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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: Uncle! Don't give up yet! | Comment 4 of 33 |
(In reply to Uncle! by Steve Herman)

Correct, there must be a jump discontinuity at every rational number. Just make sure the jumps add up to something finite. Fortunately the rational numbers are countable.

To get started, try to find a function f:[0,1]->R that is discontinuous in every number with finite binary representation. It is only a small step from there to the original problem.

  Posted by JLo on 2006-08-16 17:34:05

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