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 above
marking scheme.

But I think you've fallen victim to a common fallacy.

When trying to determine whether or not an infinite set is
countable, 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 to
fix your failed counting. I only have to provide a successful
counting.

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 all
positive integers and there's no available markers for the
missing positive integers.

Does that mean the set of positive integers is not countable?
Of course not. Sure, the above counting fails, but a failed
counting 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