Abstract
Using Riemann’s Rearrangement Theorem, Øystein Linnebo (2020) argues that, if it were possible to apply an infinite positive weight and an infinite negative weight to a working scale, the resulting net weight could end up being any real number, depending on the procedure by which these weights are applied. Appealing to the First Postulate of Archimedes’ treatise on balance, I argue instead that the scale would always read 0 kg. Along the way, we stop to consider an infinitely jittery flea, an infinitely protracted border conflict, and an infinitely electric glass rod.
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Notes
By an “infinite weight,” I mean a countably infinite weight. I will not consider the possibility of even greater, uncountable weights (if, indeed, it is a possibility). Likewise for infinite heights and other infinite quantities.
A similar discontinuity occurs in the Ross-Littlewood paradox (Allis and Koetsier 1991). Begin with an empty urn. At the nth step of the supertask, add ten balls numbered \((10n - 9)\) up to 10n, and remove ball number n. Though the total number of balls in the urn approaches infinity, no ball survives until the end.
By contrast, the Balance Principle predicts that no component weight or counterweight individually makes an essential contribution to the final net weight, since each one is only finite. So on the picture I advocate, the spring scale would stay balanced at 0 kg if any individual weight or balloon were removed.
To see that the sum of the removed weights converges to \(\frac{1}{2}\ln {(2)}\), note that:
$$\begin{aligned} \begin{array}{lrcccccccccccccccc} &{} \ln (2) &{} = &{} 1 &{} + &{} 0 &{} - &{} \frac{1}{2} &{} + &{} \frac{1}{3} &{} + &{} 0 &{} - &{} \frac{1}{4} &{} + &{} \frac{1}{5} &{} + &{} \dots \\ - &{} \frac{1}{2}\cdot \ln (2) &{} = &{} 0 &{} - &{} \frac{1}{2} &{} + &{} \frac{1}{4} &{} + &{} 0 &{} - &{} \frac{1}{6} &{} + &{} \frac{1}{8} &{} + &{} 0 &{} - &{} \dots \\ \hline \\ &{} \frac{1}{2}\cdot \ln (2) &{} = &{} 1 &{} - &{} \frac{1}{2} &{} - &{} \frac{1}{4} &{} + &{} \frac{1}{3} &{} - &{} \frac{1}{6} &{} - &{} \frac{1}{8} &{} + &{} \frac{1}{5} &{} - &{} \dots \end{array} \end{aligned}$$The top row is the familiar identity \(\ln {(2)} = \sum ^{\infty }_{n=0} \frac{(-1)^n}{n+1}\) with zeroes interspersed at regular intervals. The middle row is the result of multiplying all terms of that identity by \(- \frac{1}{2}\), and interspersing zeroes. The final row is obtained by adding the first two together.
To elaborate, here are my tentative two cents about each of these three supertasks:
The Flea. Note a crucial disanalogy with Linnebo’s scale. While the scale has infinite weights and counterweights pressing on it, there are no synchronous constraints at all on the location of the flea. That means the only thing that could settle the flea’s end point is the infinite winding path it takes to get there. In my view, this makes it unsurprising that the flea’s end point depends on its path, and so on the order of its jumps. So in this case, I find Linnebo’s prediction quite plausible, viz. that the flea ends up at \(\ln {(2)} \approx 0.69\) inch. Any alternative would involve teleportation, a violation of the cherished locality principle. This makes it reasonable to maintain the continuity of locomotion where possible, even if the more general Continuity Principle is abandoned.
Annexia. The final area of Annexia is a function of the particular regions that end up under Annexian control. Within the given parameters, Annexia could end up with 0.69 mi\(^2\) of additional territory: this happens if the trades involve an ever-shrinking border region between the two nations, so that almost every location has a final claimant. But that optimal outcome is not at all guaranteed. If the swaps are more disparate, there could be large areas that are traded back and forth indefinitely, without a final claimant. Suppose ownership of those areas remains unsettled. Then, far from gaining any land, Annexia could be left without a settled claim to any territory at all, provided its neighbour is strategic enough to organise the swaps in a suitable way.
The Rod. The positive charge of the rod is generated by the electron deficit in the rod (or, if you prefer, by the proton surplus). After the first step, the rod has an electron deficit of 1 coulomb. That deficit is subsequently compensated in part by electrons added later in the process. But how many of those compensating electrons last until the end? Assuming it makes sense to re-identify token electrons (cf. Huggett 1999), the answer depends on which individual electrons are removed at each step. For suppose the electrons exit the rod in the same order as they entered. Then each electron that enters the rod during this supertask will be removed at some later step. Consequently, the charge in the rod will then be 1 coulomb or greater at the end of the task, since nothing remains to compensate the initial deficit. Alternatively, it could be that some number of newly added electrons do stick around, which could lead to any intermediate outcome between 0.69 and 1 coulomb.
References
Allis, V., & Koetsier, T. (1991). On some paradoxes of the infinite. British Journal for the Philosophy of Science, 42(2), 187–194. https://doi.org/10.1093/bjps/42.2.187.
Benacerraf, P. (1962). Tasks, super-tasks and the modern eleatics. Journal of Philosophy, 59(24), 765–784. https://doi.org/10.2307/2023500.
Heath, T. L. (1897). The Works of Archimedes. USA: Dover Publications.
Huggett, N. (1999). Atomic metaphysics. Journal of Philosophy, 96(1), 5–24. https://doi.org/10.2307/2564646.
Linnebo, Ø. (2020). Riemann’s scale: A puzzle about infinity. Erkenntnis, online first. https://doi.org/10.1007/s10670-020-00345-x.
Acknowledgements
Thanks to Chris Scambler and a referee for comments on an earlier draft, to Dane Stocks for research assistance, and to Øystein Linnebo for his delightful paradox.
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Hoek, D. Øystein vs Archimedes: A Note on Linnebo’s Infinite Balance. Erkenn 88, 1791–1796 (2023). https://doi.org/10.1007/s10670-021-00421-w
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DOI: https://doi.org/10.1007/s10670-021-00421-w