# Some questions about Hausdorff distance

First of all the definitions I am working with:

In all the following $$(X,d)$$ will be a metric space and $$B(x,r) = \{ y \in X : d(x,y) for any $$x \in X$$ and $$r>0$$.

.Let $$C \subset X$$ and $$r>0$$, then $$C_r := \bigcup_{y \in C}B(y,r)$$.

.Let $$\mathcal{X}$$ be the collection of the nonempty, closed and bounded subsets of $$X$$. Then for every $$C,D \in \mathcal{X}$$ the Hausdorff distance is defined as $$h(C,D) := \inf \{ r | C \subset D_r , D \subset C_r \}$$

I was able to show that $$h$$ is a distance on $$\mathcal{X}$$ but I can't see how to solve this:

If $$C,D \in \mathcal{X}$$ s.t. $$h(C,D) and $$A \subset C$$. Then $$h(A, D \cap A_r) < r$$.

My attempt

First of all I suppose $$A \in \mathcal{X}$$ even if the text of the exercise doesn't specify it. I should also prove that $$D \cap A_r \in \mathcal{X}$$ but I wasn't able to do so: it is surely bounded and nonempty but why is it closed?

If $$h(C,D) then $$\exists \, 0 s.t. $$1. \, C \subset D_s$$ and $$2. \, D \subset C_s$$. I want to prove that $$s$$ also satisfy $$A \subset (D \cap A_r)_s \wedge (D \cap A_r) \subset A_s$$ from which I will have the thesis.

For the first one:

By $$A \subset C$$ and 1. I have that $$\forall z \in A \, \exists y \in D$$ s.t. $$d(z,y) and then $$y \in A_r$$ since $$s. Then I have found $$y \in (D \cap A_r)$$ s.t. $$d(z,y) i.e. $$A \subset (D \cap A_r)_s$$.

I don't know how to prove the second one.

It generally isn't. For an example, consider $X = \mathbb{R}$ (with $d(x,y) = \lvert x-y\rvert$), $C = [-2/3,2/3]$, $D = [-1,1]$ and $A = \{0\}$, with an arbitrary $r \in (1/3,1)$. Then $D \cap A_r = A_r = (-r,r)$ is not closed. We would need to consider $h(A, \overline{D \cap A_r})$, or we could extend $h$ to a pseudometric on the set of all nonempty bounded subsets of $X$ (then we'd have $h(B, \overline{B}) = 0$ for all $B$).
This example also exhibits another flaw in the statement. We can have $$\sup \: \bigl\{ \operatorname{dist}(x, A) : x \in D \cap A_r\bigr\} = r\,,$$ and when that happens we will have $h(A, \overline{D \cap A_r}) = r$. So the most we can expect to prove is $$A,C,D \in \mathcal{X}, A \subset C, h(C,D) < r \implies h(A,\overline{D\cap A_r}) \leqslant r\,. \tag{\ast}$$ (Since the conclusion is weakened to a non-strict inequality one may be tempted to relax the premise to a non-strict inequality too, but if $h(C,D) = r$ it can be that $D \cap A_r = \varnothing$.)
The proof of $(\ast)$ is easy. You have already shown one direction, and the other direction follows from $$\overline{D\cap A_r} \subset \overline{A_r} \subset A_t$$ for every $t > r$.
• I was trying to prove that $(X,d)$ totally bdd implies $(\mathcal{X},h)$ totally bounded. If $(X,d)$ is totally bdd, then $X$ is bdd, closed and nonempty then $X \in \mathcal{X}$. Let $r>0$ be fixed, then by totally bddnss there exists a finite set $S$ s.t. $h(X,S)<r$. Consider the finite family of nonempty subset of $S$, call it $\mathcal{S}$. Note $\mathcal{S} \subset \mathcal{X}$. By the proposition above we have for every $D \in \mathcal{X}$ that $h(D, \overline{S \cap D_r})<r$. But $\overline{S \cap D_r} = S \cap D_r \in \mathcal{S}$ and then $(\mathcal{X},d)$ is totally bdd. Is it ok? Commented Sep 7, 2018 at 16:19
• We have a non-strict inequality, $h(D, \overline{S \cap D_r}) \leqslant r$, so depending on the precise definition of total boundedness and lemmata about equivalent characterisations you might need to use an $r' < r$ there (or prove the equivalence of the characterisation of total boundedness using closed balls to the one using open balls, that's easy). But the argument is correct and elegant. Commented Sep 7, 2018 at 17:36