How to prove $\{y\in\mathbb R^n: |x-y|=r \text{ for some }x\in X\}$ is closed for closed $X$ and fixed positive $r$? Let $X\subset\mathbb R^n$ be a closed set and $r$ a fixed positive real number. Let $Y=\{y\in\mathbb R^n: |x-y|=r \text{ for some }x\in X\}$. Show that $Y$ is closed.
I tried to approach this problem with showing $Y^c$ is open, but I am stucked on how to use that $X$ is closed?
 A: Suppose that $y_\omega$ is a limit point of $Y$; that is, for all integers $i \geq 1$, there exists some $x_i \in X$ and $y_i \in Y$ such that $y_i \in B_{1/i}(y_\omega)$ (open ball of radius $1/i$ and centered at $y_\omega$) and $|x_i - y_i| = r$.
Note that the sequence $y_1, y_2, \dotsc$ is bounded by $B_1(y_\omega)$, so the sequence $x_1, x_2, \dotsc$ is bounded by $B_{1 + r}(y_\omega)$. By the Bolzano-Weierstrass Theorem, there is a subsequence $x_{k_1}, x_{k_2}, \dotsc$ that converges to some $x_\omega$. Since $X$ is closed, $x_\omega \in X$. Finally, note that $|x_\omega - y_\omega| = r$ so $y_\omega \in Y$.
Since $Y$ contains every limit point $y_\omega$, it must be closed.
A: Thanks a lot to @HerngYi for noticing a mistake in the first version, in which I was assuming $n=1$. I am posting this revised version only for the final remark.
Let $S = \{ t \in \mathbb{R}^n : \lvert t \rvert = r \}$, a closed (actually, compact) set.
Note that
$$
Y=\{y\in\mathbb R^n: |x-y|=r \text{ for some }x\in X\}
= \{ y \in \mathbb{R}^n: y - x \in S  \text{ for some }x\in X\} =
X + S.
$$
Now in general the sum of two closed sets need not be closed, but as remarked by Robert Israel the sum of a closed set and a compact one is closed.
A: Let $(y_n)$ denote a sequence in $Y$ which converges to some $y$. Thus $|y_n-x_n|=r$ for some sequence $(x_n)$ in $X$. Furthermore $(y_n)$ converges hence $(y_n)$ is bounded hence $(x_n)$ is bounded. That is, there exists $R$ such that, for every $n$, $x_n$ is in $X_R=X\cap\{x\in\mathbb R^n\mid|x|\leqslant R\}$. Since $X_R$ is compact, a subsequence $(x_{\varphi(n)})$ converges. Let $x$ denote its limit. Then $x$ is in $X_R$ hence in $X$, $|y_{\varphi(n)}-x_{\varphi(n)}|=r$ for every $n$ while $y_{\varphi(n)}\to y$ and $x_{\varphi(n)}\to x$. By continuity of the distance, $|x-y|=r$. Since $x$ is in $X$, this proves that $y$ is in $Y$.
A: For $y\notin Y$ let $C_1$ be the $r$-sphere around $y$ and $C_2$ the intersection of the closed $2r$-ball around $y$ with $X$. Then $C_1$ and $C_2$ are compact and do not intersect, hence have a positive distance $d$. Then $|y'-y|<\delta:=\min\{d,r\}$ implies that $y'\notin Y$ because any $x$ with $|y'-x|=r$ would have $r-\delta<|y-x|<r+\delta<2r$, hence $x\in C_2$ and then the line $yx$ intersects $C_1$ at a point $z$ with $|z-x|=\bigl||y-x|-r\bigr|<\delta\le d$, contradiciton.
EDIT: More generally, if $C$ is a compact set and $X$ is closed, then $X+C$ is closed. On the other hand, closed + closed is nott necessarily closed (take $\{(x,y)\mid xy=1\}+\{(x,y)\mid x=0\}$ in $\mathbb R^2$, for example)
