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.
I think this works:
The set of all rational numbers is countable, so order them all in some
sequence. This sequence will be used to compare any two functions.
Evaluate f(x) - g(x) , one rational number at a time, in
sequence. If f and g are different, then eventually a
rational value will be detected for which f(x) - g(x) <> 0.
For that rational, if f(x) - g(x) > 0 then declare f > g.
If f(x) - g(x) < 0 at that rational, then declare f < g.
This works because if f and g are continuous, then so is f - g.
IF f - g is not constantly 0 (in which case the functions are
identical), then there must be at least one rational at which they
differ in value. To prove this, assume that f(x) - g(x) = 0 for
all rationals. Then continuity implies that f(x) - g(x) = 0 for
all irrationals also, so f = g.
Sorting continuous functions
