On functions

Here are some examples on functions with "strange" properties.

Give an example of a function that:

1. is only continuous at a single point.
2. is continuous at isolated points.

Does there exist a function that:

1. is continuous only on the rationals?
2. is continuous only on the irrationals?

Explain your answer giving a brief explanation. 

Solution

These are very interesting questions. Function do possess interesting properties and sometimes they surprise us either happily or sadly. Let us begin with the examples:

1. The classic functions would be:
   
   $$f(x) = \left\{\begin{matrix}
   x &,  &x \in \mathbb{Q} \\
    0& , & x \in \mathbb{R} \setminus \mathbb{Q}
   \end{matrix}\right. \quad, \quad f(x)= \left\{\begin{matrix}
   x &,  & x \in \mathbb{Q} \\
   -x & , & x \in \mathbb{R} \setminus \mathbb{Q}
   \end{matrix}\right.$$
   
   which are continuous only at $x=0$. The check is immediate.


2. Now, one function that would work in this case would be the following:
   
   $$f(x) = \left\{\begin{matrix}
   \sin \pi x &,  & x \in \mathbb{Q} \\
    0& , &  x \in \mathbb{R} \setminus \mathbb{Q}
   \end{matrix}\right.$$
   
   This function is continuous at any  single specified point. The set of continuity points is actually the set formed by the roots of the equation $\sin \pi x =0$ and that is nothing else than $\mathbb{Z}$. 

Now the following answers require some knowledge of Topology, namely Baire's theorem and some knowledge of dense sets.

1.  The answer in this case would be no. There does not exist a function that is only continuous on the rationals. Reason being that the set of discontinuities must be an $F_\sigma$ one (besides being a closed set) , whereas the set of the irrationals are not since  $\mathbb{R} \setminus \mathbb{Q}$ is not a closed set which is a consequence of Baire's theorem that states that the union of closed sets with no interior is a closed set with no interior.


2. Yes, there exists. One function may be the Thomae's function defined as:
   
   $$f(x) = \left\{\begin{matrix}
   \frac{1}{q} & , & x \in \mathbb{Q} , \; q>0 \\
    0& , &  x \in \mathbb{R} \setminus \mathbb{Q}
   \end{matrix}\right.$$