# Can we construct a function $f:\mathbb{R} \rightarrow \mathbb{R}$ such that it has intermediate value property and discontinuous everywhere?

Can we construct a function $f:\mathbb{R} \rightarrow \mathbb{R}$ such that it has intermediate value property and discontinuous everywhere?

I think it is probable because we can consider $$y = \begin{cases} \sin \left( \frac{1}{x} \right), & \text{if } x \neq 0, \\ 0, & \text{if } x=0. \end{cases}$$ This function has intermediate value property but is discontinuous on $x=0$.

Inspired by this example, let $r_n$ denote the rational number,and define $$y = \begin{cases} \sum_{n=1}^{\infty} \frac{1}{2^n} \left| \sin \left( \frac{1}{x-r_n} \right) \right|, & \text{if } x \notin \mathbb{Q}, \\ 0, & \mbox{if }x \in \mathbb{Q}. \end{cases}$$

It is easy to see this function is discontinuons if $x$ is not a rational number. But I can't verify its intermediate value property.

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2
Kindly typeset your post in future. – user17762 Feb 13 '11 at 9:13

## 4 Answers

Sure. The class of functions satisfying the conclusion of the Intermediate Value Theorem is actually vast and well-studied: such functions are called Darboux functions in honor of Jean Gaston Darboux, who showed that any derivative is such a function (the point being that not every derivative is continuous).

A standard example of an everywhere discontinuous Darboux function is Conway's base 13 function.

(Perhaps it is worth noting that the existence of such functions is not due to Conway: his is just a particularly nice, elementary example. I believe such functions were already well known to Rene Baire, and indeed possibly to Darboux himself.)

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+1 for the base 13 function.  I also thought sin(1/x) was a bit of a cheat, being discontinuous only in one point. – Glen Wheeler Feb 13 '11 at 10:57

Any function that give all the reals numbers on any open interval is an example. You can find other examples of this in the answers to this question on MathOverflow.

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Please look at the problem $1.3.29$ in Problems in Mathematical Analysis Vol II, Continuity and Differentiation, by Kaczor and Nowak.

They have provided solutions also. Anyhow since one can't view it in Google books, I am Texing out the solution here.

Question: Recall that every $x \in (0,1)$ can be represented by a binary fraction $0.a_{1}a_{2}a_{3}\cdots$, where $a_{i} \in \{0,1\}$ and $i=1,2, \cdots$. Let $f: (0,1) \to [0,1]$ be defined by $$f(x) = \overline{\lim_{n \to \infty}} \ \frac{1}{n} \sum\limits_{i=1}^{n}a_{i}$$ Prove that $f$ is discontinuous at each $x \in (0,1)$ but nevertheless has the intermediate value property.

Solution. We show that if $I$ is a subinterval of $(0,1)$ with non-empty interior then, $f(I)=[0,1]$. To this end, note that such an $I$ contains a sub-interval $\bigl(\frac{k}{2^{n_0}}, \frac{k+1}{2^{n_0}}\bigr)$. So it is enough to show that, $$f\biggl(\biggl(\frac{k}{2^{n_0}},\frac{k+1}{2^{n_0}}\biggr)\biggr)= [0,1]$$

Now observe that if $x \in (0,1)$ then either $x-\frac{m}{2^{n_0}}$ with some $m$ and $n_0$ or $x \in \bigl(\frac{j}{2^{n_0}},\frac{j+1}{2^{n_0}}\bigr)$ with some $j=0,1, \cdots, 2^{n_0}-1.$ If $x = \frac{m}{2^{n_0}}$, then $f(x)=1$, and the value of $f$, at the middle point of $\bigl(\frac{k}{2^{n_0}}, \frac{k+1}{2^{n_0}}\bigr)$ is also $1$. Next if $x \in \bigl(\frac{j}{2^{n_0}}, \frac{j+1}{2^{n_0}}\bigr)$ then there is $x' \in \bigl(\frac{k}{2^{n_0}}, \frac{k+1}{2^{n_0}}\bigr)$, such that $f(x)=f(x')$. Indeed, all numbers in $\bigl(\frac{k}{2^{n_0}}, \frac{k+1}{2^{n_0}}\bigr)$ have the same first $n_0$ digits, and we can find $x'$ in this interval for which all the remaining digits are as in the binary expansion of $x$. Since $$\overline{\lim_{n\to\infty}} \ \frac{\sum\limits_{i=1}^{n} a_{i}}{n} = \overline{\lim_{n\to\infty}} \ \frac{\sum\limits_{i=n_{0}+1}^{n} a_{i}}{n}$$ we get $f(x)=f(x')$.

Consequently it is enough to show that $f((0,1))=[0,1]$, or in other words, for each $y \in [0,1]$ there is $x \in (0,1)$ such that $f(x)=y$. It follows from the above that $1$ is attained, for example at $x =\frac{1}{2}$. To show that $0$ is also attained, take $x = 0.a_{1}a_{2}\cdots,$ where $$a_{i}=\biggl\{\begin{array}{cc} 1 & \text{if} \ i=2^{k}, \ k=1,2, \cdots, \\\ 0 & \text{otherwise.}\end{array}$$

Then $$f(x) = \lim_{k \to \infty} \frac{k}{2^{k}}=0$$

To obtain $y=\frac{p}{q}$, where $p$ and $q$ are Co-prime positive integers, take $$x = \underbrace{.00\cdots 0}_{q-p} \: \underbrace{11\cdots 1}_{p} \: \underbrace{00\cdots 0}_{q-p} \cdots,$$ where blocks of $q-p$ zeros alternate with blocks of $p$ ones. Then $f(x) = \lim_{k\to\infty} \frac{kp}{kq}=\frac{p}{q}$. Now our task is to show that every irrational $y \in [0,1]$ is also attained. We know that there is a sequence of rationals $\frac{p_n}{q_n}$, where each pair of $p_n$ and $q_n$ are co-prime converging to an irrational $y$. Let $$x = \underbrace{.00 \cdots 0}_{q_{1}-p_{1}} \: \underbrace{11\cdots 1}_{p_{1}} \: \underbrace{00 \cdots 0}_{q_{2}-p_{2}} \cdots,$$

Then $f(x) = \lim_{n \to \infty} \frac{p_{1} + p_{2} + \cdots + p_{n}}{q_{1} + q_{2} + \cdots + q_{n}} = \lim_{n \to \infty} \frac{p_{n}}{q_{n}} = y$.

Since $\displaystyle\lim_{n \to \infty} q_{n} = +\infty$, the second equality follows from the Stolz theorem.

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To, the downvoter: why a downvote? – user9413 Jun 9 '11 at 10:53

Not only such functions exists, but it turns out there are `lots` of such functions.

There is actually a Theorem, I think by Sierpinski, that any real valued function is the sum of two Darboux functions...

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– Martin Sleziak Apr 1 '12 at 8:14
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