# 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?

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Is $x$ a constant? Otherwise, $Y = \mathbb{R}^n$. –  Herng Yi Feb 9 '13 at 7:26
@HerngYi: That's not correct - for example, if $X=\varnothing$, then $Y=\varnothing$. –  Zev Chonoles Feb 9 '13 at 7:28
sorry, I misunderstood the question :P –  Herng Yi Feb 9 '13 at 7:48
@JFK, it seems that when proving that a set is closed you try to prove that its complement is open. In future, you may also want to try proving that the set contains all its limit points as it is another useful technique. –  Herng Yi Feb 9 '13 at 8:01

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.

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+1, and thanks @HerngYi for letting me notice my mistake in my now deleted answer. –  Andreas Caranti Feb 9 '13 at 7:48
No problem, and I got some insight from your answer as well - for a while I tried the approach to show that $Y = \bigcup_{|v| = r}(X + v)$, but that's an infinitary union of closed sets so I got stuck. –  Herng Yi Feb 9 '13 at 7:50
I have now reposted my answer, with a reference I found useful. See also @HagenvonEitzen post in this thread. –  Andreas Caranti Feb 9 '13 at 7:53

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.

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how can you define $x + r$, when $x$ is a vector and $r$ is a scalar? –  Herng Yi Feb 9 '13 at 7:36
aha, seems like I gave up too early when I got stuck by summing over $S$! –  Herng Yi Feb 9 '13 at 7:54

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 $Y$.

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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)

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