# relatively open sets

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.

• How about $x=1$ in your own example? If $r>0$ then $B(1,r)\cap D$ will contain elements that are not in $U$. – drhab May 10 '16 at 9:03
• what if $x=1,r=\frac{1}{2}$? – miracle173 May 10 '16 at 9:04
• @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 $D$ – Yudi V May 10 '16 at 9:37
• No. If $0<r<1$ then $B(1,r)\cap D=(1-r,1+r)$ – drhab May 10 '16 at 9:39