To show a set is closed in $\mathbb{R}^m$

Question

Consider $\mathbb{R}^m$ as a normed space w.r.t $\|X\|^2 := \langle X,X\rangle$, let $D\subset\mathbb{R}^m$ be a closed set and $s>0$.

Show that the set $$A :=\left\lbrace X\in\mathbb{R}^m : (\exists Z\in D)(\| Z-X\|=s)\right\rbrace $$ is closed.

Since $A\subset \mbox{cl} A$ is obvious, it suffices to show the converse. By letting $X\in\mbox{cl} A$, there exists a sequence $X^n\in A,n\in\mathbb{N}$ s.t $X^n\xrightarrow[n\to\infty]{}X$, where for every $n\in\mathbb{N}$ one has $Z^n\in D$ with $\|Z^n-X^n \|=s$. We need a $Z\in D$ with $\|Z-X \|=s$.
Assuming first that $D$ is bounded, we have a convergent sub-sequence $(Z^{k_n})_{n\in\mathbb{N}}, k_n\in\mathbb{N}$. Since $D$ is closed $Z :=\lim\limits_{n\to\infty}Z^{k_n}\in D$ and for every $n\in\mathbb{N}$ one has $\|Z^{k_n}-X^{k_n} \|=s$. Since $(X^n)_{n\in\mathbb{N}}$ itself is convergent, its any sub-sequence converges to the same limit, therefore (?) $\|Z-X \|=s$ and we'd be done. $$\|Z-X \| = \|\lim_{n\to\infty}(Z^{k_n}-X^{k_n}) \| = \lim_{n\to\infty }\|Z^{k_n}-X^{k_n} \| = \lim_{n\to\infty} s = s $$

What happens, when $D$ is unbounded proper closed subset of $\mathbb{R}^m$? [If $D=\mathbb{R}^m$, then $E=\mathbb{R}^m$ also, so $E$ is closed].

Alternatively, we could attempt showing $\mathbb{R}^m\setminus A =: B=\left\lbrace X\in\mathbb{R}^m: (\forall Z\in D)(\|Z-X\|>s\quad\mbox{or}\quad \|Z-X \|0$ $$B(X, \varepsilon)\cap B\neq\emptyset\quad\mbox{and}\quad B(X,\varepsilon)\cap A\neq\emptyset $$ No brilliant ideas, does anyone have any hints?

Answer 1

For any element $x\in\mathbb{R}^n$ define $D_x = \{z\in D\ |\ \lVert z - x\lVert=s\}$. The nice property of $D_x$ is that it is fully contained in an open ball $B(x, s+\epsilon)$ for any $\epsilon > 0$.

Now pick a convergent sequence $(a_n)$ in $A$. We need to show that $a=\lim a_n\in A$. This sequence $(a_n)$ yields a sequence $(z_n)$ in $D$ such that $\lVert z_n-a_n\rVert = s$. In particular $z_n\in D_{x_n}$. Now since $(a_n)$ is convergent then it is bounded. And thus $\bigcup_{n}D_{x_n}$ is bounded (indeed if $(a_n)$ is contained in a ball of radius $r$ then $\bigcup_{n}D_{x_n}$ is contained in a ball of radius $r+s$). Therefore the sequence $(z_n)$ is bounded.

In particular $(z_n)$ has a convergent subsequence $(z_{n_k})$. Let $z$ be the limit. Obviously $\lVert a_{n_k} - z_{n_k}\rVert =s$ implies that $\lVert a-z\rVert = s$ (due to continuity of the norm function). And now $D$ is closed, so $z\in D$ and all in all $a\in A$. $\Box$