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Ok. This question may sound very easy, but actually i am in great need of it. I have been facing trouble in constructing functions, which are only continuous at some particular sets.

For e.g, the standard example of a function which is only continuous at one point, is the function, $f(x) = x, \ x \in \mathbb{Q}$ and $f(x) = -x, x \in \mathbb{R} \setminus \mathbb{Q}$. Similarly, i would like to know as to how to construct a function which is

  • Continuous at exactly $2,3,4$ points.

  • Continuous exactly at integers

  • Continuous exactly at Natural numbers

  • Continuous exactly at Rationals.

I would like to see many examples (with proof!), so that i can don't struggle when somebody asks me to construct such functions.

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The variety of questions you ask (and the little work you show in them) makes me think you are a whole group of people hiding behind a name, à la Bourbaki... – Mariano Suárez-Alvarez Oct 25 '10 at 16:49
@Mariano: Well, not really. I am the only one! – anonymous Oct 25 '10 at 16:54
see also:… – Isaac Oct 25 '10 at 16:56
@Mariano: This reminds me of a joke I read somewhere: "Why did Bourbaki stop writing books? They discovered that Serge Lang was a single person." – Hans Lundmark Oct 25 '10 at 17:26
@Hans: You made my day, thanks for the joke! :) @Chandru: showing people what you've tried in solving your problems would certainly dispel any doubts. – J. M. Oct 25 '10 at 23:51
up vote 18 down vote accepted
  1. One simple way of constructing a function which is continuous only at a finite number of points, $x=a_1,\ldots,a_n$, is to do a slight modification to the function you give: take a polynomial $p(x)$ that has roots exactly at $x=a_1,\ldots,a_n$ (e.g., $p(x) = (x-a_1)\cdots(x-a_n)$) , and then define $$ g(x) = \left\{\begin{array}{ll} p(x) & \text{if $x\in\mathbb{Q}$;}\\ 0 & \text{if $x\notin\mathbb{Q}$.} \end{array}\right.$$ The function is continuous at $a_1,\ldots,a_n$, and since $p(x)\neq 0$ for any $x\notin\{a_1,\ldots,a_n\}$ then $g(x)$ is not continuous at any point other than $a_1,\ldots,a_n$. Other possibilities should suggest themselves easily enough.

  2. A function that is continuous exactly at the integers: a similar idea will work: find a function that has zeros exactly at the integers, for example $f(x)=\sin(\pi x)$, and then take $$g(x) = \left\{\begin{array}{ll} \sin(\pi x) & \text{if $x\in\mathbb{Q}$;}\\ 0 & \text{if $x\notin\mathbb{Q}$.} \end{array}\right.$$

  3. A function continuous exactly in the natural numbers: take a function that is continuous at the integers, and redefine it as the characteristic function of the rationals in appropriate places(what happens at $0$ depends on whether you believe $0$ is in the natural numbers or not). Assuming that $0\in\mathbb{N}$, one possibility is: $$g(x) = \left\{\begin{array}{ll} \sin(\pi x)&\text{if $x\in\mathbb{Q}$ and $x\geq 0$;}\\ x & \text{if $x\in\mathbb{Q}$ and $-\frac{1}{2}\lt x\leq 0$;}\\ 1 & \text{if $x\in\mathbb{Q}$ and $x\leq -\frac{1}{2}$;}\\ 0 & \text{if $x\notin\mathbb{Q}$.} \end{array}\right.$$

  4. A function continuous exactly on the rationals. This one is a bit trickier. There is no such function. This follows because the set of discontinuities of a real valued function must be a countable union of closed sets.

    Perhaps then, we might anticipate the next question:

  5. A function that is continuous exactly on the irrationals. An example is the following: let $s\colon\mathbb{N}\to\mathbb{Q}$ be an enumeration of the rationals (that is, a bijection from $\mathbb{N}$ to $\mathbb{Q}$. Define $f(x)$ as follows: $$f(x) = \sum_{\stackrel{n\in\mathbb{N}}{s_n\leq x}} \frac{1}{2^n}.$$ The function has a jump at every rational, so it is not continuous at any rational. However, if $x$ is irrational, let $\epsilon\gt 0$. Then there exists $N$ such that $\sum_{k\geq N}\frac{1}{2^k}\lt \epsilon$. Find a neighborhood of $x$ which excludes every $q_m$ with $m\leq N$, and conclude that the difference between the value of $f$ at $x$ and at any point in the neighborhood is at most $\sum_{k\geq N}\frac{1}{2^k}$.

    Edit: As I was reminded in the comments by jake, in fact the "standard example" of a function that is continuous at every rational and discontinuous at every rational is Thomae's function. The example I give is a monotone function, and although it is discontinuous at every rational, it is continuous from the right at every number.

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@Arturo Magidin: Super Work! Really appreciate it, although the function which is continuous exactly at irrationals appears intricate! – anonymous Oct 25 '10 at 17:11
@Mariano: intricate, yes; the reason I say it is "tricky" is that the answer in that case is "there is no such function!". So if you go around trying to find one, you keep running into walls. – Arturo Magidin Oct 25 '10 at 17:16
@Chandru1: Speaking of irrelevant comments... Mariano is not the only one who has wondered in the past about your question-asking habits... e.g.,… So it seems hardly fair to start picking on him for something a lot of us wonder about. – Arturo Magidin Oct 25 '10 at 17:19
@Arturo Magidin: In my analysis class, we were given Thomae's function as the classic example of a function continuous exactly at all irrationals. See's_function . Is there a reason you didn't use that one? It seems much simpler. Unless it's somehow the same thing... – jake Oct 25 '10 at 20:06
@Jake: Ah; mine is the classic example of a monotonoe function that is continuous exactly at all rationals. Slightly different "class"... I did know Thomae's function, but the one I always remember is this one. – Arturo Magidin Oct 25 '10 at 20:20

Continuous at 2, 3, 4: $f(x)=(x-2)(x-3)(x-4)$ if $x$ is rational, $f(x)=0$ if $x$ is irrational.

Continuous at the integers: $f(x)=\sin(\pi x)$ if $x$ is rational, 0 if $x$ is irrational.

Continuous at the natural numbers: $f(x)=\sin(\pi x)$ if $x$ is rational and not a nonpositive integer, 0 if $x$ is irrational, 1 if $x$ is a nonpositive integer.

Continuous exactly at the rationals: Impossible, because the set of rational numbers is not a $G_\delta$.

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@Jonas: Thanks a lot Jonas: By the way can you prove the continuity of $\sin{\pi x}$ at integers! – anonymous Oct 25 '10 at 17:07
@Jonas: I think we can have a function which is discontinuous only at rationals! – anonymous Oct 25 '10 at 17:07
@Jonas. Heh; great minds and all that... – Arturo Magidin Oct 25 '10 at 17:08
@Chandru: if $x$ is not an integer, then $\sin(\pi x)\neq 0$; take $\epsilon \lt \frac{1}{2}|\sin(\pi x)|$ and in any $\delta$ nbd of $x$ you have irrationals where the value of $f$ is more than $\epsilon$ away from $f(x)$; exactly the same as in the first case of the polynomial example for the first question. – Arturo Magidin Oct 25 '10 at 17:09
@Chandru1: The limit along the irrationals is always zero, so the function is continuous precisely where the (continuous) function $\sin(\pi x)$ is zero. You can fill in the details. – Jonas Meyer Oct 25 '10 at 17:10

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