Re: ....Infinity
- From: Dave Seaman <dseaman@xxxxxxxxxxxx>
- Date: Fri, 4 Nov 2005 17:15:05 +0000 (UTC)
On 4 Nov 2005 08:41:55 -0800, zuhair wrote:
> Dave Seaman wrote:
>> On 4 Nov 2005 07:38:54 -0800, zuhair wrote:
>> > Hellow every one.
>>
>> > Amillionaire who bought a pair of socks whenever he bought a pair of
>> > boots , and never at any other time , and who had a passion for buying
>> > both that at last he had Aleph-0 pairs of boots and Aleph-0 pairs of
>> > socks . The problem is: How many boots had he,and how many socks?
>>
>>
>> > Bertrand Russell.
>>
>> > Note: Some would think that the number of boots is Aleph-0 and of socks
>> > is Aleph-0
>> > But the author of this problem says it is not necessarily the case. The
>> > solution of the problem
>> > is related to the subject of selections and the multiplicative axiom.
>>
>> > Can any body explain that?
>>
>> It's easy to show that there are aleph_0 boots, given that there are
>> aleph_0 pairs. Let P be the set of pairs, and let {p_n} be an
>> enumeration of P. Let Then UP is the union of P, which is the set of
>> boots. f: N -> UP be given by
>>
>> f(n) = the left boot of pair p_n, if n is even,
>> = the right boot of pair p_n, if n is odd.
>>
>> Then f enumerates the boots in UP, which is to say that f is a choice
>> function on UP.
>>
>> The problem is, you can't do this with the socks. Given S = set of pairs
>> of socks, and given an enumeration of the pairs, there is no way to
>> explicitly write down an enumeration of US, the set of socks. So, does
>> such an enumeration exist? It does, if you assume the axiom of choice.
>>
>>
>> --
>> Dave Seaman
> You got it, yet how if you assume the axiom of choice one can solve the
> question?
The same way I solved the problem for pairs of boots, where I used the fact
that a choice function is defined by always taking the left boot from each
pair.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>
.
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