```Date: Mar 8, 2013 4:19 PM
Author: quasi
Subject: Re: Cardinality of turning wheel

netzweltler wrote:>quasi wrote:>>>> Any infinite set of pairwise disjoint intervals on the real>> line is countably infinite since each interval contains a>> distinct rational number.>>t = 0 s: I am marking >[0, 0.5] #0, [0.5, 0.75] #-1, [0.75, 0.875] #-2, ...>>t = 0.5 s: I am marking >[0.5, 0.75] #0, [0.75, 0.875] #-1, [0.875,0.9375] #-2, ...>>t = 0.75 s: I am marking [0.75, 0.875] #0, [0.875, 0.9375] #-1,>[0.9375, 0.96875] #-2, ...>...>>Which segments have been marked #0 after 1 s?I don't understand the mechanics or the intent of the abovemarking scheme.But I think you've fallen victim to a common fallacy.When trying to determine whether or not an infinite set iscountable, a failed counting _doesn't_ disprove countability.To prove countability, only one successful counting is needed.In other words, to prove countability, I don't have to try tofix your failed counting. I only have to provide a successfulcounting.As an example, suppose I try to count the set of positive integers, starting as follows:   2 <=> #1   4 <=> #2   8 <=> #3   16 <=> #4   32 <=> #5   ...The above counting fails since it clearly doesn't count allpositive integers and there's no available markers for themissing positive integers.Does that mean the set of positive integers is not countable?Of course not. Sure, the above counting fails, but a failedcounting don't disprove countability. To disprove countability,one would have to show that no successful counting is possible. Bottom line:One failed counting doesn't disprove countability.On the other hand, one successful counting proves countability.quasi
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