You can define the terminal state to be on, off, or a plate of spaghetti and be consistent with the rules of the game. — fishfry

No you can't. I addressed this in my initial defence of Thomson here, and even more clearly below.

You're confusing yourself on this point.

You're claiming that "a plate of spaghetti" is a coherent answer to the question "is the lamp on or off after two minutes?"

So I think the confusion is yours.

That's funny, because according to SEP, he used the asymptotic density of the sequence 0, 1, 0, 1, ... to argue that the final state must be 1/2. He made that exact argument using that exact sequence. So you are mistaken, because Thompson has used the exact same reasoning I did. — fishfry

He discusses the sequence and its sum, but only to show its irrelevancy, hence the earlier quote.

From his paper:

What is the sum of the infinite divergent sequence +1, -1, +1, ...? Now mathematicians do say that this sequence has a sum; they say that its sum is 1/2. And this answer does not help us, since we attach no sense here to saying that the lamp is half-on.

If there is a sense of "know" that means "acknowledging, recognizing", then you are saying that we do know another's pain (at least, sometimes). I agree, but this is contrary to your earlier statements that we do not know another's pain. — Luke

Is there a difference between knowing someone's pain and knowing that someone is in pain?

Because there's certainly a difference between knowing someone's mother and knowing that someone has a mother.

BUT!!!!!! You have not defined the terminal state. So why do you think there should be a sensible answer for what it is?

I don't see how your expressing the problem in pseudocode adds anything. We all have agreed to it long ago, even before you wrote it down. That's the premise of the problem. But the question is about the terminal state, which is not defined. — fishfry

The terminal state isn't just undefined; any proposed terminal state is inconsistent. The lamp cannot be either on or off after two minutes even though it must be either on or off after two minutes. This is a contradiction, therefore it is impossible to have pushed the button an infinite number of times.

After all, there is no number that can serve as the limit of the sequence 0, 1, 0, 1, ... — fishfry

We're discussing the consequence of having pushed a button an infinite number of times, not the limit of some infinite sequence of numbers. These are two different things.

As Thomson says, "the impossibility of a super-task does not depend at all on whether some vaguely-felt-to-be associated arithmetical sequence is convergent or divergent."

I don’t understand your reasoning at all.

Here's an even simpler demonstration:


In this case there is no wait time between operations; the next occurs "immediately".

Does this script end in an instant, or does it continue forever? If it ends in an instant, does echo isLampOn output true or false?

But given the meaning of while (true), it must continue forever; echo isLampOn never runs.

If you think that this changes by introducing a wait time within the loop then you've obviously made a mistake.

division by 2 ... results in even number of total divisions. — SpaceDweller

I don't know what you mean by this.

The wait(i *= 0.5) simply means that pushButton is called at successively halved intervals of time, i.e. it is called for the first time after 60 seconds, for the second time after a further 30 seconds, for the third time after a further 15 seconds, and so on.

Under assumption that those 2 minutes must pass the lamp will therefore be off with console.log(isLampOn) — SpaceDweller

Given what while (true) { ... } means, it is logically impossible for console.log(isLampOn) to ever run.

2 minutes must pass, and while (true) { ... } must still be running after those 2 minutes.

@fishfry

You wanted something new so see the above.

Here's some pseudocode to demonstrate Thomson's lamp:


According to those who claim that supertasks are possible, after two minutes the echo isLampOn command will run. This is the first contradiction; the while (true) command ensures that this cannot happen.

But even assuming that echo isLampOn does run, it cannot output true and cannot output false, but also cannot output anything other than true or false. This is the second contradiction.

fishfry's solution is to inject some additional code after while (true) { ... }, assigning some arbitrary value to isLampOn, but in doing so he is no longer addressing the problem as posed.

See also this.

What is wrong with that? — Ludwig V

There's nothing wrong with defining, or performing, a recursive function. There is a problem with claiming that it is possible to have completed a recursive function.

You may say that it is possible to divided indefinitely, but that does not mean that infinite divisibility is possible. Take pi for example. You can get a computer to produce the decimal extension for pi, "indefinitely", but you never succeed in reaching an infinite extension. Divisibility is the very same principle. Some mathematical principles allow one to divide indefinitely, but you never reach infinite division. That is because infinite division, therefore infinite divisibility, is logically impossible. — Metaphysician Undercover

Thomson makes a similar point:

People have, I think, confused saying (1) it is conceivable that each of an infinity of tasks be possible (practically possible) of performance, with saying (2) that is conceivable that all of an infinite number of tasks should have been performed. They have supposed that (1) entails (2). And my reason for thinking that people have supposed this is as follows. To suppose that (1) entails (2) is of course to suppose that anyone who denies thinking (2) is committed to denying (1). Now to deny (1) is to be committed to holding, what is quite absurd, (3) that for any given kind of task there is a positive integer k such that it is conceivable that k tasks of the given kind have been performed, but inconceivable, logically absurd, that k + 1 of them should have been performed. But no-one would hold (3) to be true unless he had confused logical possibility with physical possibility. And we do find that those who wish to assert (2) are constantly accusing their opponents of just this confusion. They seem to think that all they have to do to render (2) plausible is to clear away any confusions that prevent people from accepting (1).

But there's nothing special about the lamp. It is impossible to complete any action an infinite number of times. — Ludwig V

Yes, as I further explained in this comment.

No. I'm saying that there's no natural way to define the terminal state. There are lots of ways to defined it. I define it as a plate of spaghetti. That's entirely consistent with the rules of the lamp problem, which only defines the state of the lamp at the points of the sequence — fishfry

No it's not, as explained here.

and does not appear to engage with any of the points I've made — fishfry

As far as I can see I've addressed everything you've said.

After completing the supertask the lamp must be either on or off, but as I explain here, Thomson's lamp shows that if we have pushed the button an infinite number of times then it is logically impossible for the lamp to be either on or off after the supertask is completed. This is a contradiction, therefore Thomson's lamp shows that it is logically impossible to have pushed a button an infinite number of times.

There is a fundamental problem with identifying supertasks with series limits — sime

This is the kind of mistake that Benacerraf makes in his response to Thomson, as explained here.

The lamp is not defined as being on or off at particular times; it is turned on or off at particular times by pushing a button.

It is an important distinction that some are failing to acknowledge.

P2 is false. As shown by P1. — fishfry

The argument form I am using is called modus tollens and is valid:

If P then Q. Not Q. Therefore, not P.

P = we can recite the natural numbers at successively halved intervals of time

Q = we can recite every natural number in finite time

“If P then Q” does not prove “P” and does not prove “Q”. So you are wrong to claim that P1 shows that P2 is false.

How in this great vast wonderful world of ours, does P3 justify P2? They're not even related. — fishfry

If we stop on some finite number then we don't recite every natural number. If we never stop then at no time have we recited every natural number. Therefore P3 entails P2.

(1) The sequence 1, 2, 3, 4, ... never stops. It has no last element. You can always find the next one.

(2) Under the successive halving hypothesis, all numbers are counted. Because as can be plainly seen, there is no number that isn't. — fishfry

The "successive halving hypothesis" leads to contradictions – namely Thomson's lamp and reciting every natural number in finite time – and so it is necessarily false.

But it is not inconceivable, and therefore is is not necessarily false, and therefore it is not metaphysically impossible. — fishfry

It entails contradictions. Therefore either it isn't conceivable or contradictions are conceivable. If the latter then being conceivable does not entail metaphysical possibility. If the former then you are simply mistaken in claiming it conceivable; you are failing to fully understand what it means to perform a supertask.

Thompson's lamp does not lead to a contradiction showing that supertasks are impossible. That's your interpretation, which you are failing to explain or defend to my satisfaction. — fishfry

I explained it here. I think it very clearly shows that having pushed a button an infinite number of times leads to a contradiction. And here I explain that this reasoning extends to all supertasks.

I simply do not understand why you jump to saying that means it's metaphysically impossible. — fishfry

Because it leads to contradictions as shown by Thomson's lamp, defended here and expanded on here.

Also because it's the conclusion of this sound argument:

P1. If we can recite the natural numbers at successively halved intervals of time then we can recite every natural number in finite time
P2. It is metaphysically impossible to recite every natural number in finite time
C1. Therefore, it is metaphysically impossible to recite the natural numbers at successively halved intervals of time

I justify P2 with this tautology:

P3. If we start reciting the natural numbers then either we stop on some finite number or we never stop

You could probably help me out by clearly defining metaphysically impossible. — fishfry

Metaphysical impossibilities are things which are necessarily false; e.g. see Kripke's Naming and Necessity in which he argues that "water is H₂O" is necessarily true even though not a priori (i.e. logically necessary).

But I would even go so far as to say that supertasks are logically impossible (as shown by the above argument and Thomson's lamp). I simply went for the phrase "metaphysical impossibility" because it's the weaker claim and so easier to defend.

Can you clarify which sense you mean? — fishfry

Metaphysical impossibility. Supertasks cannot be performed in any possible world. P3 is a tautology, P2 follows from P3, and so C1 is necessarily true.

When a mathematician says that 1/2 + 1/4 + 1/8 + ... = 1, they don't mean that you can perform this calculation with pencil and paper before lunchtime. They mean that the two expressions on either side of the equal sign denote the same real number. — fishfry

Here are three distinct propositions:

a) 1/2 + 1/4 + 1/8 + ... = 1
b) there is a bijection between this geometric series and the natural numbers
c) it is metaphysically possible to recite the natural numbers at successively halved intervals of time

(a) and (b) are true and (c) is false. Your argument rests on the assumption that (c) follows from (a) and (b), but it doesn't. (c) is proven false by P3, as well as arguments like Thomson's lamp.

You can continually assert that (a) and (b) are true, and I will continually agree, but until you can present actual evidence or reasoning to support (c), I will always reject it as per the above.

You should read beyond the quote to where I respond to explain the error in Benacerraf‘s reasoning.

I’ll keep it simple @fishfry. This is my argument:

P1. If we can recite the natural numbers at successively halved intervals of time then we can recite every natural number in finite time
P2. We cannot recite every natural number in finite time
C1. Therefore, we cannot recite the natural numbers at successively halved intervals of time

I justify P2 with this tautology:

P3. If we start reciting the natural numbers then either we stop on some finite number or we never stop

See also here for my defence of Thomson's lamp and here where I explain that this reasoning applies to all supertasks, further justifying C1.

What is "evidence" in a metaphysical realm? — jgill

That’s for those who assert the truth of (2) to answer. As it stands it’s just an assertion, and contradicts the tautology given in (1), so I have every reason to reject it.

How do you make this conclusion? — Metaphysician Undercover

It’s the conclusion of those who use the finite sum of a geometric series as proof that a supertask can be completed. If I recite the first number after 30 seconds, the second after 15 seconds, and so on, then I have recited them all and so stopped after 60 seconds, even though there is no largest number for me to stop on.

I think it’s nonsense as it contradicts the tautology given in (1), and so I reject (2).

To make this very simple, we have two competing claims:

1. If we start reciting the natural numbers then either we stop on some finite number or we never stop

2. It is metaphysically possible to recite the natural numbers at successively halved intervals of time

If (2) is true then we can stop without stopping on some finite number.

Some take this as proof that (1) is false. I take this as proof that (2) is false.

I think that (1) is a tautology whereas no evidence has been offered in support of (2).

But that's YOUR hypothesis, not mine. — fishfry

It's not mine. It's the hypothesis of those who claim that supertasks are possible. They try to use such things as the finite sum of a geometric series to resolve Zeno's paradox. They claim that because time is infinitely divisible it's possible for us to perform a succession of operations that correspond to a geometric series, and so it's possible to complete an infinite succession of operations in finite time.

I have been arguing firstly that it hasn't been proven that time is infinitely divisible and secondly that if we assume such a possibility then contradictions such as Thomson's lamp follow.

I was very clear on this in my reply to you on page 4, 22 days ago:

We can determine whether or not something entails a contradiction. If time is infinitely divisible then supertasks are possible. Supertasks entail a contradiction. Therefore, time being infinitely divisible entails a contradiction. — Michael

Most of the last few pages has been me trying to re-explain this to you, e.g. 10 days ago:

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. — Michael

---

You have repeatedly asked me what happens if we go backwards, saying "1" at 60 seconds, "2" at 30 seconds, and so forth. That also is a purely hypothetical thought experiment. Why on earth are you proposing hypothetical non-physical thought experiments, then saying, "Oh that's impossible!" when I attempt to engage? — fishfry

It was brought up for two reasons. The first was to address the flaw in your reasoning. That same post 10 days ago was very clear on this:

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?

...

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?

...

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. — Michael

If argument 1 is proof that it is possible to have recited the natural numbers in ascending order then argument 2 is proof that it is possible to have recited the natural numbers in descending order.

It is impossible to have recited the natural numbers in descending order.

Therefore, argument 2 is not proof that it is possible to have recited the natural numbers in descending order.

Therefore, argument 1 is not proof that it is possible to have recited the natural numbers in ascending order.

The second reason I brought it up was a proof that it is impossible to have recited the natural numbers in ascending order.

If it is possible to have recited the natural numbers in ascending order then it is possible to have recorded this and then replay it in reverse. Replaying it in reverse is the same as reciting the natural numbers in descending order. Reciting the natural numbers in descending order is impossible. Therefore, it is impossible to have recited the natural numbers in ascending order.

Or if you don't like the specific example of a recording, then the metaphysical possibility of T-symmetry might suffice.

Either way, the point is that it's special pleading to argue that it's possible to have recited the natural numbers in ascending order but not possible to have recited them in descending order. It's either both or neither, and it can't be both, therefore it's neither.

You yourself proved P2 true — fishfry

No I didn't.

Your argument is analogous to this:

If I am immortal then when will I die of old age? I won't. Therefore, I have proved that I am immortal.

Agreeing with what follows if we can recite the natural numbers at successively halved intervals of time doesn't prove that we can recite the natural numbers at successively halved intervals of time.

But you just proved P2 yourself! You agreed that under the hypothesis of being able to recite a number at successively halved intervals of time, there is no number that is the first to not be recited. — fishfry

I agreed that if P2 is true then C1 is true, as I have agreed from the beginning.

This doesn't prove that P2 is true.

In favour of eternalism there's the relativity of simultaneity and the Rietdijk–Putnam argument.

I have agreed repeatedly that we can't "count all the natural numbers backwards" since an infinite sequence has no last element. — fishfry

So we're back to my post here:

a. 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

b. 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

Here is our premise:

P1. In both (a) and (b) there is a bijection between the series of time intervals and the series of natural numbers and the sum of the series of time intervals is 60.

However, the second supertask is metaphysically impossible. It cannot start because there is no largest natural number to start with. Therefore, P1 being true does not entail that the second supertask is metaphysically possible.

Therefore, P1 being true does not entail that the first supertask is metaphysically possible.

You accept that (b) is impossible but you claim that (a) is possible. You have to prove this. P1 doesn't prove it.

Given P2, what is the first natural number not recited? I seem to remember having asked you this several times already. — fishfry

There isn't one. I've answered this several times already. That's what it means for me to accept P1.

But you need to prove P2. You haven't done so.

People disagree with the premise because we are not confident we can use such intuitions when the — unintuitive — concept of infinity is involved. — Lionino

The important part from that post is this:

The fallacy in his reasoning is that it does not acknowledge that for all t_n >= t_1/2 the lamp is on iff the lamp was off and I pressed the button to turn it on and the lamp is off iff the lamp was on and I pressed the button to turn it off.

If the lamp is on at t₁ then it must have been either turned on at t₁ or turned and left on before t₁, neither of which are allowed given the supertask, hence the contradiction.

Me and fishfry have insisted that this is a case of missing limit. — Lionino

That's why it's impossible to complete.

You think that the end of the sequence at t=1 is a temporal/logical consequence of what happens before. — Lionino

Yes, I address that here.

Any given example does not prove that supertasks in general are necessarily impossible. — Lionino

I addressed this here and here.

True. And that implies time is discrete how? — Lionino

If time is continuous then supertasks are logically possible. Supertasks are logically impossible. Therefore, time is discrete.

I'm not trying to find a solution, just to understand what's going on. Not so much why it's wrong, but why anyone would think it was right. Where does the illusion come from? — Ludwig V

They're clearly being confused by maths. They think that because a geometric series of time intervals can have a finite sum and because this geometric series has the same cardinality as the natural numbers then it is possible to recite the natural numbers in finite time. Their conclusion is a non sequitur, and this is obvious when we consider the case of reciting the natural numbers (or any infinite sequence) in reverse.

There is a far more fundamental problem, and they're just ignoring it. I have no idea why. Perhaps because they can't look beyond the maths to what it would mean for us to actually carry out the tasks. This seems to be the mistake that Benacerraf made in his response to Thomson and which I addressed here.

It isn't a physically possible task. — noAxioms

It's not just physically impossible, it's logically impossible. No physics can allow me to begin reciting the natural numbers in reverse. I can't even say one number, let alone all of them. And this is true even if we're not reciting the natural numbers in reverse but the sequence {0, 1, 0, 1, ...} in reverse, i.e. where each term, individually, can be recited in less than a second.

That there is no first number to recite is the very reason that it is logically impossible to begin reciting them in reverse and it astonishes me that not only can't you accept this but you twist it around and claim that it not having a first number is the reason that it can begin without a first number.

Though I don't quite see how your B2 follows from your B1. — Ludwig V

It was redefined as 1 at t_1/2 and never changed again, so is still defined as 1 at t₁.

You mean that we don't know the state of X at the last step before t(1), even though X must have been in one state or the other? — Ludwig V

There is no last step before t₁, hence no coherent definition of X at t₁. But also at no point between t₀ and t₁ is there a step where X goes from being defined (as either "0" or "1") to being undefined, and the definition of X is always retained until redefined to something else. It's a simple contradiction.

If you're trying to find a "solution" you won't find one. We just have to accept that supertasks are illogical. It's that easy.

This puzzles me. Is this t(1) the same t as the t(1) in C3? It can't be. There must be a typo there somewhere. — Ludwig V

No, it was three separate situations. Sorry if that wasn’t clear.

One question, then - The state of X at any t(n), depends on its predecessor state at t(n-1), doesn't it? Isn't that a definition? Why is it inapplicable to t(1)? — Ludwig V

It is applicable to t₁, but given the supertask described in P3 there’s no coherent answer to the definition of X at t₁ (no final redefinition before t₁) proving P3 to be impossible.

I have wondered whether one could replace the Thompson lamp with a question, such as whether the final number was odd or even. That would work if you start with an odd divisor and don't express everything in decimals. Perhaps it would work for all examples if you ask whether the number of steps taken is odd or even when the minute is up. I think. — Ludwig V

P1. When the letter X is given a definition it retains this definition until it is redefined.

A1. At t₀ X = 0
A2. Therefore, at t₁ X = 0

B1. At t₀ X = 0 and then at t_1/2 X = 1
B2. Therefore, at t₁ X = 1

C1. At t₀ X = 0 and then at t_1/2 X = 1 and then at t_3/4 X = 0, and so on ad infinitum
C2. Therefore, at t₁ X = ?

In all cases the definition of X at t₁ must be a logical consequence of what occurs between t₀ and t₁.

Given that in C2 X cannot be defined as either "0" or "1" but must be defined as either "0" or "1" then C1 is necessarily false. The supertask described in C1 is impossible.

This addresses the very logic of a supertask without some dependency on a physical performance.

Consider the infinite sequence {0, 1, 0, 1, 0, 1, ...}.

Now consider reciting its terms in reverse.

To recite its terms in reverse I am only allowed to say "0" or "1" but I cannot start by saying "0" and I cannot start by saying "1". Therefore I cannot start.

No appeal to a geometric series of time intervals can save you from this.

The lack of a first step does not prevent the beginning of the task — noAxioms

It literally does.

I described exactly how to do that — noAxioms

No you didn't. You ignored it and just said "when the time comes I say the next number". That doesn't explain how the recitation can begin without a first number to say.

I am right now trying to recite the natural numbers in descending order but am silent because I cannot begin. It's been 60 seconds. Not only have I failed to recite them all, but I have failed to recite even one. Help me out here.

According to this, "many philosophers have argued that relativity implies eternalism. Philosopher of science Dean Rickles says that, "the consensus among philosophers seems to be that special and general relativity are incompatible with presentism." Christian Wüthrich argues that supporters of presentism can salvage absolute simultaneity only if they reject either empiricism or relativity."