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 February 10th, 2015, 02:08 AM #231 Senior Member     Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 155 Math Focus: Abstract algebra Yeah, so you got rid of $\omega^2$ coins from $t=0$ to $t=2$. Now repeat the whole thing twice as fast, then you can get rid of the next $\omega^2$ from $t=2$ to $t=3$. Speed up again, and you can get rid of the third $\omega^2$ from $t=3$ to $t=3.5$. And so on, ad infinitum. Oh well, maybe the poor gambler will be stuck in hell for all eternity after all. At least he won’t be bored.
 February 10th, 2015, 04:29 AM #232 Senior Member     Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 155 Math Focus: Abstract algebra Okay, let’s do this slightly differently. Put the first coin in at $t=0$, the next at $t=0.5$, then at $t=0.75$, etc. By $t=1$, $\aleph_0$ coins will be disposed of. Let’s write $\omega$ for $\aleph_0$; it’s easier. So by $t=1$, $\omega$ coins disappear. Now speed things up twice as fast. Then by $t=2$, $\omega^2$ coins will have disappeared. But now we can adjust the speed in such a way that the next $\omega^2$ coins will disappear by $t=2.25$, the next after by $t=2.375$, and so on … so this time $\omega^3$ coins disappear by $t=2.5$. Now carry on adjusting the speed so that $\omega^4$ coins disappear by $t=2.75$, $\omega^5$ by $t=2.875$, and so on. Result: by $t=3$, $\omega^{\omega}$ coins will be gone. But $\omega^{\omega}$ is uncountable! For $\omega^{\omega}>2^{\omega}=\left|\frak P(\mathbb N)\right|$ and we know that the power set of the natural numbers is uncountable. Since the gambler only has a countable amount of coins to get rid of, he might be able to do that in three seconds! What do you think? Last edited by Olinguito; February 10th, 2015 at 04:38 AM.
 February 10th, 2015, 06:42 AM #233 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms It doesn't work -- $\omega^\omega$ is countable. (And yes, $\omega$ is the one you want here, not $\aleph_0$.)

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