# Any two points of path connected domain in $R^n$ has a path that the distance between path and boundary of domain is strictly positive?

I am a Graduate student! I have a question.

Let's have our domain $\Omega$ as an open, path-connected subset in $R^n$. Let $x,y$ be two points in $\Omega$. Then surely there exists at least a path from $x$ to $y$. So here's my question. For such set of paths, is there any path $\gamma$ such that $dist(\gamma,\partial\Omega)$ is strictly positive?

Motivation:(which you do not have to understand to answer my question) I was struggling to prove Strong Maximum Principle of weak solution of generalized elliptic divergence form. Since the problem I'm considering is not continuous, the set $\Omega_{M}=\{x\in\Omega|u(x)=M\}$ where $M=sup_{\Omega}u$ could not be closed. What I'm having now is that for some ball $B\subset\subset \Omega$ if we have $sup_{B}u=sup_{\Omega}u\geq 0$, then the function $u$ must be constant in ball $B$ almost everywhere. so ignoring measure zero set, $\Omega_{M}$ is open in $\Omega$ and I need the closedness of $\Omega_{M}$ to get the desired result.

My reference of motivation is Gilbarg, Trudinger Elliptic PDE of Second Order Thm 8.19.

Thank you so much!

• Hint: A continuous function on a compact set attains its minimum. Apply it to the image of any path in $\Omega$ connecting $x$ to $y$. – Moishe Kohan Feb 9 '17 at 7:57
• Oh yes you are definitely right. shame on me.. :) Thanks! – Dongjun Kim Feb 10 '17 at 9:22

Yes. If $f:[0,1]\to \Omega$ is continuous, with image $P,$ then $P$ is compact because $[0,1]$ is compact. There are many ways to show that $\inf \{d(p,q):p\in P\land q\in \partial \Omega \}>0.$
Note that $\partial \Omega \subset R^n$ \ $\Omega$ because $\Omega$ is open, so it suffices to show that $$\inf \{d(p,q): p\in \Omega \land q \in R^n \backslash \Omega \}>0.$$
Method (1). The function $g(p)=\inf \{d(p,q):q\in R^n$ \ $P\}$ for $p\in P$ is continuous, and $P$ is compact, so the image of $g$ is compact, so $g$ attains its minimum value. Which must be positive.
Method (2). Assume $\Omega$ is not empty. For $p\in P$ let $r(p)>0$ such that the open ball $B_d(p, r(p))$ is a subset of $\Omega.$ Let $C=\{B_d(p, \frac {1}{2}r(p_i)): p\in P\}.$Then C is an open cover of the non-empty compact $P.$ So $C$ has a finite subcover $\{B_d(p_i),\frac {1}{2}r(p_i)):i=1,...,n\}.$
Let $r=\min (\frac {1}{2}r(p_i):i=1,...,n).$ Any $p\in P$ satisfies $d(p,p_i)<\frac {1}{2}r(p_i)$ for some $i,$ and we have $B_d(p_i,r(p_i))\subset \Omega.$ So the triangle inequality implies $\inf d(p,R^n$ \ $\Omega)\geq \frac {1}{2}r(p_i)\geq r.$