0

  1. Can Dirichlet's function be modified in such a way that it is continuous at some real number? For instance, as $xD(x)$ is continuous at $x=0$, is it possible that $(x-1)D(x)$ is continuous at $x=1$? Here $D(x)$ is the Dirichlet function.

  2. Can it be modified to be continuous at finitely many points in $\mathbb{Q}$?

  3. I am aware that there can't be a function continuous only on rational numbers, but with the same approach as $2$, if possible, why isn't it possible to construct such a function?

Thanks in advance!

mathematics2x2life
  • 13,322

1 Answers1

0

Given numbers $x_1,\ldots,x_n$, you can define an "accordion function" $g(x)$ by piecing together absolute values so that $g(x_j)=0$ and $g((x_j+x_{j+1})/2)=1$. Then $g(x)D(x)$ is continuous at $x_1,\ldots,x_n$ and discontinuous everywhere else.

The same approach wouldn't work for all the rationals because they are dense, and so there is no room for the function to "raise up" and "come down".

Martin Argerami
  • 205,756
  • I don't understand the part "piecing together absolute values." –  Nov 11 '13 at 01:56
  • A picture would be easier. But I'm talking about a function is zero at $x_1$; then it raises linearly until it is $1$ at $(x_1+x_2)/2$; the decreases linearly until it is zero at $x_2$; then it raises linearly... – Martin Argerami Nov 11 '13 at 02:11
  • Why do you want it to raise till 1, before it decreases. Are you constructing a triangle between every consecutive x? I don't understand why? –  Nov 13 '13 at 02:53
  • Well, it has to raise and come down somehow. Any continuous function $g(x)$ that is zero precisely at $x_1,\ldots,x_n$ will do. I just gave an easy concrete example. – Martin Argerami Nov 13 '13 at 03:27
  • I think I now see how that would work for countable terms n. But rational numbers are countable yet dense in every interval; therefore, we wouldn't be able to make a function go up and down enough times to touch all rationals. Is that right? –  Nov 13 '13 at 03:35
  • Exactly. It can still be done if you have countably many points with a single accumulation point (${1/n}$, say). But it will not be possible whenever the countable set is dense in any interval. – Martin Argerami Nov 13 '13 at 03:42
  • Um, what is an accumulation point? –  Nov 13 '13 at 03:47
  • An accumulation point for a sequence ${x_n}$ is a point $y$ such that for any $\delta>0$, the interval $(y-\delta,y+\delta)$ contains infinitely many points of the sequence. In the example mentioned above, the sequence ${1/n}$ has $0$ as its only accumulation point. – Martin Argerami Nov 13 '13 at 03:50
  • How can I continue this discussion? Is it okay to go on this comment section? –  Nov 14 '13 at 02:53
  • We can continue here or in the chat, it's the same for me. – Martin Argerami Nov 14 '13 at 15:13