Continuity (close to Dirichlet)

Question

Lets have look at $f:\mathbb{R} \rightarrow \mathbb{R}$ with $f(x) = \left\{\begin{matrix} x &,& \text{ if } x \in \mathbb{Q}\\ a \in \mathbb{R} & ,& \text{ else } \end{matrix}\right.$

Prove where $f$ is continuos depending on the constant $a \in \mathbb{R}$.

I think $f$ could be continuos where $a$ equals $x_0 \in \mathbb{Q}$. But there are $a$'s that wont work on this. Im not sure how to handle this $a$.

Further i want to show that $f$ is nowhere else continuos. I thought about it this way. I would like to use the $\epsilon - \delta $ Criterium. Let me give you an $\epsilon>0$ with $\epsilon = 1$ than theres always a irrational value for $a \in \mathbb{R}-\mathbb{Q}$ such that for every $x,y \in \mathbb{Q}$ with $x>y$ and $x>a>y$ so that $a$ is not in the $\delta$ enviroment.

This is only a draft and not a try of a proof. Please help me out on this. First i wont to know if there is a continuos point on $f$ depending on $a$ and then i would like to know how to show that $f$ is nowhere (else) continuos. Thank you!

Answer 1

Your function is continuous at $x = a$. Indeed for any $\epsilon > 0$, choosing $\delta = \epsilon$ guarantees that $$ |f(x) - f(a)| \leq \epsilon \quad \text{if} \quad |x-a| \leq \delta,$$ because $f(x) - f(a) = x -a $ if $x$ is rational $f(x) - f(a) = 0$ if $x$ is irrational.

Let us now prove that it is not continuous anywhere else. Let $y \neq a$. If $y$ is rational, consider a sequence $x_n$ in $\mathbb R - \mathbb Q$ such that $x_n \to y$. Clearly, $\lim_{n\to \infty} f(x_n) = a \neq y = f(y),$ so $f$ is not continuous at $y$. If $y$ is irrational, take a sequence $x_n$ in $\mathbb Q$ such that $x_n \to y$. In this case $\lim_{n\to\infty} f(x_n) = y \neq a = f(y)$, so $f$ is not continuous at $y$ either.