27 - This gives us an argument that can be set out like...

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Unformatted text preview: This gives us an argument that can be set out like this: a. We know that no Z-run reaches G. b. Suppose R makes all the Z-runs and no other runs. c. Then R has not reached G. [From (a). Reason: no Z-run reaches G, and R has not made any other runs.] d. But R cannot be to the left of G. [From (b). Reason: R has made all the Z runs, and if R were to the left of G, there would still be Z-points between R and G, so not all the Z-runs would have been made.] e. So R has reached G. [From (d)] Since (b) leads to a contradiction [(c) contradicts (e)], it is logically impossible for (b) to be true. Therefore, f. It is impossible for R to make all the Z-runs. 0 Does the argument work? Thomson thinks that it does not, because it does not prove (f), but only a weaker conclusion: It is impossible for R to make all the Z -runs without making a run that is not a Z -run. So Thomson’s response to Zeno is that one can make all the Z-runs simply by making a run (e.g., a G-run) that is not a Z-run. To think this is impossible requires an equivocation on the word ‘run’. This turns out to be a weak reply. For although it shows that Zeno doesn’t get his contradiction from the assumption that R makes all the Z-runs, it concedes too much to Zeno. For it supposes that there is some description of a super-task (“making all the Z-runs and no other runs”) that does lead to a contradiction. That is, Thomson maintains both: i. “R makes all the Z-runs and no other runs” entails “R reaches G” and ii. “R makes all the Z-runs and no other runs” entails “R does not reach G.” But as Paul Benacerraf has shown (in the article “Tasks, Super-tasks, and the Modern Eleatics,” on e-reserve) neither of these entailments holds. That is, we cannot derive a contradiction even from the assumption that R makes all the Z-runs and no others. [ Benacerraf’s key claim: From a description of the Z-series, nothing follows about any point outside the Z-series. We can apply this point to both the lamp and the race course: a. The lamp: Nothing about the state of the lamp after two minutes follows from a description of the lamp’s behavior during the two-minute interval when the super-task was being performed. It does not follow that the lamp is on; it does not follow that the lamp is off. It could be either. b. The race course: Nothing about whether and when the runner reaches G follows from the assumption that he has made all the Z-runs and no others. This is because G is the limit point of the infinite sequence of Z-points. It is not itself a Zpoint. If we assume that the runner makes all the Z-runs and no other runs, we have the following options about G. It can be either: a. The last point R reaches, or b. The first point R does not reach. It must be one of these, but it does not have to be both. Benacerraf explains why (“Tasks, Super-Tasks and the Modern Eleatics,” p. 117-118): “… any point may be seen as dividing its line either into (a) the sets of points to the right of and including it, and the set of points to the left of it; or into (b) the set of points to the right of it and the set of points to the left of and including it: That is, we may assimilate each point to its righthand segment (a) or to its left-hand segment (b). Which we choose is entirely arbitrary …” Consequently, both of the following situations are possible: a. R makes all the Z-runs and no others, and reaches G. b. R makes all the Z-runs and no others, and does not reach G. All that “R makes all the Z-runs and no others” entails is that R reaches every point to the left of G, and no point to the right of G. It entails nothing about whether G itself is one of the points reached or one of the points not reached. So while Thomson thinks that both entailments (i) and (ii) hold, Benacerraf shows that neither of them holds. T Benacerraf’s vanishing genie suppose the runner is a genie who vanishes as soon as he makes all the Z-runs. There is a temptation to say that there must be a last point he reaches before he vanishes. And that would have to be G. So how is it possible for him to make all the Z-runs without reaching G? Benacerraf gives us a beautiful illustration of this possibility by adding one new wrinkle — a shrinking genie: [“Tasks, Super-Tasks and the Modern Eleatics,” p. 119]: “Ours is a reluctant genie. He shrinks from the thought of reaching 1. In fact, being a rational genie, he shows his repugnance against reaching 1 by shrinking so that the ratio of his height at any point to his height at the beginning of the race is always equal to the ratio of the unrun portion of the course to the whole course. He is full grown at 0, half-shrunk at ½, only 1/8 of him is left at 7/8, etc. His instructions are to continue in this way and to disappear at 1. Clearly, now, he occupied every point to the left of 1 (I can tell you exactly when and how tall he was at that point), but he did not occupy 1 (if he followed instructions, there was nothing left of him at 1). Of course, if we must say that he vanished at a point, it must be at 1 that we must say that he vanished, but in this case, there is no temptation whatever to say that he occupied 1. He couldn’t have. There wasn’t enough left of him.” ...
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This note was uploaded on 01/31/2011 for the course PHILOSOPHY 101 taught by Professor Markelwin during the Summer '09 term at UC Davis.

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