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Definition of a relatively open set:

$ D \subset K^N $ is a set. $U \subseteq D $ is relatively open in D if $$ U = \emptyset \quad$$ or $$ \forall x \in U \quad \exists \quad r > 0 \quad | \quad B(x,r) \cap D \subseteq U$$

What I want to know is: is there a set U with $x \in U \subseteq D $ such that $ B(x,r) \cap D \nsubseteq U$.

Example:

If $D = (0,2] $ and $ U =[1,2]$ and $x = 2$, then $B(2,r) \cap D = (2-r,2] \subseteq U$.

In the above example I dont see for any $x \in U$ where $ B(x,r) \cap D \subseteq U $ is not satisfied. Can someone please give examples of $D $ and $ U $ where $ B(x,r) \cap D \subseteq U $ is not satisfied.

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    $\begingroup$ How about $x=1$ in your own example? If $r>0$ then $B(1,r)\cap D$ will contain elements that are not in $U$. $\endgroup$ – drhab May 10 '16 at 9:03
  • $\begingroup$ what if $x=1,r=\frac{1}{2}$? $\endgroup$ – miracle173 May 10 '16 at 9:04
  • $\begingroup$ @drhab, if $x =1$, then $B(1,r) \cap D = (1-r, 1+r) \subseteq U$ I was reading $B(1,r) \cap D $ as take elements that are common to both $B(1,r) $ and $U $ instead of $B(1,r) $ and $D $ $\endgroup$ – Yudi V May 10 '16 at 9:37
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    $\begingroup$ No. If $0<r<1$ then $B(1,r)\cap D=(1-r,1+r)$ $\endgroup$ – drhab May 10 '16 at 9:39
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Definition of a relatively open set, U must be in D, but D=(0,2] and U=[1,2] , U is not in D, this is D in U, your example is not a counterexample

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