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I need the following as a lemma for another proof I'm working on:

"Let $E$ be a set such that $E\subseteq Y\subseteq X$, where $(X,d)$ is a metric space and also $E=K\cap Y$, with $K\subseteq X$ closed set (i.e. contains all of its boundary points). Let also $x$ be a boundary point of $E$ (thus we know that $B(x,r)\cap K\cap Y\neq\emptyset, B(x,r)$);

show that $\forall r>0, B(x,r)\cap K\neq\emptyset, B(x,r)$."

I showed that $B(x,r)\cap K\neq\emptyset$ arguing by contradiction ($B(x,r)\cap K=\emptyset\Rightarrow B(x,r)\cap K\cap Y=\emptyset\cap Y=\emptyset$, contradiction) but I haven't been able to show that $B(x,r)\cap K\neq B(x,r)$ so I'd appreciate any help in proving this last part of the lemma.

lorenzo
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    Let $X=K$, where $X$-closed set, then $E=Y$, and your statement can be wrong. you should improve your question. – serg_1 Jun 26 '17 at 10:40
  • @serg_1 I was trying to use this result to show that if $E=K\cap Y$ defined as above then $E$ is a closed set but now I see that I should come up with another proof. Thanks. – lorenzo Jun 26 '17 at 10:42

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I think this statement is wrong

Choose $X=\mathbb R, K=[0,2], Y=[1,3], x=1, r=\frac12$.

Then $x$ is a boundary point of $E$, but not a boundary point of $K$.

supinf
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