Interior ball condition

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.

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1 Answer

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.

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