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Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be given by $f(x) = \left\{ \begin{array}{ll} 2x & \quad x \text{ is rational} \\ -2x & \quad x \text{ is irrational } \end{array} \right.$

Show that $f$ is discontinuous at every point $x_0 \neq 0$.

My idea to show this is to consider when $x_0$ is rational and then consider when it is irrational. Then using the density of the rationals we have sequences which converge to $x_0$. Then use The Divergence Criterion for Functional Limits Theorem to finish the proof. Is this along the right track?

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    $\begingroup$ For reference, you'll probably find the proof that the Dirichlet function is discontinuous everywhere somewhat fruitful with this. I haven't tried the proof for myself, but I imagine showing the discontinuity will have a similar flavor, and then you have the special case of $0$ to handle. On the other hand, here's a similar exercise, a bit closer to the flavor of this one: math.stackexchange.com/questions/2369895/… $\endgroup$
    – PrincessEev
    Commented Mar 8, 2019 at 6:32

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Let $x_0 \in \mathbb R$. Take a sequence $(r_n)$ of rational numbers and a sequence $(s_n)$ of irrational numbers with $r_n \to x_0$ and $s_n \to x_0$.

Then: $f(r_n) =2r_n \to 2x_0$ and $f(s_n) =-2s_n \to -2x_0$.

This shows that $f$ is not continuous in $x_0$, if $x_0 \ne 0.$

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  • $\begingroup$ You get the final line by the Divergence Criterion for Functional Limits? $\endgroup$
    – babynewton
    Commented Mar 8, 2019 at 6:46
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    $\begingroup$ If $f$ is continuous at $x_0$, then we have $ \lim f(r_n)= \lim f(s_n)$, hence $2x_0=-2x_0$, thus $x_0=0$. $\endgroup$
    – Fred
    Commented Mar 8, 2019 at 6:50

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