Find the set of real numbers where $f(x)$ is continuous.

Question

The function is defined as

• $f(x) = x+\sqrt{2}$ if $x$ is rational
• $f(x) = x-3$ if $x$ is irrational

The question:
Find the set of real numbers at which $f$ is continuous.

Note : It presumably wants me to justify it with an epsilon-delta proof.

Any advice would be appreciated, I can't quite see how to proceed.

EDIT: For anyone looking for help on similar problems, try here -->

Proving Discontinuity using Epsilon Delta for Function

I eventually found it after further hunting, and it provides a greater degree of mathematical instruction than the answers given here.

Answer 1

With sequences the problem is easy to solve: let $x_0 \in \mathbb R$.

Take a sequence $(a_n) $ in $\mathbb Q$ with $a_n \to x_0$. Then

$f(a_n)=a_n + \sqrt{2} \to x_0+\sqrt{2}$.

Next, take a sequence $(b_n) $ in $\mathbb R \setminus\mathbb Q$ with $b_n \to x_0$. Then

$f(b_n)=b_n -3\to x_0-3$.

Now suppose that $f$ is continuous in $x_0$. Then we would have

$\lim_{n \to \infty}f(a_n)=\lim_{n \to \infty}f(b_n)$, hence $x_0+\sqrt{2}=x_0-3$, a contradiction.