Re: ....Infinity

12. Dave Seaman
Nov 4, 10:01 pm show options

Newsgroups: sci.math
From: Dave Seaman <dsea...@xxxxxxxxxxxx> - Find messages by this author

Date: Fri, 4 Nov 2005 19:01:47 +0000 (UTC)
Local: Fri, Nov 4 2005 10:01 pm
Subject: Re: ....Infinity
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On 4 Nov 2005 09:48:55 -0800, zuhair wrote:




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> Dave Seaman wrote:
>> 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
> In reality you didn't show the enumeration for socks explicitely you
> only refered to the axiom of choice.


That's because there is no way to write down an enumeration explicitly.

That's exactly what I said when I was explaining the need for the axiom

of choice in the first place.


> Why not put one sock in each boot. and make an enumeration of sockes
> marked by the boot they are put into.


How do you decide which sock to put in the left boot, and which in the
right one? You can decide that for one pair, or for any finite number
of
pairs, but not for all the pairs at once. That's why the axiom of
choice
is needed.
----------------------------------------------------------------------------------

What do you mean how I decide, I will put ONE sock in ONE boot. the
matter of left
or right is not that important.

Since their is ONE-ONE correspondance between
boots and socks, then the number of all socks should be the same as all
boots.

This is obvious.

I want to ask you ? then how Cantor worked out
his diagonal proof? didn't he say if we change
the n-th decimal in the n-th raw? .Then I can
say the same thing against this argument
we can only do this operation for finite number
of terms ie for n. and not infinitely.

In a similar manner we can put ONe sock in ONe
boot.(like the n-th decimal in n-th raw).

Zuhair

.

Relevant Pages

  • Re: ....Infinity
    ... >> only refered to the axiom of choice. ... >> Why not put one sock in each boot. ... >> marked by the boot they are put into. ... An explicit formula is exactly what is lacking in constructing a mapping ...
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  • Re: ....Infinity
    ... > In reality you didn't show the enumeration for socks explicitely you ... > only refered to the axiom of choice. ... > Why not put one sock in each boot. ...
    (sci.math)
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