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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: Still in Deep Water, cont. | Comment 7 of 14 |
(In reply to Still in Deep Water by Richard)

The previous case has a simple extension to the case of all  continuous real functions on the whole real line.  Just use the previous recipe on the restrictions of the functions to [0,infinity) when they do not agree everywhere there.  If they do agree everywhere on [0,infinity), then use the previous recipe on the restrictions of f(-x) and g(-x) to [0,infinity).
  Posted by Richard on 2006-08-24 01:46:10

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