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Here is Prob. 18, Chap. 4 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Every rational $x$ can be written in the form $x = m/n$, where $n > 0$, and $m$ and $n$ are integers without any common divisors. When $x=0$, we take $n=1$. Consider the function $f$ defined on $\mathbb{R}^1$ by $$ f(x) = \begin{cases} 0 &x\notin\mathbb Q, \\ \frac{1}{n} &x\in\mathbb Q. \end{cases}$$ Prove that $f$ is continuous at every irrational point, and that $f$ has a simple discontinuity at every rational point.

My effort:

Let $x$ be a rational number. Then $f(x) = \frac{1}{n} \neq 0$. But $f(t) \to 0$ as $t \to x$ through irrational numbers. What if $t \to x$ through rational numbers? [Is this way of reasoning rigorous enough for Rudin, I wonder?] How to show that $f(x-) < f(x+)$? Finally how to show that $$f(x) = \lim_{t\to x} f(t)$$ for every irrational $x$?

Simply Beautiful Art
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Saaqib Mahmood
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  • Have you tried reading http://math.stackexchange.com/questions/530097/proof-of-continuity-of-thomae-function-at-irrationals – Juniven Acapulco Feb 04 '17 at 00:16
  • Yep, just noticed what I have is basically the same as that :) – Wen Feb 04 '17 at 00:18
  • I'm a little surprised that Rudin would refer to pairs of integers "without any common divisors," as opposed to pairs whose greatest common divisor is $1$. – Barry Cipra Feb 04 '17 at 00:19

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Firstly, your line of reasoning for the first part is fully correct. The second part though, using this definition for continuous functions (which is equivalent) for a irrational $y$ we can see that since there are only finitely many values of $x$ such that $f(x)>\epsilon$ (within, say $y\pm 1$) we just take a value of $\delta$ such that the closest value of $x$ for which $f(x)>\epsilon$ satisfies $x>\delta$.

Wen
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