Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.

Topic: wellordering?
Replies: 4   Last Post: Dec 1, 2001 2:18 PM

 Messages: [ Previous | Next ]
 Andy Averill Posts: 14 Registered: 12/13/04
Re: wellordering?
Posted: Nov 27, 2001 8:37 PM

"Elaine Jackson" <elainejackson7355@home.com> wrote in message
news://%JRM7.70869\$Ud.3453794@news1.rdc1.bc.home.com...
> Let F be the set of all functions f with ran(f)<=dom(f)=(the set of
positive
> integers), and consider the lexicographical ordering of F.
(Lexicographical
> ordering: f1<f2 iff (1) f1,f2 are different functions, and (2)
f1(n)<f2(n),
> where n is the number with f1(k)=f2(k) for all k<n.) Clearly this is a
> linear order, but is it a wellordering? The only thing I can think of is
> this: given a nonempty subset S of F, first throw out all the functions
> whose value at 1 is not minimal (in S), then throw out all the functions
> whose value at 2 is not minimal (in the set remaining from the previous
> step), and so on. If f belongs to every set in the resulting sequence,

then
> f is the smallest element of S, but how can you prove that some f belongs
to
> every set in the sequence?

By the way you've constructed these sets. If we let

S1 = the set of all f's in S whose value at 1 is minimal
S2 = the set of all f's in S1 whose value at 2 is minimal
Sn = the set of all f's in S(n-1) whose value at n is minimal

then if the intersection of the Sn's is empty, there must be a least k for
which Sk is empty (since they're nested). But that means that among the f's
contained in S(k-1), none has a minimal value at k, which is not possible,
since the range of the function is the positive integers.

-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 80,000 Newsgroups - 16 Different Servers! =-----

Date Subject Author
11/27/01 Elaine Jackson
11/27/01 magidin@math.berkeley.edu
11/27/01 Andy Averill
11/28/01 magidin@math.berkeley.edu
12/1/01 Herman Rubin