Does there exist a function that is continuous at every rational point and discontinuous at every irrational point? And vice versa?

Question

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

Continuous at every rational point and discontinuous at every irrational point?
Continuous at every irrational point and discontinuous at every rational point?

No, for the reason that $\bar{\mathbb{Q}}=\mathbb{R}$ and also $\bar{\mathbb{R}-\mathbb{Q}}=\mathbb{R}$ — Martigan, Oct 27 '14 at 17:57
Yes for continuous only at the irrationals, no for continuous only at the rationals. See http://en.wikipedia.org/wiki/Gδ_set where all is explained. — Henno Brandsma, Oct 27 '14 at 18:02
@KBusc The $\delta$ symbol behaves weird in links, apparently. — Henno Brandsma, Oct 27 '14 at 18:07
@HennoBrandsma try this one http://en.wikipedia.org/wiki/G%CE%B4_set — KBusc, Oct 27 '14 at 18:08

score 5 · Answer 1 · answered Oct 27 '14 at 18:01

5

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

answered Oct 27 '14 at 18:01

Barry Cipra

79,832

score 3 · Answer 2 · answered Nov 07 '18 at 12:45

3

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.

answered Nov 07 '18 at 12:45

Can you clear why this happens? – Ppp Oct 21 '19 at 13:57
@Ppp See https://math.stackexchange.com/a/67626/1128456 – Aman Feb 02 '23 at 23:41

score 0 · Answer 3 · answered Oct 24 '23 at 17:08

The set of discontinuities of any real function must be the countable union of closed sets.

The rationals, Q, is such a set, and so your Case #2 does exist (and should be easily Googled). But the irrationals are not the countable collection of closed sets, so your Case #1 does not exist.

Does there exist a function that is continuous at every rational point and discontinuous at every irrational point? And vice versa?

3 Answers3

Linked