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