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To my shame, I must admit that I have no solution to the original problem... We know that there must be such an order since every set can even be well-ordered. (A set is well-ordered if every non-empty subset has a smallest element.) However I suspect that it is in fact impossible to specify a particular order through some well-defined mathematical characterization. Unfortunately it is beyond me to provide a mathematically accurate formulation for my suspicion, let alone to prove it.
BTW if you have a particular well-order of all real numbers (which must exist), it is easy to derive an order for all functions. It goes like this:
Assume two (unequal) functions f and g. Now take the "smallest" x where f(x) unequal g(x) (Smallest with respect to our well-order of all real numbers, of course). Then define
f < g when f(x) < g(x)
and
g < f when g(x) < f(x)
The second <'s are of course with respect to the given well-order of the real numbers.
Richard and vswitchs hit the nail on the head when they observed that the problem has to do with cardinality. The set of all functions just seems to be too big to be ordered easily. The set of continuous functions however is no bigger than the set of all real numbers (even if that seems paradox, the set of all continuous functions can indeed be mapped bijectively to all real numbers) and can be sorted rather easily (Credits to Steve who found this solution):
Say r1,r2,... is a counting of all rational numbers. Note that a particular counting can be constructed, exercise for the reader ;-). Take arbitrary f and g, f unequal g. Then say ri is the first rational number (ie. with minimal i) where f(ri) unequal g(ri). Then we define f < g if f(ri) < g(ri) and g < f if g(ri) < f(ri). Obviously this is an order. It remains to show that the order is well-defined, i.e. that for two uneqal functions f, g, such a "first" ri does indeed exist. OK, assume it does not exist, then f and g are identical on all rational numbers. But since f and g are continuous and the rational numbers are dense within the real numbers, this means f and g must be identical for all real numbers (another little exercise...), which contradicts our assumption.
An alternative solution for continuous functions follows Eric's idea to use integrals. Again we reduce our sorting task to a countable set that we can handle, only that now we pick on integration intervals rather than rational numbers:
Consider all intervals of type Im,n=[2mn,2m(n+1)] where m and n are integers. For every fixed m these intervals cover all real numbers. By decreasing m, we can also make these intervals arbitrarily small. The set of intervals Im,n is also countable. We can therefore simply go through the intervals one by one until we determine an Im,n such that the integral of f and g over this interval produces two different results. This must happen for one of the intervals, because when f and g are different functions, they must differ at on point x, say e.g. f(x)<g(x). Because of continuity, we have f(x)<g(x) on a whole interval around x. Then there must also be an interval Im,n where we have f(x)<g(x). Now we declare f<g.
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