinguish this case from your definition? — fishfry

If supertasks are impossible and motion is possible then motion isn't a supertask.

* You have not convinced me or even made me understand your reasoning that supertasks are "metaphysically impossible" or that they entail a logical contradiction. — fishfry

By definition supertasks are non-terminating processes, therefore you've gone wrong somewhere if you conclude that they can terminate after 2N seconds.

Also I think the clearest example I gave was that of having counted down from infinity. We can assert (explaining what happened in reverse) that I recited 0 after 60 seconds, recited 1 after 30 seconds, recited 2 after 15 seconds, recited 3 after 7.5 seconds, etc., and we can say that we can sum an infinite series with terms that match the described (and implied) time intervals, but it doesn't then follow that we can have counted down from infinity; we can't even start such a count. The mathematics is evidently a non sequitur, and so it's a non sequitur in the case of having counted up to infinity as well (and so for any proposed supertask).

In the case of Thomson's lamp, nothing ever happens to the lamp except as described by this process: I turn it on after 30 seconds, turn if off after 15 seconds, turn it on after 7.5 seconds, etc. It cannot be on after 60 seconds because I always turn it off after turning it on and it cannot be off after 60 seconds because I always turn it on after turning it off, but it must be either on or off after 60 seconds, and so therefore there is a contradiction.

If you want to say that such a supertask is possible then the burden is on you to explain the state of the lamp after 60 seconds, and your answer must follow from the description of the supertask. If nothing follows then the supertask is impossible.

Well ok, then why don't I complete a supertask when I walk across the room, first going halfway, etc.? Can you distinguish this case from your definition?
— fishfry

If supertasks are impossible and motion is possible then motion isn't a supertask. — Michael

I don't find that satisfactory. It only casts doubt on the premise "if supertasks are impossible."

I agree with you that the lamp and staircase and other related puzzles are qualitatively different than Zeno's paradoxes of motion, so perhaps in that sense you want to reserve the word supertask for the former. But your definition is "completing a countably infinite number of tasks in finite time," and walking across the room seems to satisfy that definition.

Not to mention that, if we take the real numbers as a model of space, we pass through uncountably many points in finite time. That's another mystery.

* You have not convinced me or even made me understand your reasoning that supertasks are "metaphysically impossible" or that they entail a logical contradiction.
— fishfry

By definition supertasks are non-terminating processes, therefore you've gone wrong somewhere if you conclude that they can terminate after 2N seconds. — Michael

I don't know what you mean that supertasks are nonterminating by definition. Just thinking mathematically for a moment, limits "terminate" a sequence in the sense that 1 is the terminus of the sequence 1/2, 3/4, 7/8, ... The limit 1 is not part of the sequence, but we can imagine the 1 stuck at the end of an ordered set, as I have been doing, and it's perfectly sensible.

In other words supertasks are nonterminating, but they definitely may have a terminal state; just as a convergent mathematical sequence has no final term, yet has a limit. Is my analogy unsatisfactory with respect to your conception of supertasks?

Also I think the clearest example I gave was that of having counted down from infinity. We can assert (explaining what happened in reverse) that I recited 0 after 60 seconds, recited 1 after 30 seconds, recited 2 after 15 seconds, recited 3 after 7.5 seconds, etc., and we can say that we can sum an infinite series with terms that match the described (and implied) time intervals, but it doesn't then follow that we can have counted down from infinity; we can't even start such a count. — Michael

I don't follow how you are counting down from infinity. In fact when you count down from infinity, it's always only finitely many steps back. If I take the ordered set <1/2, 3/4, 7/8, ..., 1> and I start at 1, my first step backwards jumps over all but finitely many elements of the sequence, and it's always only finitely many steps back from 1 to 1/2.

[Per my recent convo w/@No Axioms I am using angle brackets to denote ordered sets].

You did lose me when you said that counting 0, 1, 2, ... is "counting down from infinity." I did not understand that example when you gave it earlier. Mathematically, the ordered set <1, 2, 3, ...> exists, all at once. Its counting is completed the moment it's invoked into existence by the axiom of infinity.

But let me ask you this. Suppose I say 0 at 60 seconds, and 1 at 30 seconds, and 3 at 15 seconds, and so forth.

Now I claim that after 120 seconds (the sum of the series) I have counted all the natural numbers!

Yes I claim that. And as proof, I challenge you to name the first number that I did not count.

Since you can not do that, I have indeed counted all the natural numbers.

The mathematics is evidently a non sequitur — Michael

I either don't understand what you mean, or I strenuously disagree.

Explain please?

, and it's a non sequitur in the case of having counted up to infinity as well. — Michael

I just proved to you, using a very standard inductive argument, that I can indeed count all the natural numbers as you described, in intervals of 60, 30, etc. Because you can not name any number I can't count. Did I count 47? Sure. Googolplex? Sure. Graham's number? Sure. There is no number that I didn't eventually count. Therefore I counted them all.

I don't know what you mean that supertasks are nonterminating by definition. — fishfry

Tasks are performed ad infinitum. I never stop counting. There's always another number to count.

You did lose me when you said that counting 0, 1, 2, ... is "counting down from infinity." I did not understand that example when you gave it earlier. Mathematically, the ordered set <1, 2, 3, ...> exists, all at once. Its counting is completed the moment it's invoked into existence by the axiom of infinity. — fishfry

I'm talking about reciting the numbers. So imagine someone reciting the natural numbers up to infinity. Now imagine that process in reverse. That's what I mean by someone counting down from infinity.

It is a non sequitur to argue that because we can sum an infinite series with terms that match the proposed time intervals that it is possible to have counted down from infinity. It is impossible, even in principle, to start such a count. The maths of an infinite series doesn't change this.

And it is a non sequitur to argue that because we can sum an infinite series with terms that match the proposed time intervals that it is possible to have counted up to infinity. It is impossible, even in principle, to stop such a count. The maths of an infinite series doesn't change this.

I don't know what you mean that supertasks are nonterminating by definition.
— fishfry

Tasks are performed ad infinitum. I never stop counting. There's always another number to count. — Michael

Did I not move you, surprise you, convince you, that if you count 1, 2, 3, ... successively halving the time intervals, that you will indeed count every single natural number in finite time? If not, why not?

I am still waiting for you to name the first number I didn't count.

This is a standard inductive argument. To prove that a property holds for all natural numbers, I show the impossibility of there being a first number where the property fails.

I'm talking about reciting the numbers. So imagine someone reciting the natural numbers up to infinity. Now imagine that process in reverse. That's what I mean by someone counting down from infinity. — Michael

But counting backward from infinity is always finite! I showed you how that works, counting backward from 1 in the ordered set <1/2, 3/4, 7/8, ..., 1>

In fact this is true of all the transfinite ordinals. It's only finitely many steps backward from any transfinite ordinal, no matter how large. That's because stepping back from any limit ordinal (defined as an ordinal without an immediate predecessor) necessarily jumps over all but finitely many elements of the sequence that led up to it.

It is a non sequitur to argue that because we can sum an infinite series with terms that match the proposed time intervals that it is possible to have counted down from infinity. It is impossible, even in principle, to start such a count. The maths doesn't change this. — Michael

It's easy, I'll count backward from infinity right here on a public Internet forum, in plain view of the world.

1, 15/16, 7/8, 3/4, 1/2. Done. My first step necessarily jumped over all but finitely many elements of the infinite sequence. It must be that way.

That's because the first step backward from any limit ordinal necessarily jumps over all but finitely members of the sequence whose limit it is.

Counting backward from infinity is easy, and always finite!

Did I not move you, surprise you, convince you, that if you count 1, 2, 3, ... successively halving the time intervals, that you will indeed count every single natural number in finite time? If not, why not? — fishfry

Because it begs the question.

But counting backward from infinity is always finite! I showed you how that works, counting backward from 1 in the ordered set <1/2, 3/4, 7/8, ..., 1> — fishfry

What number do you recite after 1?

It's easy, I'll do it right here on a public Internet forum.

1, 15/16, 7/8, 3/4, 1/2. Done.

That's because the first step backward from any limit ordinal necessarily jumps over all but finitely members of the sequence whose limit it is. — fishfry

That's not counting down from infinity. That's just reciting five rational numbers.

Because it begs the question. — Michael

I go 1 at 60, 2 at 30, etc.

Name the first number that I fail to count

Third time I'm asking you the question. (At ever decreasing intervals of time!)

This is a standard inductive argument. If it's impossible to name the first natural number at which a property fails to hold, the property must hold for all natural numbers. Think back to when you learned inductive proofs in school. I can name 1. And if I name n, I can name n + 1. Therefore I can name all the numbers. Counterintuitive though it may be, it's true. You learned this in high school.

Please give this argument some thought.

That's not counting down from infinity. — Michael

You have no proof or evidence. On the contrary, the mathematics is clear.

I go 1 at 60, 2 at 30, etc.

Name the first number that I fail to count

Third time I'm asking you the question.

This is a standard inductive argument. If it's impossible to name the first natural number at which a property fails to hold, the property must hold for all natural numbers.

Please give this argument some thought. — fishfry

It begs the question. Your premise is necessarily false. Such a supertask is impossible, even in principle, to start.

In your opinion. But you have no proof or evidence. On the contrary, the mathematics is clear. — fishfry

You just listed five rational numbers and are claiming that this is proof of you reciting all the natural numbers in descending order? You're talking nonsense.

But counting backward from infinity is always finite! I showed you how that works, counting backward from 1 in the ordered set <1/2, 3/4, 7/8, ..., 1> — fishfry

What number do you recite after 1?

It begs the question. Your premise is necessarily false. Such a supertask is impossible, even in principle, to start. — Michael

Did you learn mathematical induction in school? Please review that. Please take the time to understand the argument I made.

Under the premises of the problem you posted, there is no number that does not get spoken.

It's imperative that you understand that. It's pointless for you to disagree. You must show that there's a number that did not get spoken. If you can't do that, then every number gets spoken.

You just listed five rational numbers and are claiming that this is proof of you reciting all the natural numbers in descending order? — Michael

I did not make that claim. I said I counted backward from a limit ordinal. That's easy. It's always a finite number of steps.

You're talking nonsense. — Michael

I'm counting backward from a limit ordinal. Very standard math.

What number do you recite after 1? — Michael

7/8 will do just fine. I necessarily had to jump over all but finitely members of the sequence.

Of course I can not count ALL the numbers backward. That's impossible. That's because limit ordinals do not have predecessors. That's the definition of a limit ordinal, an ordinal that does not have an immediate predecessor. So it's your challenge that's nonsense.

But please, I'm asking you to sit down and think about the inductive argument I made.

Counting forward with your 60 second idea, which number won't be spoken?

We can certainly say "1". And if we say n, we can say n + 1. This is high school mathematical induction. Please tell me you learned this. If not, that would explain your confusion. But if you made it through high school math (do they still teach induction in high school? What do I know) then you have the means to understand the argument.

Please take the time to think it through. What number can't be spoken?

ps -- I looked it up. Perhaps induction is not universally taught in high school, and it doesn't come up in calculus.

Do you know mathematical induction? It's a row of dominos.

https://en.wikipedia.org/wiki/Mathematical_induction

7/8 will do just fine. I necessarily had to jump over all but finitely members of the sequence. — fishfry

No, we're reciting the numbers in descending order. It's impossible to do, even in principle. The fact that we can baselessly assert that I recite the first number in N seconds and the second number in N/2 seconds and the third number in N/4 seconds, and so on ad infinitum, and the fact that the sum of this infinite series is 2N, doesn't then entail that the supertask is possible.

That we can sum this infinite series is evidently a red herring.

No, we're reciting the numbers in descending order. It's impossible to do, even in principle. The fact that we can assert that I recite the first number in N seconds and the second number in N/2 seconds and the third number in N/4 seconds, and so on ad infinitum, and the fact that the sum of this infinite series is 2N, doesn't then entail that the supertask is possible.

That we can sum such an infinite series is a red herring. — Michael

You're right that we can't "name all the numbers" going backward. but that's obvious. There's no largest number and limit ordinals don't have immediate predecessors.

It's pointless for you to snap back a minute later arguing with well established mathematical facts. I gave a solid inductive argument that by the premises of your 60 second puzzle, all the numbers will be spoken. That's because there's no first number that won't be spoken. If you doubt that, then name a number that's not spoken.

I ask you to read carefully what I'm writing, and think about it.

Did you ever learn mathematical induction? If yes, I gave a standard inductive argument. If no, that's a good starting point and I'll be happy to give a summary. I gave you the Wiki link. I can't argue well established facts with you.

In the puzzle you gave, every number must be spoken. In fact we could calculate, if we cared to, the exact time at which it's spoken.

Please give this some thought.

What number won't be spoken?

I've given solid a mathematical argument that your 60 second puzzle guarantees that all the numbers will be spoken. — fishfry

No you haven't. Your premise begs the question and simply asserts that all the natural numbers have been recited within 60 seconds.

No you haven't. Your premise begs the question and simply asserts that all the natural numbers have been recited within 60 seconds. — Michael

Name the first one that's not. It's a trivial exercise to identify the exact time at which each natural number is spoken. "1" is spoken at 60, "2" at 90, "3" at 105, "4" at 112.5, and so forth.

Can you not see that we can calculate the exact time at which each number is spoken?

I did not "simply assert" all the numbers are spoken. I proved it logically. Induction works in the Peano axioms, I don't even need set theory.

If you work through this example you will obtain insight.

I love when people agree with me. It happens so seldom around here — fishfry

So do I. There's a paradox about agreement, that it is the purpose, but also the end, of the discussion. So people tend to focus on disagreements.

And in fact we have a name for that. In ortdinal theory, an ordinal with a predecessor is a successor ordinal. And an ordinal without a predecessor is a limit ordinal. So your intuitions are spot on. — fishfry

I found that discussion very helpful.

But in the staircase problem, if 1 is "walker is on the step" and 0 otherwise, then we have the sequence 1, 1, 1, 1, ... which has the limit 1. So 1, the walker is on the step, is the natural state at the end of the sequence. — fishfry

Have I understood right, that 0 means "walker is not on the step", and that "the step" means "the step that is relevant at this point" - which could be 10, or 2,436? So 0 would be appropriate if the walker is on the floor from which the staircase starts (up or down)
My instinct would have been to assign 0 also to being on the floor at which the staircase finishes (up or down). It makes the whole thing symmetrical and so more satisfying.

That's because the first step backward from any limit ordinal necessarily jumps over all but finitely members of the sequence whose limit it is. — fishfry

I don't like that way of putting it, at least in the paradoxes. Doesn't the arrow paradox kick in when you set off in the.reverse direction? Or perhaps you are just thinking of the numbers as members of a set, not of what the number might be measuring. I suppose that's what "ordinal" means?

I confess to not knowing the answer to Zeno. It's a clever argument. Unless the answer is that we satisfy Zeno and execute a supertask every time we walk across the room. But Michael objects to that, for reasons I don't yet understand. — fishfry

Yes ok but then ... how is walking across the room by first traversing 1/2, then half of the remaining half, etc., not a supertask? I don't understand this point. — fishfry

Michael's way of putting the point is, IMO, a bit dramatic. The boring truth for me, is that the supertask exists as a result of the way that you think of the task. If you think of it differently, it isn't a supertask. It's not about reality, but about how you apply mathematics to reality.

Not to mention that, if we take the real numbers as a model of space, we pass through uncountably many points in finite time. That's another mystery. — fishfry

Well, if you insist on describing things in that way .... I'm not sure what you mean by "model". I think of what we are doing as applying a process of measuring and counting to space - or not actually to space itself, but to objects in space. A geometrical point has no dimensions at all. So it is easy to see how we can pass infinitely many points in a finite time. (I'm not quite sure how this would apply to numbers, but they do not have any dimensions either.) This doesn't apply to the paradoxes we are considering, which involve measurable lengths, but it may help to think of them differently.

A supertask is "a countably infinite sequence of operations that occur sequentially within a finite interval of time." — Michael

That's all very well. But it also takes us back to the question what this "operation" actually is. If you think of it as an action that takes a measurable amount of time, you can't, by definition. When we perform a calculation, that is an action in physical time. But a mathematical operation isn't quite like that, and somewhere in that is the answer (possibly).

Name the first one that's not. It's a trivial exercise to identify the exact time at which each natural number is spoken. "1" is spoken at 60, "2" at 90, "3" at 105, "4" at 112.5, and so forth.
I did not "simply assert" all the numbers are spoken. I proved it logically. Induction works in the Peano axioms, I don't even need set theory. — fishfry

Yes, but you didn't speak all the natural numbers, and indeed, if induction means what I think it means, your argument avoids the need to deal with each natural number in turn and sequence.

I'm sorry this is a bit scrappy, but there are lot of issues going on at the same time here. Great fun!

After 60 seconds I said "0", 30 seconds before that I said "1", 15 seconds before that I said "2", 7.5 seconds before that I said "3", and so on ad infinitum.

What natural number did I not say?

You can't answer, therefore it is metaphysically possible to have recited the natural numbers in descending order.

---

Obviously the above is fallacious. It is metaphysically impossible to have recited the natural numbers in descending order. The fact that we can sum an infinite series with terms that match the described and implied time intervals is irrelevant. The premise begs the question. And the same is true of your version of the argument.

Obviously the above is fallacious. It is metaphysically impossible to have recited the natural numbers in descending order. — Michael

I already agreed with this, because limit ordinals do not have immediate predecessors.

The fact that we can sum such an infinite series is irrelevant. And the same is true of your version of the argument. — Michael

If you would engage in your private time with the 60 second puzzle, you would see that each number is spoken at a specific, calculable time; that there is no first number that's not spoken; and therefore every number is spoken.

It's not productive for me to give a high-school level inductive argument and for you to say "nonsense" and "metaphysically impossible" without ever engaging with the argument.

Please read the Wiki page on mathematical induction and ask questions as necessary, and challenge yourself to engage with the argument.

Ask yourself: What is the first number not spoken? If you ask yourself that enough times, you may have an epiphany.

Argument 1
Premise: I said "0", 30 seconds after that I said "1", 15 seconds after that I said "2", 7.5 seconds after that I said "3", and so on ad infinitum.

What natural number did I not recite? There is no answer. Therefore I have recited the natural numbers in ascending order.

Argument 2
Premise: I said "0", 30 seconds before that I said "1", 15 seconds before that I said "2", 7.5 seconds before that I said "3", and so on ad infinitum.

What natural number did I not recite? There is no answer. Therefore I have recited the natural numbers in descending order.

---

In both cases for any given natural number I can calculate how long it took me to reach it.

These arguments only show that if I recite the natural numbers as described then I have recited all the natural numbers, but this does nothing to prove that the antecedent is possible, and it is the possibility of the antecedent that is being discussed. As it stands you're begging the question.

Now let's assume that it's metaphysically possible to have recited the natural numbers in ascending order and to have recorded this on video/audio. What happens when we replay this video/audio in reverse? It's the same as having recited the natural numbers in descending order which you admit is metaphysically impossible. Therefore having recited the natural numbers in ascending order must also be metaphysically impossible.

Both Argument 1 and Argument 2 are unsound. The premises are necessarily false. It is impossible in principle for us to recite the natural numbers in the manners described.

That's all very well. But it also takes us back to the question what this "operation" actually is. — Ludwig V

It could be anything. The problem has nothing to do with the operation being performed and everything to do with continually halving the time between operations.

At 0s A ≔ 1, at 30s A ≔ red, at 45s A ≔ turtle, at 52.5s A ≔ 1, at 56.25s A ≔ red, and so on ad infinitum.

Or:

At 60s A ≔ 1, at 30s A ≔ turtle, at 15s A ≔ red, at 7.5s A ≔ 1, at 3.75s A ≔ turtle, and so on ad infinitum.

That an infinite series with terms that match the described and implied time intervals has a finite sum isn't that it makes sense for either set of tasks to have actually been carried out. This should be self-evident in the second case. You're being deceived by maths if you think the first case is different.

Very interesting. If only I knew what "metaphysically possible" means? Can you help?

(I do know what "I said <x>" meant and what seconds are)

https://en.wikipedia.org/wiki/Subjunctive_possibility#Types_of_subjunctive_possibility

It happens so seldom around here :-) — fishfry

Such is the nature of philosophy.

If only I knew what "metaphysically possible" means? — Ludwig V

I had a thread on that a while ago if you care https://thephilosophyforum.com/discussion/14855/metaphysically-impossible-but-logically-possible/p1

strictly with respect to order, they are two different representations of the same ordered set. — fishfry

Agree.

Transfinite ordinal numbers are numbers.

Are they? Does √ω have meaning? It does for numbers. It's a serious question. I am no expert on how transfinite ordinal numbers are treated. It seems like a different species, like having a set {1, 2, 3, ... , green} which is also a valid set, and countable.

Yes, ordered set. I have been casually using the curly braces, but you are absolutely correct. {1/2, 3/4, 7/8, ..., 1} has no order, I could stick the 1 in the middle or at the beginning and it would be the same set, but I'd lose the order that I consider important.

Ordering irrelevant. The set supposedly needs to be countable, and it is. Michael's definition of supertask came from wiki, and that definition says it is countable, else it's a hypertask. The SEP definition of supertask omits the 'countable' part and seemingly groups the two categories under one word.

The definition also includes 'sequential', meaning parallel execution of multiple steps is not allowed.

Yes ok but then ... how is walking across the room by first traversing 1/2, then half of the remaining half, etc., not a supertask?

Clearly it isn't a supertask if it is impossible to go only half the remaining distance for some intervals. If that is possible, then it must be a supertask.

It violates thebijunction
— noAxioms

I take that back. It doesn't violate the bijection. And I spelled it wrong too. So many errors.

Note that I no longer have an order-preserving bijection.

That's fine. The rational numbers are both ordered and countable, but they cannot be counted in order.

Ah yes, why am I doing all this?

It solves the lamp problem. The lamp state is a function on <1/2, 3/4, 7/8, ..., 1> defined as "on" at 1/2, "off" at 3/4, "on" at 7/8, and so forth.

But now we see (more clearly, IMO) that the state at 1 is simply undefined. The statement of the problem defines the lamp state at each element of the sequence; but does NOT define the state at the limit.

Sounds like the lamp problem is unsolved. It is still 'undefined'.

Another note: The paradox of the gods that I occasionally bring up is fun to ponder, but it isn't a supertask since it cannot be completed (or even started). Progress is impossible. Ditto with the grim reaper 'paradox' where I die immediately and cannot complete the task.

Note that the staircase is different. The walker is on step 1, on step 2, etc. So the natural, continuous way of completing the sequence is to say that the walker is at the bottom of the stairs.

There is no bottom, and the OP did not suggest a bottom step. He is done, and no stairs are observable. It's mathematical only, but framed with a physical sounding analogy, which makes it fall apart.
Your ω might help with the stairs. The guy is at 'the bottom' and there is but the one step there, labeled ω. No steps attached to it, but step on that one step and up you go, at some small finite numbered step after any arbitrarily small time.

Unless the answer is that we satisfy Zeno and execute a supertask every time we walk across the room. But Michael objects to that, for reasons I don't yet understand.

His assertion isn't justified, I agree.

Some speculative physicists (at least one, I believe) think the world is a large finite grid

So much for the postulates of relativity then. I kind of thought we demolished that idea with some simple examples. It seems to be a 'finite automata' model, and the first postulate of SR is really hard (impossbile) to implement with such a model, so a whole new theory is needed to explain pretty much everything if you're going to posit something like that. I haven't read it of course, so any criticism I voice is a strawman at best.

The chessboard universe sounds very classical, and it's been proven that physics is not classical, so I wonder how this model you speak of gets around that.

Well ok, then why don't I complete a supertask when I walk across the room, first going halfway, etc.? Can you distinguish this case from your definition?
— fishfry

If supertasks are impossible and motion is possible then motion isn't a supertask. — Michael

This evaded the question ask. Sure, we all agree that if supertasks are impossible, then supertasks are impossible. He asked how you justify the impossibility of a supertask. All your arguments seem to hinge on a variant that there isn't a largest natural number.

By definition supertasks are non-terminating processes — Michael

The wiki definition you gave made no mention of 'terminate'. If you mean that it doesn't complete, it by definition does in a finite time. If you mean that it has no terminal step, then you're making the mistake I identify just above since the definition does not require one.

Tasks are performed ad infinitum. I never stop counting. — Michael

You also wield the term 'ad infinitum', which typically means 'going on forever', which also violates the definition which explicitly requires a finite time to the task You very much do stop counting at time 1. There is at that time not another number, so by counterexample, your assertion that you will never stop counting is false.

If you mean that it doesn't complete, it by definition does in a finite time. If you mean that it has no terminal step, then you're making the mistake I identify just above since the definition does not require one. — noAxioms

How can a sequence of operations in which each operation is performed only after the previous operation is performed complete without there being a final operation?

You just seem to hand-wave this away with no explanation.

You also wield the term 'ad infinitum', — noAxioms

Well, yes. That's how to define it as an infinite sequence of operations rather than a finite sequence of operations.