Actually there are 2 questions, but they are closely related. Does it exist a function that is:

  1. Continuous at every rational point and discontinuous at every irrational point?
  2. Continuous at every irrational point and discontinuous at every rational point?
  • $\begingroup$ No, for the reason that $\bar{\mathbb{Q}}=\mathbb{R}$ and also $\bar{\mathbb{R}-\mathbb{Q}}=\mathbb{R}$ $\endgroup$ – Martigan Oct 27 '14 at 17:57
  • $\begingroup$ Yes for continuous only at the irrationals, no for continuous only at the rationals. See en.wikipedia.org/wiki/Gδ_set where all is explained. $\endgroup$ – Henno Brandsma Oct 27 '14 at 18:02
  • $\begingroup$ @HennoBrandsma your link is broken $\endgroup$ – KBusc Oct 27 '14 at 18:06
  • $\begingroup$ @KBusc The $\delta$ symbol behaves weird in links, apparently. $\endgroup$ – Henno Brandsma Oct 27 '14 at 18:07
  • $\begingroup$ @HennoBrandsma try this one en.wikipedia.org/wiki/G%CE%B4_set $\endgroup$ – KBusc Oct 27 '14 at 18:08

For part 2, let $f(p/q)=1/q$ for rational points $p/q$ (in reduced form) and $f(x)=0$ for irrational $x$.


For first part there does not exist any function Because f is discontinuous on only F-sigma set. F-sigma set is defined as the set which can be written as countable union of closed set.

  • $\begingroup$ Can you clear why this happens? $\endgroup$ – Ppp Oct 21 at 13:57

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