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I wrote such definition at the class. I understand the definition. But also my teacher said " needs provoing!" I underlined it with blue pencil. I could not understand what my teacher wants us and why he wrote such proof requirment. Please explain me. How do I prove this? Ans ıs this important?

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    $\begingroup$ How do you define $E^\circ$? $\endgroup$ – copper.hat Jun 19 '13 at 19:02
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    $\begingroup$ You probably missed writing "$E°$ is open". That's what you need to prove, I think. $\endgroup$ – Giuseppe Negro Jun 19 '13 at 19:05
  • $\begingroup$ I definied as the union of the set {$V :$ $V \subseq E$ and $V$ is open in $\Bbb R^n$} @copper.hat $\endgroup$ – user315 Jun 19 '13 at 19:05
  • $\begingroup$ Hmm okay Maybe.. @GiuseppeNegro $\endgroup$ – user315 Jun 19 '13 at 19:08
  • $\begingroup$ Can you see from that definition that the set of interior points must be in $E^\circ$? So you just need to show that if $x \in E^\circ$ then it must be an interior point. But if $x\in E^\circ$, then $x \in V$ for some open set contained in $E$, hence... $\endgroup$ – copper.hat Jun 19 '13 at 19:09

To prove that $E^o$ is open, note that if $x \in E^o$, there exists $\epsilon > 0$ such that $B_\epsilon(x) \subseteq E$. For any $y \in B_\epsilon(x)$, let $$\delta_y = \frac{\epsilon - d(x,y)}{2}$$

Then $B_{\delta_y}(y) \subseteq B_{\epsilon}(x)$, so $B_{\epsilon}(x) \subseteq E^o$.

  • $\begingroup$ $d(x,y)$ means $||x-y||$. Is this right? $\endgroup$ – user315 Jun 19 '13 at 19:17
  • $\begingroup$ @B11b Yes, in the case of Euclidean space, but the result holds in any metric space. $\endgroup$ – Ink Jun 19 '13 at 19:21

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