Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\Omega\subset\mathbb{R}^n$ be a open set. We say that $y\in\partial \Omega$ satisfies the interior ball condition, if there is $x\in \Omega$ and $r>0$ such that $$B(x,r)\subset\Omega,\ y\in \partial B(x,r),$$

where $B(x,r)=\{z\in\mathbb{R}^n:\ \|z-x\|<r\}$. I am trying to prove that (I don't know if it is true) the set of points in $\partial \Omega$ which satisfies the interior ball condition are dense in $\partial\Omega$.

If we go by contradiction then, there is $y\in\partial\Omega$ (which we can assume to not be isolated) and a neighborhood $F$ of $y$ ($F\subset\partial\Omega$) such that $$d(x,\partial\Omega)<d(x,F),\ \forall x\tag{1}$$

Now I am trying to get a contradiction with $(1)$. Any idea is appreciated.

Update: I think I have a answer, please verify if it is correct.

Assume ad absurdum that there is $y\in \partial\Omega$, $y$ is not isolated in $\partial\Omega$, such that in the set $V_r=\overline{B(y,r)\cap \partial\Omega}$ (for some $r>0$) there is no point which satisfies IBC.

If there is $x\in \Omega$ with $d(x,\partial\Omega)= d(x,V_r)$ we are done, hence, assume that $(1)$ is satisfied for $F=V_r$. Take $z\in V_r$ with $\|z-y\|<\delta$ and $0<\delta<r$.

As $d(z,\partial\Omega)<d(z,V_r)$, the infimum of $d(z,\partial\Omega)$ is achieved in $\partial\Omega\setminus V_r$ in some point $w_{\delta}$. By choosing $\delta$ sufficiently small, we must have that $\|z-y\|=d(z,y)<d(z,w_{\delta})$, because $w_{\delta}$ is outside the ball $B(x,r)$. Therefore, we have a contradiction.

share|cite|improve this question
up vote 1 down vote accepted

Your proof is correct, but there is no need to present it as an argument by contradiction. The goal is to show that for every $y\in \partial\Omega$ and every $r>0$ there is $z\in\partial \Omega$ which satisfies the interior ball condition and $|z-y|<r$.

So, pick $x\in \Omega$ such that $|x-y|<r/2$. Let $z$ be the closest point of $\partial \Omega $ to $x$. Note that $|x-z|\le |x-y|<r/2$. Hence $|z-y|<r$. The interior ball condition holds for $z$, by its construction.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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