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Weird function challenge (Posted on 2006-08-15) Difficulty: 4 of 5
Find a function f:R->R (R the set of real numbers), such that

1. f has a discontinuity in every rational number, but is continous everywhere else, and
2. f is monotonic: x<y → f(x)<f(y)

Note: Textbooks frequently present examples of functions that meet only the first condition; requiring monotonicity makes for a slightly more challenging problem.

See The Solution Submitted by JLo    
Rating: 4.3000 (10 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Hints/Tips re(2): Oops... Maybe another hint? | Comment 16 of 33 |
(In reply to re: Oops... by Steve Herman)

In fact there is a solution and you'll be surprised how simple it is, once you found it. You and Ken's observations made me think a bit more about this problem and I am realizing that my previous hint was maybe overly complicated, although the  w-function is quite interesting to study, don't you think?

So here is another hint if you care for it:

1. Given a rational number r, what is the simplest function f_r you can think of, that is discontinuous at r and continuous elsewhere?

2. Now imagine that you have a whole bunch of rational numbers, how could you use the corresponding f_r's to construct a function that is discontinuous at all your rational numbers?

3. Can that be extended if the "whole bunch" consists of ALL rational numbers?

  Posted by JLo on 2006-08-20 10:42:45

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