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Question: Let $(M,d)$ be a metric space and $\Omega$ be a bounded open subset of $M.$ For every positive real number $\epsilon,$ let $$\Omega_{\epsilon}:=\{x\in\Omega \mid d(x,\partial\Omega)>\epsilon\}.$$ Assume $x\in \Omega_{\epsilon},$ and consider the open ball $B(x,\epsilon):=\{y\in M\mid d(x,y)<\epsilon\}.$ Is it true that the closure $\bar{B}(x,\epsilon)$ of $B(x,\epsilon)$ still in $\Omega,$ that is, $\bar{B}(x,\epsilon)\subset \Omega\, ?$

I think that the answer is true. But I have no clue to prove it. So, can anyone help me to give a proof, or to give a counterexample to disprove it?

nuage
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1 Answers1

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Since $d(x, \partial \Omega) > \epsilon$, we can find an $\epsilon ' > \epsilon$ such that $d(x, \partial \Omega) > \epsilon'$ too. Hence $B(x, \epsilon') \subset \Omega$ and $\overline{B}(x, \epsilon) \subset B(x, \epsilon')$. This proves the claim.

Alex G.
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  • Why $B(x,\epsilon')\subset\Omega?$ – nuage Jun 12 '14 at 04:39
  • There's a few ways to do this. I'm not sure if the following is the cleanest possible proof. Let $D$ be the maximum $\epsilon$ such that $B(x, \epsilon) \subset \Omega$. Then $\overline{B}(x, D)$ contains a point in $\partial \Omega$ (I'll explain why in another comment if you wish; you should try to prove it yourself). Thus $d(x, \partial \Omega) = D$. Hence for any $0 < \epsilon < d(x, \partial \Omega)$, $B(x, \epsilon) \subset B(x, D) \subset \Omega$. – Alex G. Jun 12 '14 at 13:38