6
$\begingroup$

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?
$\endgroup$
6
  • $\begingroup$ 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. $\endgroup$ 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. $\endgroup$ 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
4
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Can you clear why this happens? $\endgroup$
    – Ppp
    Oct 21 '19 at 13:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.