Construct an increasing function f on R that is continuous at every irrational number an

Maliyah Robles 2022-07-07 Answered
Construct an increasing function f on R that is continuous at every irrational number and is discontinuous at every rational number.
Solution: Let r n be a sequence with distinct terms whose range is Q . Let f : R R be given by
f ( x ) = r n < x 1 2 n
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Answers (1)

Alexzander Bowman
Answered 2022-07-08 Author has 19 answers
Let
f 0 ( x ) = { 0  if  r 0 x 1  if  r 0 < x .
It's increasing. Besides, it is discontinuous at r 0 and only at r 0 .
Now, let
f 1 ( x ) = { 0  if  r 1 x 1 2  if  r 1 < x .
It's increasing and it is discontinuous at r 1 and only at r 1 . So, f 0 + f 1 is increasing and it is discontinuous at r 0 and at r 1 and only at those points.
More generally, for each n Z + , let
f n ( x ) = { 0  if  r n x 1 2 n  if  r n < x .
Then f is increasing, since it is equal to n = 0 f n . And it is not hard to see that it is discontinuous at x if and only if x { q n n Z + } = Q (this follows from the fact that the convergence of the series n = 0 f n is uniform, by the Weierstrass M-test). The reason why I told you in the comments that it should be 1 2 n rather than 1 2 n was so that the expression n = 0 f n makes sense, that is, so that it converges, for every x R .
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