4
$\begingroup$

For every open subset $U\subset \mathbb{R}^n$ does there exists a function $f :\mathbb{R}^n \to \mathbb{R}$ such that $f$ is discontinuous at each point of $U$ and continuous on $\mathbb{R}^n\setminus U$?

As I know, given any open set $U\subset \mathbb{R}^n$ there exists a function $f :\mathbb{R}^n\to \mathbb{R}$ such that $f$ is continuous at each point of $U$ and discontinuous on $\mathbb{R}^n\setminus U.$

$\endgroup$
  • $\begingroup$ If $f : U \to \mathbb R^n$ then $f$ is not defined on $\mathbb{R}^n\setminus U$. Can you clarify your intent? $\endgroup$ – user4894 Sep 17 '17 at 2:34
  • $\begingroup$ That was typos. I changed it. Thanks :) $\endgroup$ – Sachchidanand Prasad Sep 17 '17 at 2:39
4
$\begingroup$

Let $g(x)$ be the distance from $x$ to $\mathbb R^n\setminus U$. Let $h$ be the characteristic function of $\mathbb Q^n$. Then $f(x)=g(x)h(x)$ has this property. (This is assuming $U\neq \mathbb R^n$; otherwise take $f=h$.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.