I'm working on the following problem:

Consider the function $$f(x) = \begin{cases} x + x^2 & x \in \mathbb{Q} \\ x - x^2 & x \notin \mathbb{Q} \\ \end{cases} $$ Prove that $f$ is nowhere continuous except at $x=0$.

My attempt:

First, we show that $f$ is continuous at $x=0$. Accordingly, fix $\epsilon > 0$, and choose $\delta < \frac{-1 + \sqrt{1 + 4 \epsilon}}{2}$ (note that since the quantity on the left hand side is strictly positive, the Archimedean Principle guarantees the existence of such a $\delta$). Then, if $$ |x - 0| = |x| < \delta$$ then $$ |f(x) - f(0)| = |f(x)| = \begin{cases} |x + x^2| & x \in \mathbb{Q} \\ |x - x^2| & x \notin \mathbb{Q} \\ \end{cases} $$ In either case, the triangle inequality gives that \begin{align*} |f(x)| &\leq |x| + |x|^2 \\ &< \delta + \delta^2 \\ &= \frac{-2 + 2\sqrt{1+4 \epsilon} + 1 - 2\sqrt{1 + 4 \epsilon} + 1 + 4 \epsilon}{4} \\ &= \frac{4\epsilon}{4} \\ &= \epsilon \end{align*} Thus, for arbitrary $\epsilon>0$, $\exists \delta > 0$ s.t. $$ |x - 0| < \delta $$ implies $$ |f(x) - f(0)| < \epsilon,$$ so $f$ is continuous at $x=0$. Now, consider some real $x_0 \neq 0$. It follows from the Sequential Density of Irrationals/Rationals that there exists some sequence $u_n$ of rational numbers that converges to $x_0$, and there exists some sequence $v_n$ of irrational numbers that also converges to $x_0$. Now, since the limits $\lim_{n \rightarrow \infty} u_n = x_0$ and $\lim_{n \rightarrow \infty} v_n = x_0$ exist, we conclude that

$$\lim_{n \rightarrow \infty} f(u_n) = \lim_{n \rightarrow \infty} u_n + u_n^2 = x_0 + x_0^2$$ and $$\lim_{n \rightarrow \infty} f(v_n) = \lim_{n \rightarrow \infty} v_n + v_n^2 = x_0 - x_0^2.$$ Since these limits are distinct for $x_0 \neq 0$, it follows that either the image of $u_n$ or the image of $v_n$ under $f$ does not converge to $f(x_0)$. Thus, $f$ is not continuous for any $x_0 \neq 0$.

Is this a valid proof? I'm concerned I jumped in a bit too eagerly - is it necessary to include the former proof of continuity? Or does it suffice to say the images limit to the same value only when $x_0$ is identically 0?

  • 2
    $\begingroup$ The proof is nice, very nice indeed. I don't see any issues, and the elaboration of continuity at $x = 0$ is excellent. $+1$. $\endgroup$ – астон вілла олоф мэллбэрг May 19 '18 at 5:18
  • $\begingroup$ Yes, this is a good proof! $\endgroup$ – Fimpellizieri May 19 '18 at 5:19

The example given to you generalizes to what is a great class of problems posed in the topic of continuity, to students in their first year or so(and in plenty of competitive exams, including one I had seen recently). That is,

Let $f,g : \mathbb R \to \mathbb R$ be two continuous functions. Define for $S \subset \mathbb R$ the indicator function $1_S : \mathbb R \to \{0,1\}$ by $1_S(x) = 1$ if $x \in S$ and $0$ otherwise. Define the function $h(x) = f1_{\mathbb Q} + g1_{\mathbb R \setminus \mathbb Q}$. Then $h$ is continuous precisely at the points when $f = g$.

For example, in the situation you had above, we have $f = x+x^2$ and $g = x - x^2$, and these are equal exactly at $x = 0$.

The proof is very simple. Indeed, at points where $f(x) \neq g(x)$, we may approach via the rationals and irrationals separately(as you did in the second part of your proof) to get discontinuity at $x$, since $f(x)$ and $g(x)$ will be the limits of the values of the rational and irrational sequences respectively (by continuity) and these are unequal at the point.

However, if $f(x) = g(x)$ then fix $\epsilon > 0$. Since $\lim_{t \to x} f(t) = \lim_{t\to x} g(t) = f(x) = g(x)$, there are $\delta_1$ and $\delta_2$ corresponding to $f$ and $g$, and we may take the minimum of these to get a $\delta > 0$, which will work out.

A common example of this is with $f \equiv 0$ and $g \equiv 1$, the characteristic function of the rationals which is continuous nowhere.

| cite | improve this answer | |
  • $\begingroup$ This is a great reference! I have seen many variations of this type of problem, so it's very nice to have such a concise generalization. $\endgroup$ – OGBerglemir May 19 '18 at 20:13
  • $\begingroup$ Thank you for the acceptance. $\endgroup$ – астон вілла олоф мэллбэрг May 20 '18 at 10:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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