27
$\begingroup$

Can we construct a monotonic function $f : \mathbb{R} \to \mathbb{R}$ such that there is a dense set in some interval $(a,b)$ for which $f$ is discontinuous at all points in the dense set? What about a strictly monotonic function?

My intuition tells me that such a function is impossible.

Here is a rough sketch of an attempt at proving that such a function does not exist: we could suppose a function satisfies these conditions. Take an $\epsilon > 0$ and two points $x,y$ in this dense set such that $x<y$. Then, $f(x)<f(y)$ because if they are equal, then the function is constant at all points in between, and there is another element of $X$ between $x$ and $y$, which would be a contradiction. Take $f(y)-f(x)$. By the Archimedean property of the reals, $f(y)-f(x)<n\epsilon$ for some $n$.

However, after this point, I am stuck. Could we somehow partition $(x,y)$ into $n$ subintervals and conclude that there must be some point on the dense set that is continuous?

$\endgroup$
41
$\begingroup$

Such a function is possible.

Let $\Bbb Q=\{q_n:n\in\Bbb N\}$ be an enumeration of the rational numbers, and define

$$f:\Bbb R\to\Bbb R:x\mapsto\sum_{q_n\le x}\frac1{2^n}\;.\tag{1}$$

The series $\sum_{n\ge 0}\frac1{2^n}$ is absolutely convergent, so $(1)$ makes sense. If $x<y$, there is some rational $q_n\in(x,y)$, and clearly $f(y)\ge f(x)+\frac1{2^n}$, so $f$ is monotone increasing. However, $f$ is discontinuous at every rational:

$$\lim_{x\to {q_n}^-}f(x)=\sum_{q_k<q_n}\frac1{2^k}<\sum_{q_k\le q_n}\frac1{2^k}=f(q_n)\;.$$

Thus, $f$ is discontinuous on a set that is dense in $\Bbb R$ (and in every open interval of $\Bbb R$).

$\endgroup$
  • 1
    $\begingroup$ Seriously? I was just typing that... $\endgroup$ – Asaf Karagila Jul 19 '12 at 7:09
  • 2
    $\begingroup$ @Asaf: It’s a pretty standard example, I think. $\endgroup$ – Brian M. Scott Jul 19 '12 at 7:10
  • $\begingroup$ True.${}{}{}{}$ $\endgroup$ – Asaf Karagila Jul 19 '12 at 7:11
  • 2
    $\begingroup$ @AndrewSalmon: Of course there is not. It also does not assume that. $\endgroup$ – Asaf Karagila Jul 19 '12 at 7:15
  • 1
    $\begingroup$ @BrianM.Scott you mentioned in one of comments: "You can use the same trick to build a function that’s discontinuous at each point of any countably infinite X⊆R." Can we build a function defined on E⊆R that is discontinuous only at countable infinite subset C⊆E $\endgroup$ – Sushil Nov 25 '14 at 1:49
5
$\begingroup$

For those interested, here is a slightly more rigorous proof that the construction exists using measure theory.

We will define a function $H$ which is continuous from the left at ever $x \in [0,1]$ but is only continuous from the right at the irrationals. To build such a function we need the folliowing scaffolding.

First let $\psi: \mathbb{N} \to \mathbb{Q} \cap [0,1]$ be a bijection enumerating the rationals in the interval $[0,1]$. Then define $B(x) = \{n : \psi(n)< x\}$ or equivalently $B(x) = \psi^{-1}([0,x))$. Finally let $\mu: P(\mathbb{N}) \to \overline{\mathbb{R}}$ be the counting measure. We additionally define a measure $\nu: P(\mathbb{N}) \to \mathbb{R}$ such that $$\nu(A) = \int_A 2^{-n} d \mu(n).$$ The measure $\nu$ has the additional property that $\nu \ll \mu$ and $\nu(\mathbb{N}) = \sum_{n \in \mathbb{N}} 2^{-n} \mu({n}) = 1 < \infty.$

We claim that the function $H(x) = \nu(B(x))$ has the properties of $f$ in the statement of the problem. We will first show that for every $x \in [0,1]$ the function $H(x)$ is left continuous. Take a sequence of $x_k \to x$ from the left, we can then rearrange the sequence to be strict monotonic. It follows that if $k > m$ then $$B(x_m) = \{n: \psi(n) < x_m < x_k < x\} \subset \{n: \psi(n) < x_k < x\} = B(x_k).$$ By the finiteness of $\nu$ we have that by upward measure continuity $$\lim_{k \to \infty} H(x_k) = \lim_{k \to \infty} \nu(B(x_k)) = \nu\left(\bigcup_{k=1}^\infty B(x_k) \right) = \nu(\left \{n: \psi(n) < x\}\right ) = H(x).$$ Note that if $ n \in \bigcup B(x_k)$ there is an $K$ so that $\psi(n) < x_k < x $ so any $n$ with $\psi(n) < x$ is in $\bigcup B(x_k).$

Next we claim that $H$ is only right continuous only when $x$ is irrational. Take a sequence of $x_k \to x$ from the right and rearrange the sequence to be strict montonic. It follows that if $k > m$ then $$B(x_m) = \{n: \psi(n) < x_m\} \supset \{n: \psi(n) < x_k < x_m\} = B(x_k).$$ By finiteness of $\nu$ and downard measure continuity $$\lim_{k \to \infty} H(x_k) = \lim_{k \to \infty} \nu(B(x_k)) = \nu\left(\bigcap_{k=1}^\infty B(x_k) \right) = \nu(\left \{n: \psi(n) < x_k\ \forall k\}\right ).$$

If $x$ is irrational then $m \in \{n: \psi(n) < x_k\ \forall k\}$ implies that $\psi(m) < x$ and if $\psi(m) < x$ then $\psi(m) < x_k$ for all $k$ so $\{n: \psi(n) < x_k\ \forall k\} = B(x)$ and $H(x_k) \to H(x)$ from the right. If $x$ is rational then $x= \psi(q)$ for some $q \in \mathbb{N}.$ Thus $x < x_k \forall k$ implies that $\{n: \psi(n) < x_k\ \forall k\} = B(x) + \{q\} = D$. It follows that $\nu(D) = \nu(B(x)) + 2^{-q} > H(x)$. So $H(x_k) \to H(x) + 2^{-q} \neq H(x)$ from the right, and so $H$ is not right continuous at the rationals.

We have thus shown that for any $x \in [0,1] \setminus \mathbb{Q}$, any sequence $x_k \to x$ has the property $\lim H(x_k) = x$ from the left and the right, and if $x \in [0,1] \cap \mathbb{Q}$ then if $x_k \to x$, $\lim H(x_k)$ does not exist. Therefore $H$ is continuous at every irrational and discontinuous at every rational.

You can then repeat this construction along the whole real line by adding the nearest integer each time.

$\endgroup$

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.