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 &lt; x 1 2 n

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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            &amp;lt;      x            1          2      n

Alexzander Bowman · Accepted Answer

Let      f    0    (  x  )  =      {                            0                           if                       r            0                    ⩾          x                                      1                           if                       r            0                    &amp;lt;          x          .                        It&#039;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                    &amp;lt;          x          .                        It&#039;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                    &amp;lt;          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  .

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

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