All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Weird function challenge (Posted on 2006-08-15)
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: Needless to say this is very confusing | Comment 12 of 33 |
(In reply to Needless to say this is very confusing by Charlie)

Charlie,

it doesn't matter that there are infinitely many discontinuities in a given epsilon-neighborhood of the irrational (which there are), so long as all of function values in the neighborhood can be made arbitrarily close to the function value of the irrational by picking an appropriately small epsilon.

That is the situation with the weird function that has been identified.  As we approach the irrational from the left and from the right, the limit of the function values equal the function when evaluated at the irrational.

 Posted by Steve Herman on 2006-08-19 18:34:11

 Search: Search body:
Forums (0)