The Race Course: Part 2

1. Our look at the plurality argument suggests that Zeno may have thought that to run all the Z-runs would be to run a distance that is infinitely long. If this is what he thought, he was mistaken.

   The reason the sum of all the Z-intervals is not an infinitely large distance is that there is no smallest Z-interval. And Zeno does not establish that there is some smallest Z-run. (If there were a smallest Z-run, he wouldn't have been able to show that R had to make infinitely many Z-runs.) 
   
   

2. What about Aristotle's understanding of Zeno? Here is what he says:

   "Zeno's argument makes a false assumption when it asserts that it is impossible to traverse an infinite number of positions or to make an infinite number of contacts one by one in a finite time" (Physics 233a21-24). 
   
   

3. Aristotle points out that there are two ways in which a quantity can be said to be infinite: in extension or in divisibility. The race course is infinite in divisibility. But, Aristotle goes on, "the time is also infinite in this respect."

   Hence, there is a sense in which R has an infinite number of distances to cross. But in that sense he also has an infinite amount of time to do it in. (A finite distance is infinitely divisible, then why isn't a finite time also infinitely divisible?) 
   
   

4. So Zeno cannot establish (2) for either of the first two reasons we considered: to make all the Z-runs, R does not have to run infinitely far. Nor does R have to keep running forever.

Logical Impossibility: Infinity Machines & Super-Tasks

1. On this reading, Zeno's argument attempts to show that it is logically impossible for R to reach G. That is, Zeno's puzzle is not that the runner has to run too far, or that the runner has to run for too long a time, but that the claim that the runner has completed all the Z-runs leads to a contradiction.
   
   
2. Following James Thomson ["Tasks and Super-Tasks," on reserve], let us define a super-task as an infinite sequence of tasks. Can one perform a super-task? Bertrand Russell thought that one could:

   "Russell suggested that a man's skill in performing operations of some kind might increase so fast that he was able to perform each of an infinite sequence of operations after the first in half the time he had required for its predecessor. Then the time required for all of the infinite sequence of tasks would be only twice that required for the first. On the strength of this Russell said that the performance of all of an infinite sequence of tasks was only medically impossible."  ["Tasks and Super-Tasks," p. 93]

   But Thomson argues that to assume that a super-task has been performed in accordance with Russell's "recipe" leads to a logical contradiction. 
   
   

   1. Thomson's Lamp example ["Tasks and Super-Tasks," pp. 94-95]:

      "There are certain reading lamps that have a button in the base. If the lamp is off and you press the button the lamp goes on, and if the lamp is on and you press the button the lamp goes off. So if the lamp was originally off, and you pressed the button an odd number of times, the lamp is on, and if you pressed the button an even number of times the lamp is off. Suppose now that the lamp is off, and I succeed in pressing the button an infinite number of times, perhaps making one jab in one minute, another jab in the next half minute, and so on, according to Russell's recipe. After I have completed the whole infinite sequence of jabs, i.e. at the end of the two minutes, is the lamp on or off? It seems impossible to answer this question. It cannot be on, because I did not ever turn it on without at once turning it off. It cannot be off, because I did in the first place turn it on, and thereafter I never turned it off without at once turning it on. But the lamp must be either on or off. This is a contradiction." 
      
      

   2. Applying this to the race course ["Tasks and Super-Tasks," pp. 97-98].

      … suppose someone could have occupied every Z-point without having occupied any point external to Z. Where would he be? Not at any Z-point, for then there would be an unoccupied Z-point to the right. Not, for the same reason, between Z-points. And, ex hypothesi, not at any point external to Z. But these possibilities are exhaustive. The absurdity of having occupied all the Z-points without having occupied any point external to Z is exactly like the absurdity of having pressed the lamp-switch an infinite number of times….
      
      

3. This gives us an argument that can be set out like this:
   
   
   1. Suppose R makes all the Z-runs.
   2. Then R cannot be to the left of G. [Reason: if R is to the left of G, there are still Z-points between R and G, and so not all of the Z-runs have been made.]
   3. So R has reached G.
   4. But, since no Z-run reaches G, R has not reached G.

      Since (a) leads to a contradiction [(c) contradicts (d)], the argument continues, it is logically impossible for (a) to be true. Therefore,
      
      

   5. It is impossible for R to make all the Z-runs.
      
      
4. Does the argument work? There are two parts:
   
   
   1. Does (a) "R makes all the Z-runs" entail (c) "R reaches G"?
   2. Does (a) "R makes all the Z-runs" entail (d) "R does not reach G"?

      It turns out that (as Paul Benacerraf has shown, see "Tasks, Super-tasks, and the Modern Eleatics", on reserve) neither of these entailments holds.
      
      

5. We'll start with (ii). The reason for supposing that R does not reach G is that no Z-run reaches G. So we must be assuming (as Thomson actually says) that R makes all the Z-runs and no others. So now we ask: how can R reach G if the only runs he makes are Z-runs, and no Z-run reaches G? How can one run to G without making a run that reaches G?

   Now it appears that what leads to a contradiction is the assumption that R makes all the Z-runs and no others. This allows for two possible replies to Zeno. 
   
   

   1. The weak reply: Zeno is entitled to assume that R makes all the Z-runs. But he is not entitled to assume that R makes all the Z-runs and no others. So he doesn't get his contradiction.
   2. A stronger reply (Benacerraf): we cannot derive a contradiction even from the assumption that R makes all the Z-runs and no others.
      
      
6. 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: 
   
   

   1. 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.
      
      
   2. 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.
      
      
7. This is because G is the limit point of the infinite sequence of Z-points. It is not itself a Z-point. 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:
   
   
   1. The last point R reaches, or
   2. 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 right-hand segment (a) or to its left-hand segment (b). Which we choose is entirely arbitrary …"

      Consequently, both of the following situations are possible:
      
      

   1. R makes all the Z-runs and no others, and reaches G.
   2. 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.
      
      

8. Consider 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 the second situation possible?

   Benacerraf illustrates this beautifully 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; only1/8 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." 
   
   

9. The mistake in Thomson's argument (which tries to show that a contradiction can be derived from the assumption that the runner makes all the Z-runs and no others) is to assume that one and the same point, G, has to be both the last one that R reaches and the first one that he doesn't reach.

   But this assumption is mistaken. G divides the space R traverses from the space that he does not traverse. But G itself cannot be said to belong to both spaces (even though it is arbitrary which of the two we associate it with). Indeed, if there is such a thing as the last point R (or anyone) reaches, then there cannot be a first point that he does not reach.

   The reason is that (as Zeno is assuming) space is a continuum; points in space do not have next-door neighbors. There is no next point after G. Therefore, if G is last point R reaches, then there is no first point R does not reach. Consequently, G cannot be that point. So Thomson's argument fails.

   Movement through a continuum, through infinitely divisible space, is indeed a puzzling phenomenon. But it does not lead to Zeno's paradox.

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• This page was last updated on 11/12/00.