I do not find how to prove: When A is closed, $[A]_r$ is closed for every $r \in \mathbb{R}^+$.

$[A]_r = \{ x \in \mathbb{R}^p \mid \exists a \in A: d(x,a)\leq r\}$

My first idea was to prove that every limit of a row that converges was in $[A]_r$, but this didn't work, so I tried to prove: if it is not closed, than we have a contradiction, but I also get stuck here because it seems that I have to little information.

Can anybody help me?


2 Answers 2


Suppose $x_m\in [A]_r$ and $x_m\to x.$ Then for each $m,$ there exists $a_m\in A$ such that $d(x_m,a_m)\le r.$ Argue that $(a_m)$ is a bounded sequence, hence there is a subsequence $a_{m_k}$ converging to some $a\in A.$ Thus

$$d(x,a)\le d(x,x_{m_k}) + d(x_{m_k},a_{m_k}) + d(a_{m_k},a)$$ $$ \le d(x,x_{m_k}) + r + d(a_{m_k},a).$$

As $k\to \infty,$ we get $0+r+0$ in the last line, and therefore $d(x,a)\le r.$ This implies $x\in [A]_r$ as desired.

  • $\begingroup$ In our classes we need to say that for all $n > n_0$ for a random $\epsilon$>0: $d(x,x_n)< \epsilon$. Than we have to say that $x \in [A]_{r+2\epsilon}$. How can I fix this? $\endgroup$ May 6, 2021 at 19:23
  • $\begingroup$ You let $\epsilon>0$ and say that the last line is $<2\epsilon+r$ for $k$ large enough. It follows that $d(x,a)\le 2\epsilon +r$ for any $\epsilon.$ That implies $d(x,a)\le r.$ $\endgroup$
    – zhw.
    May 6, 2021 at 20:09

Fix $r>0$. Assume that $y\notin [A]_r$. Then $y\notin A$. Let $a\in A$ be the point so that $$ d(a, y) = \min_{z\in A} d(z, y).$$ Since $A$ is closed, such $a$ exists. Note $d(a, y) >r$ since $y\notin [A]_r$. Now we show that the open ball $B$ of radius $d(a, y) -r$ centered at $y$ does not intersect $[A]_r$ (This will show that the complement of $[A]_r$ is open).

Let $z\in B$. If $z\in [A]_r$, then there is $a'\in A$ such that $d(a', z)\le r$. But by triangle inequality,

$$ d(a', y)\le d(a', z) + d(z, y) < r + (d(a, y)-r) = d(a, y),$$ this contradicts to the choice of $a\in A$.


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