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Study continuity of the following function: $$ f(x) = \begin{cases} |x|,\ \text{if $x$ is irrational}\\ \frac{qx}{q+1},\ \text{if}\ x = {p\over q}, q\in\Bbb N, p\in\Bbb Z, p\perp q \end{cases} $$

I've been recently studying some similar functions. The usual trick was to consider different sequences $x_n$ and then study the behavior of $f(x)$ as $x_n$ approaches some point $x_0$.

I wasn't able to apply the same trick for this function but here some intuition though, which I want to formalize somehow. If we take any sequence $\{x_n\}_{n\in\Bbb N}$ of irrational numbers such that: $$ \lim_{n\to\infty}x_n = x_0 $$ Then: $$ \forall x_n \in\Bbb R\setminus \Bbb Q:f(x_n) = |x_n| $$ In such case: $$ \lim_{n\to\infty} f(x_n) = |x_0| $$

So it looks like the function is continuous at every irrational point.

For the rational ones, I was trying to use a similar approach. Let $\{y_n\}$ be a sequence of rational numbers such that for $y_n \in\Bbb Q$ and $y_0\in\Bbb R\setminus\Bbb Q$: $$ \lim_{n\to\infty}y_n = y_0 $$ But: $$ \lim_{n\to\infty}f(y_n) \ne f(y_0) = |y_0| $$

Now every neighborhood of a given point in $\Bbb R$ contains infinitely many rationals and irrationals. So we might approximate $y_0$ with points from $y_n$ closer and closer to $y_0$ so if we introduce a $\{q_n\}$ denoting consequent denominators from $p\over q$ then it is going to grow and eventually: $$ \lim_{n\to\infty}{q_n\over q_n + 1} = 1 $$

So looks like every rational point is a point of removable discontinuity at least for $x \ge 0$. My problem is with putting down a rigorous proof behind that intuition.

Could you please help me with that?

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    $\begingroup$ How did you deduce that $f$ is continuous at every irrational point $x_0$? Through your approach, you would have to prove that, for every sequence $(x_n)_{n\in\mathbb N}$ converging to $x_n$, the sequence $\bigl(f(x_n)\bigr)_{n\in\mathbb N}$ converges to $f(x_0)$. But you only proved it for sequences of irrational numbers. $\endgroup$ Jul 15, 2019 at 16:51
  • $\begingroup$ @JoséCarlosSantos well, yes you are right $\endgroup$
    – roman
    Jul 15, 2019 at 16:58
  • $\begingroup$ @roman:$\;$The conclusion you want is:$\;f\;$is continuous at $x=a$ if and only if $a=0$ or $a$ is a positive irrational number. $\endgroup$
    – quasi
    Jul 15, 2019 at 17:19

2 Answers 2

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$f$ is continuous exactly at $\{0\} \cup (\langle 0, \infty\rangle \setminus \mathbb{Q})$.

  • Let $x < 0$ rational. Then for a sequence of irrational numbers $(x_n)_n$ converging to $x$ we have $f(x_n) = |x_n| \to |x| \ne f(x)$ because $f(x) < 0$ so $f$ is not continuous at $x$.
  • Let $x < 0$ irrational. Then for a sequence of rational numbers $(x_n)_n$ converging to $x$ we have $f(x_n) \not\to |x| = f(x)$ because $f(x_n) < 0$ for all $n \in \mathbb{N}$ so $f$ is not continuous at $x$.
  • Let $x > 0$ rational. Write $x = \frac{p}{q}$ with $p \in \mathbb{Z}, q \in \mathbb{N}, \gcd(p,q) = 1$. Then for a sequence of irrational numbers $(x_n)_n$ converging to $x$ we have $$f(x_n) = |x_n| \to |x| = \frac{p}{q} \ne \frac{p}{q+1} = f(x)$$ so $f$ is not continuous at $x$.
  • Let $x = 0$. Notice that $|f(x)| \le |x|$ for all $x \in \mathbb{R}$ so continuity at $0$ follows.
  • Let $x > 0$ irrational. Notice that for any interval $I \subseteq \mathbb{R}$ and any $N \in \mathbb{N}$ there are only finitely many rational numbers $\frac{p}{q}$ of the form $p\in\mathbb{Z}, 1 \le q \le N, \gcd(p,q) = 1$ inside $I$.

    Let $\varepsilon > 0$ and let $N \ge \frac{x+\frac\varepsilon2}{\frac\varepsilon2}$. Set $\delta > 0$ as $$\delta := \min\left\{\min_{\substack{p \in \mathbb{N}, 1 \le q \le N, \\\gcd(p,q) = 1, \frac{p}q \in \left\langle \frac{x}2, \frac{3x}2\right\rangle}} \left|x-\frac{p}q\right|, \frac{x}2, \frac\varepsilon2\right\}$$

    Then for $\left|x-\frac{p}{q}\right| < \delta$ we have $q \ge N+1$ and $\frac{p}{q} < x+\delta \le x+\frac\varepsilon2$ so $$\left|f(x) - f\left(\frac{p}{q}\right)\right| = \left|x-\frac{p}{q+1}\right| \le \left|x-\frac{p}{q}\right| + \frac{p}{q(q+1)} < \frac\varepsilon2 + \frac{x+\frac\varepsilon2}{q+1} \le \frac\varepsilon2 + \frac{x+\frac\varepsilon2}{N} \le \varepsilon$$ Furthermore, if $|x-y| < \delta$ for $y$ irrational then $|f(x) - f(y)| = |x- y| < \frac\varepsilon2 < \varepsilon$. We conclude that $f$ is continuous at $x$.

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  • $\begingroup$ Thank you for the answer, the last portion of it ($x>0,\ x\in\Bbb R\setminus \Bbb Q$ ) is pretty convoluted, what is the idea behind choosing $\epsilon$ and $\delta$ in such form? Could you please elaborate a bit? $\endgroup$
    – roman
    Jul 16, 2019 at 10:36
  • $\begingroup$ @roman To prove that $f$ is continuous at $x$ for every $\varepsilon > 0$ we have to provide $\delta > 0$ such that for all $y\in\mathbb{R}$ we have $|x-y| < \delta \implies |f(x) - f(y)| < \delta$. This is easy when $y$ is irrational, but for rational $y = \frac{p}{q}$ we have to bound $$|f(x)-f(y)| = \left|x-\frac{p}{q+1}\right| \le \left|x-\frac{p}q\right| + \left|\frac{p}{q} - \frac{p}{q+1}\right| \le \delta + \frac{p}{q(q+1)}$$ We want this to be less than $\varepsilon$. We have noticed that $\frac{p}{q} < x+\delta$ so $$|f(x)- f(y)| \le \delta + \frac{x+\delta}{q+1}$$ $\endgroup$ Jul 16, 2019 at 10:41
  • $\begingroup$ @roman We can set $\delta < \frac\varepsilon2$ to obtain $|f(x) - f(y)| < \frac\varepsilon2 + \frac{x+\frac\varepsilon2}{q+1}$ Now we see that we must bound the denominator $q$ from below so that $ \frac{x+\frac\varepsilon2}{q+1} \le \frac\varepsilon2$. We can do this by cleverly picking $\delta$ small enough so that $\langle x-\delta, x+\delta\rangle$ contains only rationals $\frac{p}{q}$ such that $q$ is larger than some $N \in \mathbb{N}$. We set $N $ such that $\frac1N \le \frac{\varepsilon/2}{x+\varepsilon/2}$ and pick $\delta > 0$ as noted above. $\endgroup$ Jul 16, 2019 at 10:45
  • $\begingroup$ @roman See here how I picked $\delta$. Finally, I wanted $\delta < \frac{x}2$ so that $\langle x-\delta, x+\delta\rangle \subseteq \left\langle \frac{x}2, \frac{3x}2\right\rangle \subseteq \langle 0, +\infty\rangle$ so that we only pick positive rationals, making notation easier. $\endgroup$ Jul 16, 2019 at 10:48
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    $\begingroup$ @roman I believe that $\delta < \frac{x}2$ is necessary: if $\left|x-\frac{p}{q}\right| < \delta$ then $\left|x-\frac{p}{q}\right| < \frac{x}2$ so $\frac{p}{q} \in \left\langle \frac{x}2, \frac{3x}2\right\rangle$. Now since $\left|x-\frac{p'}{q'}\right| < \delta < \min_{\substack{p' \in \mathbb{N}, 1 \le q' \le N, \\\gcd(p',q') = 1, \frac{p'}q' \in \left\langle \frac{x}2, \frac{3x}2\right\rangle}} \left|x-\frac{p'}{q'}\right|$ it follows that $q \ge N+1$. $\endgroup$ Jul 17, 2019 at 18:50
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Claim:$\;f\;$is continuous at $x=a$ if and only if $a=0$ or $a$ is a positive irrational number.

Proof:

Fix $a\in\mathbb{R}$, and suppose $(x_n)$ is an infinite sequence of rational numbers such that

  • ${\displaystyle{\lim_{n\to\infty}x_n=a}}$$\\[4pt]$
  • $x_n\ne a$, for all $n$.

For each positive integer $n$, write $x_n={\large{\frac{p_n}{q_n}}}$ where $p_n,q_n$ are relatively prime integers, and $q_n > 0$.

For each positive integer $d$, let $S_d$ be the set of positive integers $n$ such that $q_n=d$.

Suppose $S_d$ is infinite, for some $d$.

But the sequence $(x_n)$ is a Cauchy sequence (since it converges), hence for sufficiently large $m,n\in S_d$, we must have $|x_m-x_n| < {\large{\frac{1}{d}}}$. But this is impossible unless $x_m=x_n$.

It follows that the infinite subsequence of $(x_n)$ with $n\in S_d$ is eventually constant, with constant value $c$ say. But then since the sequence $(x_n)$ converges to $a$, we must have $c=a$, contrary to the assumption that $x_n\ne a$, for all $n$.

Therefore $S_d$ must be finite.

Thus, for any positive integer $d$, there are at most finitely positive integers $n$ such that $q_n =d$.

It follows that $\displaystyle{\lim_{n\to\infty}q_n=\infty}$, hence $$ \lim_{n\to\infty}f(x_n) = \lim_{n\to\infty}\frac{q_nx_n}{q_n+1} = \lim_{n\to\infty}\left(\frac{q_n}{q_n+1}\right)x_n = \lim_{n\to\infty}x_n = a $$

As a consequence, if $f(a)\ne a$, then $$ \lim_{n\to\infty}f(x_n) = a \ne f(a) $$ so $f$ is not continuous at $x=a$.

Suppose $a$ is a nonzero rational number.

Write $a=\frac{p}{q}$, where $p,q$ are relatively prime integers, and $q > 0$.

Then $f(a) = \left({\large{\frac{q}{q+1}}}\right)a\ne a$, so $f$ is not continuous at $x=a$.

Next suppose $a$ is a negative irrational number.

Then $f(a) = |a| \ne a$, so $f$ is not continuous at $x=a$.

Thus if $a < 0$ or if $a$ is a positive rational number, $f$ is not continuous at $x=a$.

Next suppose $a=0$ or $a$ is a positive irrational number.

Note that both cases, we have $f(a)=a$.

We want to show that in both cases, $f$ is continuous at $x=a$.

Let $(w_n)$ be a sequence of real numbers such that

  • ${\displaystyle{\lim_{n\to\infty}w_n=a}}$$\\[4pt]$
  • $w_n\ne a$, for all $n$.

If the sequence $(w_n)$ has only finitely many irrational terms, then as previously shown, we have $$ \lim_{n\to\infty}f(w_n) = a = f(a) $$ and if the sequence $(w_n)$ has only finitely many rational terms, then we have $$ \lim_{n\to\infty}f(w_n) = \lim_{n\to\infty}|w_n| = |a| = a = f(a) $$ Finally, if the sequence $(w_n)$ has infinitely many rational terms and infinitely many irrational terms, then since $$\lim_{n\to\infty}f(w_n)=a$$ on each of those two subsequences, it follows that $$\lim_{n\to\infty}f(w_n)=a$$ for the whole sequence $(w_n)$.

Thus, we have $$\lim_{n\to\infty}f(w_n)=a=f(a)$$ so $f$ is continuous at $x=a$.

This completes the proof.

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  • $\begingroup$ Thank you for your answer. Do I understand correctly that by introducing a set $S_d$ and obtaining some its properties you are actually showing that for a given finite interval there is a finite set of rational numbers with coprime numerator and denominator? If I'm not mistaken I've seen something similar for proving continuity/discontinuity of the Thomae's function. $\endgroup$
    – roman
    Jul 17, 2019 at 17:11
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    $\begingroup$ @roman: What I showed was that if $(x_n)$ is an infinite sequence of rationals converging to some $a\in\mathbb{R}$, with $x_n\ne a$ for all $n$, then for any positive integer $d$, there are at most finitely terms of the sequence $(x_n)$ with denominator $d$. As a consequence, for such a sequence $(x_n)$, the sequence $(q_n)$ of denominators must approach infinity. $\endgroup$
    – quasi
    Jul 17, 2019 at 17:18

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