Comments on: Making Money Disappear Through Infinite Iteration http://xorshammer.com/2008/08/22/supertasks-and-ordinals/ Some things in mathematical logic that I find interesting Sat, 03 Mar 2012 11:53:40 +0000 hourly 1 http://wordpress.com/ By: mkoconnor http://xorshammer.com/2008/08/22/supertasks-and-ordinals/#comment-112 Thu, 02 Oct 2008 11:23:35 +0000 http://xorshammer.wordpress.com/?p=100#comment-112 One way to think about the g(\alpha) is the following. Suppose that at some stop \alpha, you have exactly two dollar bills. The devil destroys one, but since we’re supposing that he’s helping you is determined to preserve the other. How might the devil decide to do it? When he gives you countably infinitely new bills at \alpha, he could choose a mapping from \omega (representing the bills he just gave you) onto some countable interval above \alpha, say \lbrack \alpha,\beta). Then you are safe at least until \beta.

However, you’ve gotten many new bills at the intervening stops and the devil can use them to preserve your dollar until some higher countable ordinal \beta_1. Using the dollar bills gained there, you will be safe until \beta_2, etc. But by stop \beta_\omega = \sup\{\beta_i\}, all of your “safety” dollar bills will be gone, and you will have just the one dollar bill left which the devil will have to destroy. This stop will be g(\alpha) under this strategy of the devil’s.

As you can see, there is not necessarily any particular structure on the g(\alpha)‘s, in that they can be unboundedly high up in the countable ordinals.

]]> By: Kenny Easwaran http://xorshammer.com/2008/08/22/supertasks-and-ordinals/#comment-110 Thu, 02 Oct 2008 05:31:46 +0000 http://xorshammer.wordpress.com/?p=100#comment-110 I follow the argument, but it’s still a bit confusing. Let’s assume for the sake of contradiction that at every stop you have at least two bills. Then consider the following strategy for the devil. Fix one of the bills you have at stop 0, and at any stop, choose a bill other than this one to destroy (this strategy just needs AC, and the fact that you always have at least two bills). Then you must have this bill at the end. So by the above proof, the assumption must be false, so there must be some stop at which you have only one bill (or none).

Clearly, such a stop can’t be at a successor ordinal, because at any successor ordinal you have countably many bills. I suppose it must be at one of those g(\alpha), though I guess I don’t really know much about what the structure of those ordinals is like.

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