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Sort of a sorting problem (Posted on 2006-08-22) Difficulty: 5 of 5
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.

See The Solution Submitted by JLo    
Rating: 4.3333 (6 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: Sorting continuous functions | Comment 13 of 14 |
(In reply to Sorting continuous functions by Steve Herman)

You beat me to it, Steve.  For what they're worth, some additional observations: More than just the continuous real functions can be compared with this method.  In fact, one can compare every two general real functions which differ only at rational arguments.  So it remains to compare groups of real functions which within one group are identical on the irrationals.  The catch: there are as many irrationals as reals, so finding a lexicographical order for the mentioned groups is no easier than solving the general problem right away.  Too bad.
  Posted by vswitchs on 2006-08-25 13:30:33

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