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.

(In reply to re(3): Still in Deep Water by Steve Herman)

No need to feel taunted, this was certainly not my intention. When I said the problem is much, much easier for continuous functions, I only indicated that the general problem is very, very hard. I mean, really, really, really hard!

BTW Eric's first posting contains the right idea for a possible order for continuous functions (although it is different from mine), it just needs a little refinement to work.

Posted by JLo on 2006-08-24 18:20:01