$\Omega$ is open subset of $\mathbb{R}^n$.let $A\subset \Omega$ be the compact subset of $\mathbb{R}^n$.

Prove distance $d(\partial \Omega,A)>0$

My proof,assume $d(\partial \Omega,A) = \inf_{x\in A}d(\partial\Omega,x) = 0$,so exist a sequence $(x_n)\in A$ such that $d(\partial \Omega,x_n) <\frac{1}{n}$ .Now since $A$ is bounded,it exist convergence subsequence, $x_{n_k}\to x\in A$(since $A$ is closed.)

So by continunity of distance function we have $d(\partial\Omega,x) = 0$ which implies $x\in \partial\Omega$ (since $\partial \Omega$ is closed).Which is contradiction since no intersection of $\partial\Omega$ and $A$ due to $\Omega$ is open.

The question is I use assumption that $A$ is bounded set to construct the sequence,but in fact this bounded condition seems not necessary,so how to prove if we only assume $A$ is closed subset of $\mathbb{R}^n$ that contains in $\Omega$?


1 Answer 1


This is false if we drop compactness: Let $\Omega=\bigcup (2n,2n+1)$ and $A=\{2n+\frac 1 n: n \geq 1\}$. Then $A$ is a closed subset of the open set $\Omega$ and $d(\partial \Omega, A) \leq |2n-(2n+\frac 1n) |=\frac 1n $ for all $,n$. So $d(\partial \Omega, A)=0$


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