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Take the set $\{(x,y) : (x-2)^2 + y^2 < 2\}$ . Is this set relatively open or relatively closed in the subspace $B(2,0)$ radius $\sqrt2$. (The open ball centered at $(2,0)$ with radius $\sqrt2$) ?

I want to say it is both relatively open and relatively closed although I've been told it is just relatively closed. I don't understand how it is one and not the other. You can get an open ball around every point in the set and it is still in the subspace. Since the subspace is equal to the set, you can find any closed or open set in $\mathbb R^2$ s.t. the set is equal to the intersection.

Thanks!

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  • $\begingroup$ Write down which exactly the open ball centered at $(2,0)$ is. Then all will be clear. $\endgroup$ – leo May 21 '13 at 5:04
  • $\begingroup$ You're mixing up the meaning of relatively open. To be relatively open in the subspace $B_{\sqrt 2}(2,1)$, for each point in the set you would need to find an open ball containing that point whose intersection with $B_{\sqrt 2}(2,0)$ lies entirely within the set. $\endgroup$ – dfeuer May 21 '13 at 5:06
  • $\begingroup$ this is getting very sloppy. What on Earth is $E$? $\endgroup$ – dfeuer May 21 '13 at 5:09
  • $\begingroup$ @Andy: Stop and take a breath before you edit the question any more. I think you might want to review some of the definitions and perhaps step back from thinking about relatively open sets to get a better understanding of open ones. $\endgroup$ – dfeuer May 21 '13 at 5:15
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Given a topological space, a subset $A \subset X$ becomes a topological space of its own by defining the open sets in $A$ to all sets of the form $U \cap A$ where $U \subset X$ is open (same for closed sets). So $A$ is open and closed in $A$.

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