2
$\begingroup$

Goal: Show that there exists a strictly increasing function on $\mathbb{R}$ discontinuous at all points of $\mathbb{Q}$ and continuous at all irrational numbers.

Attempt:

  1. Let $f(x)$ denote the popcorn function on $\mathbb{R}$ s.t.

    $$ f(x) = \begin{cases} \frac{1}{q} &\text{if }x\text{ is rational, }x=\tfrac{p}{q}\text{ in lowest terms and } q > 0\\ 0 &\text{if }x\text{ is irrational.} \end{cases} $$

    The popcorn function $f(x)$ can be visualized on $(0,1)$ as follows:

    enter image description here

  2. Let $g(x): \mathbb{R}_{\ge 0}\rightarrow \mathbb{R}$ denote the following function:

    $$ g(x) = \begin{cases} 0 &\text{if }x=0\\ \sup\{g(y) : y < x\} + \frac{1}{q} &\text{if }x>0\text{ is rational, }x=\tfrac{p}{q}\text{ in lowest terms and } q > 0\\ \sup\{g(y) : y < x\} &\text{if }x>0\text{ is irrational.} \end{cases} $$

  3. Extend $g(x)$ to all of $\mathbb{R}$ by making $g(-x) = -g(x)$. We'll refer to this extended function as $g(x)$ as well (abusing the name $g$ without harm).

  4. $g$ is strictly increasing: Let $x < y$. Then there exists a fully reduced rational $p/q \in (x,y)$ so that

    $$ g(x) < g(x) + p/q \le g\left(p/q\right) \le g(y) $$

    which implies $g(x) < g(y)$ as desired.

  5. $g$ is discontinuous at all points of $\mathbb{Q}$: Let $x = {p \over q}$ be a fully reduced irrational. We can further assume $x$ is positive without loss of generality. Let $\epsilon = {1 \over q}$. Then for any $\delta > 0$, we have that $(x- \delta, x)$ contains at least one irrational $y$ so that

    $$ g(y) < g(y) + {1 \over q} \le g(x) $$

    This in particular implies that $g(y) \notin (g(x)- \epsilon, g(x)+\epsilon)$ so that $g$ fails to be continuous at $x$ as desired.

Question: How can I show that $g$ is continuous at all irrational points?

$\endgroup$
  • $\begingroup$ You mean to use $\sup f(x)$ in the definition of $g(x)$? $\endgroup$ – Hagen von Eitzen Apr 26 '14 at 13:17
  • $\begingroup$ I meant to use $\sup g(x)$ so that the function can be strictly increasing. $\endgroup$ – user1770201 Apr 26 '14 at 13:22
  • $\begingroup$ Then the definitin of $g$ is circular. In fact from $1>\frac12>\frac13>\ldots$, you would require $g(1)\ge 1+g(\frac12)\ge 1+\frac12+g(\frac13)\ge 1+\frac12+\frac13+\ldots$ $\endgroup$ – Hagen von Eitzen Apr 26 '14 at 13:23
2
$\begingroup$

My suggestion: Try $$ g(x)=\sum_{0\le q\le x}f(q)^3$$ or $$ g(x)=\sum_{n\in\mathbb N}\frac{\lfloor nx\rfloor}{n^3}$$ for $x>0$ and then $$h(x)=\begin{cases}g(x)&\text{if }x>0\\0&\text{if }x=0\\-g(-x)&\text{if }x<0\end{cases}$$

$\endgroup$
  • $\begingroup$ How does one establish that $\sum_{0\le q\le x}f(q)^3$ is continuous on irrational points? I'm assuming this is where the power of $3$ will come in handy. $\endgroup$ – user1770201 Apr 26 '14 at 14:29
  • $\begingroup$ Hagen von Eitzen Can you help Connection of complex $e^z$ and real Dirichlet please? $\endgroup$ – BCLC Aug 19 '18 at 6:26

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.