Real valued function which is continuous only on transcendental numbers

Question

First of all, I am sorry for asking this question.

We know that $R$ is uncountable. And also the set of all transcendental numbers is uncountable. How can I construct a function $f(x)$ on $R$ which is continuos only at transcendental numbers? Is it possible?

Thanks in advance.

Answer 1

You have had some strong hints. Here's an even stronger one. Start with Thomae's function:

f(x) =

\begin{cases} \frac{1}{q}, & \text{if $x \in \mathbb{Q}$ and $x = \frac{p}{q} \land p \in \mathbb{Z} \land q \in \mathbb{N} \land p,q$ coprime} \\ 0, & \text{if $x \notin \mathbb{Q}$} \end{cases}

You need to understand why this works first. The idea is for $\epsilon > 0$ there are only finitely many values of $q$ such that $\frac{1}{q} \geq \epsilon$.

Now you need to add a case for the algebraic numbers.

f(x) =

\begin{cases} \frac{1}{q}, & \text{if $x \in \mathbb{Q}$ and $x = \frac{p}{q} \land p \in \mathbb{Z} \land q \in \mathbb{N} \land p,q$ coprime} \\ ?, & \text{if $x \notin \mathbb{Q} \land x \in \mathbb{A}$} \\ 0, & \text{if $x \notin \mathbb{A}$} \end{cases}

So, you need a suitable value for that ?, what value will work for algebraic numbers? You can use an idea similar the rationals. First, remember what an algebraic number is: it is the root of a non-zero polynomial with rational coefficients. Let $n$ be the lowest degree of any such polynomial. Can you see how to use $n$ similarly to how $q$ is used for the rational case?

Answer 2

I think one can also proceed as follows: Define the step function $\psi: \mathbb{R} \rightarrow \mathbb{R}$ by \begin{equation*} \psi(x) = \begin{cases} 1 \text{ if } x \geq 0\\ 0 \text{ otherwise} \end{cases} \end{equation*} Let $a_{1},a_{2},\dots$ be an enumeration of all algebraic integers and let \begin{equation*} f(x) = \sum_{n = 1}^{\infty}2^{-n}\psi(x-a_{n}). \end{equation*}