2
$\begingroup$

$\Omega$ 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 $\Omega$?

$\endgroup$
4
$\begingroup$

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 $d(\partial \Omega, A)=0$

$\endgroup$

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.