Re: ....Infinity
- From: "zuhair" <zaljohar@xxxxxxxxx>
- Date: 4 Nov 2005 08:41:55 -0800
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?
Zuhair
.
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