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Let $C$ be a closed subset of $\Bbb R^n$ and $r$ be a positive real. Consider the set

$$ D = \{ y \in \Bbb R^n : \exists\ x \in C\ \text {such that}\ \|x-y\| = r \}.$$

Show that $D$ is a closed subset of $\Bbb R^n$. (where $\|\cdot\|$ is the usual or Euclidean norm on $\Bbb R^n$).

EDIT $:$

I have proved it by using sequences. But I want to prove it using continuous functions. Can we say that $D$ is the inverse image of a closed set under certain continuous function?

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  • $\begingroup$ I have proved it by using sequences. But I want to prove it using continuous functions. Can we say that $D$ is the inverse image of a closed set under certain continuous function? $\endgroup$ – math maniac. Feb 4 at 13:18
  • $\begingroup$ Thanks for your comment. Please include this in the question body so that I can replace my downvote with an upvote. My vote for your question is now locked unless someone edits it. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 4 at 13:21
  • $\begingroup$ @mathmaniac. Closed subsets of metric spaces are functionally closed...$D$ is closed iff there is a continuous $f:\Bbb R^n\to [0,1]$ with $D=f^{-1}\{0\}.$ $\endgroup$ – DanielWainfleet Feb 4 at 20:51
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Let $(y_n)$ be a convergent sequence in $D$ with limit $y$. Then there is a sequence $(x_n)$ in $C$ such that $||x_n-y_n||=r.$

Now show that $(x_n)$ is bounded. Hence $(x_n)$ containes a convergent subsequence $(x_{n_k})$ with limit $x \in C$ $\quad$ ($C$ is closed !).

Then we have $||x_{n_k}-y_{n_k}||=r$ for all $k$.

With $k \to \infty$ we get $||x-y||=r$, thus $y \in D.$

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  • $\begingroup$ I have already done this problem by considering sequences. Any other approach will be appreciated. $\endgroup$ – math maniac. Feb 4 at 13:31
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Suppose $y\in \overline D.$

For every $s\in (0,r)$ let $E(s)= C\cap \{x: r-s\le d(x,y)\le r+s\}.$ Then $E(s)\ne \emptyset.$ Proof: There exists $y_s\in D\cap B(y,s)$ and $c_s \in C$ with $d(y_s,c_s)=r.$ So $$d(y,c_s)\le d(y,y_s)+d(y_s,c_s)<s+r.$$ And $$d(c_s,y)\ge d(c_s, y_s)-d(y,y_s)>r-s.$$ So $c_s\in E(s).$

Now $H=\{E(s):0<s< r\}$ is a non-empty family of closed subsets of the compact space $\overline {B(y,2r)},$ and $H$ has the Finite Intersection Property (FIP). So $\cap H\ne \emptyset.$ So there exists $c\in \cap H.$ Clearly $c\in \cap H\implies d(y,c)=r.$ And since every $E(s)\subset C ,$ we have $c\in \cap H\implies c\in C.$

So $y\in D.$

Remarks.(Background check). Consider (1). (Heine-Borel). A closed bounded real interval is compact. (2). A product of finitely many compact spaces is compact. (3). If $S$ is a compact Hausdorff space then every closed subset of $S$ is compact. (4). A space $S$ is compact iff every non-empty family $H$ of closed sets, where $H$ has the FIP, satisfies $\cap H\ne \emptyset$.... Proving (1) does not require sequences... (2),(3),and (4) apply to all compact (or compact Hausdorff) spaces, including those whose topologies cannot be described in terms of sequences... By (1) and (2), $[-m,m]^n$ is a compact subset of $\Bbb R^n$ for every $m\in \Bbb R^+,$ and hence by (3), every closed bounded subset of $\Bbb R^n$ is compact. (In particular, $\overline {B(y,2r)}$ is compact.)

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