Math Q&A Part 2

Apr 2014
320
156
Greater London, England, UK
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. :p
 
Apr 2014
320
156
Greater London, England, UK
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:

CRGreathouse

Forum Staff
Nov 2006
16,046
936
UTC -5
It doesn't work -- $\omega^\omega$ is countable. (And yes, $\omega$ is the one you want here, not $\aleph_0$.)
 
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