Hausdorff metric:When A is closed, $[A]_r$ is closed for every $r \in\mathbb{R}^+$. 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?
 A: 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.
A: 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 $a\in A$.
