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Let $f\colon \mathbb{R} \to \mathbb{R}$ be a function dfined by $$ f(x) = \begin{cases} 0,\quad x\not\in \mathbb{Q}\\ \frac{p}{p+1}, \quad x=\frac{p}{q} \end{cases} $$ Where is $f$ continuous?

I know that a function cannot be continuous and non constant at just the set of rationals but i am unable to prove its continuity at the irrational points.

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  • $\begingroup$ p and q are integers and g.c.d(p,q)=1 $\endgroup$ – Souvik Deb Apr 17 at 18:07
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Presumably when you say the rational $x = p/q$ you mean $p$ and $q$ are integers and $p/q$ is in lowest terms, and $p \ge 0$ (you need to allow $p$ or $q$ to be negative to allow negative rationals, but your formula would be undefined if $p=-1$).

It is not continuous anywhere. In any nonempty open interval there are both rational and irrational numbers; at an irrational $x$ you have $f(x)=0$ and at a nonzero rational $y = p/q$ you have $f(y) = p/(p+1) \ge 1/2$.

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I believe this function is everywhere discontinuous.

When $x$ is rational, $f(x)=\frac{p}{p+1}>0$. However, there exists irrational numbers $x_0$ arbitrarily close to $x$ where $f(x)=0$, hence $f$ is discontinuous at rational points.

When $x$ is irrational, $f(x)=0$. However, there exists rational numbers $x_0=\frac pq$ that are arbitrarily close to $x$ where $f(x_0) = \frac{p}{p+1} \geq \frac 12$, hence $f$ is discontinuous at irrational points.

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