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
Still in Deep Water by Richard)
Well, I seem to be Mr. Negative counter-example here.
Richard, I don't think that your idea in comment 5 works.
Consider the first function from comment 2 (repeated here):
f(x) = 0 if x=0
xsin(1/x) if x <> 0
as compared to
g(x) = 0
The largest closed interval [0,T] containing 0 such that
f(x)=g(x) for all x in [0,T] is [0,0]. But there is no nonempty open interval
(0,S) such that f(x)-g(x) has the same sign for all x in (0,S).
I actually had exactly the same idea that you did when I posted the
example in comment 2. I posted the example because I couldn't
make our idea work for all functions.
re: Still in Deep Water
