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I know that this kind of problem is asked on this site earlier but I put this question here to clear my doubt related to my conceptual confusion about these kinds of functions.

Consider the function

$$ f(x)= \begin{cases} 1/q : x\in p/q \cap [0,1]; p,q \in \mathbb{N}, (p,q)=1 \\ 0 :~~~~~~~~~~~~~~~~~ else \end{cases} $$

Then How to check the continuity of $f(x) $ in the given domain

I know how to check the continuity of function by sequential criteria like the Dirichlet function. But in this function, how do select how sequences of a rational and irrational number??

I have studied about continuity of these kinds of functions which are defined by rationals and irrationals; I think there are a maximum of four possible types of sequences to check about continuity of function in the given domain that is we choose

the sequence of rationals that converges to irrationals or rational and the sequence of irrationals that converges to irrationals or rational and these four types of sequences satisfy the sequential criteria of continuity then we say the function is continuous or not??

Please correct me if I am not correct and please give some hints to solve the above question.

Thanks!

Manish Saini
  • 570

1 Answers1

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Hint

Any sequence $\{x_n\}$ is composed of a subsequence of rational points $\{x_n^r\}$ and a sequence of irrationals $\{x_n^i\}$. And to prove that such a sequence converges, it is sufficient to prove that both $\{x_n^r\}$ and $\{x_n^i\}$ converges to the same limit.

$f$ is obviously discontinuous at any $a \in [0,1] \cap \mathbb Q$ as $f(a) \neq 0$ as for any sequence of irrationals $\{a_n\}$ converging to $a$, $\{f(a_n)\}$ is the always vanishing sequence.

Now for $a \in [0,1] \cap (\mathbb R \setminus \mathbb Q)$, and a sequence $a_n \to a$, separate the rational part and the irrational part of the sequence. The image of the irrational subsequence under $f$ is the always vanishing sequence. You have to prove that the image of the rational subsequence of $a_n \to a$ under $f$ also converges to zero. And for that, the idea is to prove that if a sequence of rationals converges to an irrational, then the denominators have to go to infinity.

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