Sort the set of functions f:R→R, with R being the set of real numbers. This means you have to find an
order "«" that lets you compare any two pairs of unequal functions f and g; unequal means, f(x)≠g(x) for at least one x.
More precisely, these are the requirements for the order "«" you are challenged to find:
1. If f≠g, either f«g or g«f.
2. If f«g and g«h then f«h
You might be tempted to declare f«g when f(x)<g(x) for all x but that would of course fail because e.g. f(x)=x and g(x)=-x would not be comparable with respect to your order.
For a much, much easier challenge, start by finding an order for all continuous functions.
Here is an attempt at a simpler case. Let us only consider continuous functions from [0,infinity) to R.
If f(0) and g(0) are unequal, then say f and g compare the same as f(0) and g(0).
If f and g are unequal and f(0)=g(0), then consider the T such that
[0,T] is the largest closed interval containing 0 and such that
f(x)=g(x) for all x in [0,T]. There is some nonempty open interval
(T,S) such that f(x)-g(x) has the same sign for all x in (T,S). Say f
< g if that sign is negative and f > g if it is positive.
I believe that this generalizes lexicographic order.
(T,S) need not exist, as Steve Herman points out above, so this method does not really work.
Edited on August 24, 2006, 2:28 am
|
Posted by Richard
on 2006-08-23 18:28:02 |