# Establishing the existence of a strictly increasing real function, discontinuous at all rationals and continuous at all irrationals

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:

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

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

1. 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).

2. $$g$$ 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.

1. $$g$$ is discontinuous at all points of $$\mathbb{Q}$$: Let $$x = {p \over q}$$ be a fully reduced rational. 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 rational $$y$$ 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?

• You mean to use $\sup f(x)$ in the definition of $g(x)$? Apr 26, 2014 at 13:17
• I meant to use $\sup g(x)$ so that the function can be strictly increasing. Apr 26, 2014 at 13:22
• Then the definitin of $g$ is circular. In fact from $1>\frac12>\frac13>\ldots$, you would require $g(1)\ge 1+g(\frac12)\ge 1+\frac12+g(\frac13)\ge 1+\frac12+\frac13+\ldots$ Apr 26, 2014 at 13:23

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 $$h(x)=\begin{cases}g(x)&\text{if }x>0\\0&\text{if }x=0\\-g(-x)&\text{if }x<0\end{cases}$$
• 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. Apr 26, 2014 at 14:29
• Hagen von Eitzen Can you help Connection of complex $e^z$ and real Dirichlet please?