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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.)
re(3): Oops... Maybe another hint? | Comment 20 of 33 |
(In reply to re(2): Oops... Maybe another hint? by JLo)


Maybe I'm being dense, but this hint doesn't get me any closer.

1) The simplest monotonic function which is discontinuous at a point and continuous elsewhere is a step function, because of the jump at the point.

2) If we have a finite number of rational numbers, we can clearly construct a function that jumps at each rational number.

3) But I'm still not figuring out how to construct a function that jumps at every rational number.  We already knew that the weird function would have a jump at every rational number.  And this means, incidentally, that if it is monotonic increasing then it is strictly monotonic increasing,  because there are rational numbers between every pair of irrational numbers.  But this hint still leaves me where I was at my second post.

Except now I am hoping VERY MUCH that this function really works as advertised.  Because I am prepared to be impressed.


  Posted by Steve Herman on 2006-08-21 12:50:07

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