Let $X$ be a metric space and let $A \subset X$ be an arbitrary subset. We define $$A_r=\{x \in X : B_r(x) \subseteq A\}.$$ Prove that $A_r$ is closed for every radius $r$.

Maybe this is easy but I am totally stuck. First I've tried to prove it directly, i.e., take a convergent sequence in $A_r$ and prove that its limit is in $A_r$. Well, I didn't got anywhere. Then I've tried to prove that $A_r$'s complement is open. $A_r$'s complement is the set $$X \setminus A_r=\{x \in X : B_r(x) \not\subseteq A\}$$ Given $x \in X \setminus A_r$, I have to take $\epsilon$ such that $B(x,\epsilon) \subseteq X \setminus A_r$. Can anyone give me a hint?

I add the proof:

To prove $A_r$ is closed is equivalent to prove its complement is open. So let $X \setminus A_r$={$y\in X\mid \exists u\notin A, y\in B_r(u)$}. Take $\epsilon=r-d(y,u)$ and consider $B(y,\epsilon)$. Let $z \in B(y,\epsilon)$, $d(z,u)\le d(z,y)+d(y,u)<r-d(y,u)+d(y,u)=r$. It follows that $z \in B_r(u)$, so $B_r(u)$ is open. $X \setminus A_r=\bigcup_{u\in X-U}B_r(u)$, as union of open sets gives an open set, $X \setminus A_r$ is open, which implies $A_r$ is closed.

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    $\begingroup$ Pick a point $y \in B_r(x)\setminus A$. $\endgroup$ – Daniel Fischer Oct 6 '13 at 15:05
  • $\begingroup$ Now, if we choose $\epsilon=r-d(x,y)$, then $B_{\epsilon}(x)\subset X\setminus A_r$. As $z\in B_{\epsilon}(x)$ implies that $d(z,x)<r-d(x,y)$. Thus $d(z,y)<r$. This means $y\in B_r(z)$. Thus $B_r(z)$ is not contained in $A$. So, $z\notin A_r$. $\endgroup$ – Anupam Oct 6 '13 at 15:46
  • $\begingroup$ Thanks for the help, I think this leads exactly to what Stefan suggested me to do. $\endgroup$ – Andrew Oct 6 '13 at 15:48
  • $\begingroup$ @ Andrew: Yes, you are right. $\endgroup$ – Anupam Oct 6 '13 at 15:52

The set $A_r$ is the complement of the set $\{y\in X\mid \exists u\notin A, y\in B_r(u) \}$. Can you show that this set is open?

  • $\begingroup$ I'm still trying to convince myself that the set you've defined is exactly $A_r$'s complement. If I take $\epsilon$=$r-d(y,u)$, then $B(y,\epsilon)$ is contained in $A_r$'s complement. So, $A_r$'s complement is open, which means, $A_r$ is closed. Is this correct? $\endgroup$ – Andrew Oct 6 '13 at 15:32
  • $\begingroup$ Actually we have $$X-A_r=\bigcup_{u\in X-U}B_r(u)$$ What you do in your comment is essentially the proof that $B_r(u)$ is open. $\endgroup$ – Stefan Hamcke Oct 6 '13 at 15:38
  • $\begingroup$ I've thought about it a little more time and now I understand that it is in fact $A_r$'s complement $\endgroup$ – Andrew Oct 6 '13 at 15:40
  • $\begingroup$ Ok, but the union of open sets is still open, couldn't we conclude then that $X-A_r$ is open? $\endgroup$ – Andrew Oct 6 '13 at 15:42
  • $\begingroup$ @Andrew: Yes, that correct. $\endgroup$ – Stefan Hamcke Oct 6 '13 at 15:42

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