Supertasks Last All Summer Long

Suppose I have a supertask. I ask about it: what are the tasks that compose it? The answer for this particular task is: an infinity of supertasks. That is, each individual task in the supertask is itself a supertask. So I ask about these "2nd level" supertasks, what tasks are these composed of? The answer: supertasks. In fact, for any level of supertasks, those supertasks are each composed of an infinity of supertasks. So we get an infinite regress. There are a couple of puzzling things about this: first, it seems that we end up with an uncountable infinity of individual tasks being done in a finite amount of time - we have an infinity (an infinite number of individual tasks in the first supertask) ^infinity (the infinite regress of levels of supertasks, each level generating an infinity of tasks). So it seems that this was a hypertask (what the Stanford Encyclopedia of Philosophy says is "a non-numerable infinite sequence of actions or operations carried out in a finite interval of time") in disguise. The second weird thing: at the end of the day (or hour, or whatever extended temporal interval you choose) just what has been done? Are supertasks like these ever done? Are they always done? Is it just incomprehensible to talk about supertasks like these? There seems to be nothing incoherent about any of the constituent concepts, but I don't know what to think when they're put together in this way.

Comments

[Matt Weiner]

I'm not sure that the number of subtasks is uncountable--each level generates an infinite number of subtasks, but it seems to me these multiply rather than exponentiate.

This reminds me of the proof that the ordinal w^w^w^w^... is countable (where all those w's are omegas).

Explanation (er, sort of) of this number: w is the limit of the integers, w + 1 is its successor, w + 2 is its successor...
2w is the limit of that set, 2w + 1 is its successor, 2w + 2 is its successor
3w is the limit of that set..., 4w is the limit of that one...
w^2 is the limit of that set, w^2 + 1 is its successor... w^2 + w..., w^2 + 2w..., w^2 + 3w..., ..., 2w^2,...
3w^2..., 4w^2 ..., ... w^3,... w^4,... w^5...,
gets you to w^w--then you can do it again to get to w^w^w--and thence to w^w^w^w, and so on--and take the limit of those ordinals, and you've got a new ordinal (epsilon I think)--and even that is countable.

So maybe we have--level 1 task correspond to the members of w--level 2 tasks correspond to the members of w^2--etc., which gets you only to w^w, I think.

There is a very elegant proof which I forget that actually shows you how to map the elements of epsilon onto the natural numbers. I think maybe the powers of the first prime are the members of w, the powers of the second prime are the members of 2w, ..., then for the members of w^2 you take powers of the first prime times powers of the second prime, but I'm not able to reconstruct it all at the moment.

(And may I humbly suggest changing the title of the post to "Supertasks last all summer long"?--expl. at http://www.wired.com/wired/archive/5.01/ffsupertoys_pr.html)

[Matt Weiner]

http://www.wired.com/wired/archive/5.01/ffsupertoys_pr.html or just click on my name for this post--I'm not doing well with the HTML here.