Does exist a function f that discontinuous at rational and continuous at every irrational but the restriction f to the set of all irrational numbers is not constant

Celeste Barajas

Celeste Barajas

Answered question

2022-11-18

Does exist a function f that discontinuous at rational and continuous at every irrational but the restriction f to the set of all irrational numbers is not constant and f ( q n ) is convergent where { q n } is a sequence of rationals.
Thomae function is one of an example of function that satisfies above condition if the restriction f to the set of all irrational numbers is constant.

Answer & Explanation

luthersavage6lm

luthersavage6lm

Beginner2022-11-19Added 22 answers

Denote by T ( x ) Thomae's function. Then
f ( x ) = T ( x ) + x
is a function satisfying your condition.

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