# 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?

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.

• 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? May 6, 2021 at 19:23
• 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 $d(x,a)\le r.$
– 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 $$a\in A$$.