0
$\begingroup$

Construct an increasing function f on R that is continuous at every irrational number and is discontinuous at every rational number.

Solution: Let ($r_n$) be a sequence with distinct terms whose range is $\mathbb{Q}$. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be given by $$f(x)= \sum_{r_n<x} \frac{1}{2^n}$$

If $x_1 < x_2$, then the series yielding $f(x_2)$ has additional positive terms than the series whose sum is $f(x_1)$. Thus f is increasing.

I don't understand this function at all. Can anyone tell me about the construction of function?

$\endgroup$
8
  • 1
    $\begingroup$ It means that you have to consider those indices for $n$ for which $r_n\lt x$. For example: suppose $r_1,r_3,r_5 \lt x$ and for all other $r_n$ we have $r_n\ge x$ then $f(x)=\frac 12+\frac 1{2^3}+\frac 1{2^5}$. Also refer this: math.stackexchange.com/questions/4126663/… $\endgroup$
    – Koro
    Jun 15, 2021 at 12:08
  • $\begingroup$ Are you sure that it's $\frac1{2n}$? I think that it should be $\frac1{2^n}$. $\endgroup$ Jun 15, 2021 at 12:08
  • 1
    $\begingroup$ The function is a sum of step functions of decreasing amplitude $2^{-n}$, with a discontinuity at $r_n$. Hence by construction, the sum converges everywhere, but has a discontinuity at every rational. $\endgroup$
    – user65203
    Jun 15, 2021 at 12:15
  • 1
    $\begingroup$ One special thing about this function is: it is discontinuous only at $r_n$'s and continuous elsewhere. $r_n$ is a rational no. Note that set of rationals is countable so can be indexed (enumerated). $\endgroup$
    – Koro
    Jun 15, 2021 at 12:15
  • 1
    $\begingroup$ $r_n$ is any enumeration of the rationals (they are countable). $\endgroup$
    – user65203
    Jun 15, 2021 at 12:15

2 Answers 2

2
$\begingroup$

Let$$f_0(x)=\begin{cases}0&\text{ if }r_0\geqslant x\\1&\text{ if }r_0<x.\end{cases}$$It's increasing, right!? Besides, it is discontinuous at $r_0$ and only at $r_0$.

Now, let$$f_1(x)=\begin{cases}0&\text{ if }r_1\geqslant x\\\frac12&\text{ if }r_1<x.\end{cases}$$It's increasing and it is discontinuous at $r_1$ and only at $r_1$. So, $f_0+f_1$ is increasing and it is discontinuous at $r_0$ and at $r_1$ and only at those points.

More generally, for each $n\in\Bbb Z_+$, let$$f_n(x)=\begin{cases}0&\text{ if }r_n\geqslant x\\\frac1{2^n}&\text{ if }r_n<x.\end{cases}$$Then $f$ is increasing, since it is equal to $\sum_{n=0}^\infty f_n$. And it is not hard to see that it is discontinuous at $x$ if and only if $x\in\{q_n\mid n\in\Bbb Z_+\}=\Bbb Q$ (this follows from the fact that the convergence of the series $\sum_{n=0}^\infty f_n$ is uniform, by the Weierstrass $M$-test). The reason why I told you in the comments that it should be $\frac1{2^n}$ rather than $\frac1{2n}$ was so that the expression $\sum_{n=0}^\infty f_n$ makes sense, that is, so that it converges, for every $x\in\Bbb R$.

$\endgroup$
7
  • $\begingroup$ One missing comment: the sum is always convergent. $\endgroup$
    – user65203
    Jun 15, 2021 at 12:19
  • $\begingroup$ Fantastic! Thank you for this :-) I would also like to add that instead of $\sum \frac 1{2^n}$, any convergent series of positive terms will also work :) $\endgroup$
    – Koro
    Jun 15, 2021 at 12:21
  • $\begingroup$ @YvesDaoust Thank you. I've added that to my answer. $\endgroup$ Jun 15, 2021 at 12:22
  • $\begingroup$ @Koro Indeed.${}$ $\endgroup$ Jun 15, 2021 at 12:22
  • $\begingroup$ In fact it seems worth mentioning that the sum converges uniformly (hence it's continuous at every point where all the terms are continuous). $\endgroup$ Jun 15, 2021 at 13:05
1
$\begingroup$

First you need a sequence covering all the rational numbers, such as $$0,-1,1,-2,-\frac12, \frac12, 2, -3, -\frac13, \frac13, 3, -4, -\frac32, -\frac23, -\frac14,\frac14, \ldots$$ and then a corresponding sequence of powers of $\frac12$ $$\tfrac12,\tfrac14,\tfrac18,\tfrac1{16},\tfrac1{32},\tfrac1{64},\tfrac1{128},\tfrac1{256},\tfrac1{512},\tfrac1{1024},\tfrac1{2048},\tfrac1{4096},\tfrac1{8192},\tfrac1{16384},\tfrac1{32768}, \tfrac1{65536},\ldots$$

So if you want to find for example $f(-1)$ you would take those powers of $\frac12$ corresponding to less than $-1$ in the sequence of rationals (to $-2,-3,-4,-\frac32,\ldots$), and add them up, i.e. $$f(-1)=\frac1{16}+\frac1{256}+\frac1{4096}+\frac1{8192}+\cdots \approx 0.006678$$

This is clearly discontinuous at $x=-1$ as any $x> -1$ has $f(x)$ at least $\frac14$ larger. Similarly with any rational.

$\endgroup$
6
  • $\begingroup$ What the properties of this sequence? Any sequence whose terms are distinct and every element is rational? I am stuck by how to confirm such a sequence. But many thanks for your concrete example. This is very helpful. $\endgroup$
    – Mariana
    Jun 15, 2021 at 13:23
  • 1
    $\begingroup$ The properties are that this is a strictly increasing sequence, taking values on $(0,1)$, discontinuous (though left-continuous) on the rationals and continuous (and differentiable with derivative $0$) on the irrationals. With a slight adjustment to make it right-continuous, e.g. using $1-f(x)$, it could be a cumulative distribution function for a discrete random variable on the rationals $\endgroup$
    – Henry
    Jun 15, 2021 at 13:32
  • $\begingroup$ no, I was asking the sequence $r_n$. $\endgroup$
    – Mariana
    Jun 15, 2021 at 13:35
  • 1
    $\begingroup$ $r_n$ takes each rational number as $\frac{a}{b}$ with $b$ a positive integer coprime to integer $a$, then orders the rationals by $|a|+|b|$ and within those by $\frac{a}{b}$. So $\frac{-3}{1}, \frac{-1}{3}, \frac{1}{3}, \frac{3}{1}$ are those where $|a|+|b|=4$, and we can ignore $\frac{2}{2}$ because $2$ is not coprime to $2$ (in fact we already listed that as $\frac11$) $\endgroup$
    – Henry
    Jun 15, 2021 at 13:40
  • 1
    $\begingroup$ Any complete list of the rationals is fine. If you leave a rational out, your $f(x)$ will be continuous there $\endgroup$
    – Henry
    Jun 15, 2021 at 13:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .