8
$\begingroup$

I have been trying to find a function $f : \mathbb R \to \mathbb R$ such that $\lim_{x \to c} f(x)$ exists when $c$ is irrational and the limit doesn't exist when $c$ is rational.

I tried variations of the Dirichlet function and Thomae's function, but I couldn't get anywhere. I also tried proving that such a function cannot exist, using the fact that both the rationals and the irrationals are dense in real numbers. But I couldn't get a satisfying proof that way either.

$\endgroup$
2
  • $\begingroup$ Why doesn't Thomae's function do it? $\endgroup$
    – wlad
    Nov 16, 2020 at 11:20
  • 1
    $\begingroup$ @ogogmad Because for Thomae's function the limit at every point is zero. $\endgroup$
    – Wojowu
    Nov 16, 2020 at 11:22

1 Answer 1

10
$\begingroup$

Arrange rationals in a sequence $q_n$, and set $$f(x) = \sum_{n = 1}^\infty 2^{-n} \mathbb{1}_{[q_n,\infty)}(x),$$ where $$\mathbb{1}_{[q_n,\infty)}(x) = \begin{cases} 1 & \text{if $x \geqslant q_n$,} \\ 0 & \text{if $x < q_n$.} \end{cases} $$ In other words, $$f(x) = \sum_{n : q_n \leqslant x} 2^{-n} .$$ By the dominated convergence theorem, we have $$\lim_{x \to a^-} f(x) = \sum_{n : q_n < a} 2^{-n}$$ and $$\lim_{x \to a^+} f(x) = \sum_{n : q_n \leqslant a} 2^{-n} = f(a) .$$ It follows that $$\lim_{x \to q_n^+} f(x) = 2^{-n} + \lim_{x \to q_n^-} f(x)$$ and hence $f$ has no limit at each $q_n$, but $f$ is continuous at every irrational point.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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