First of all the definitions I am working with:

In all the following $ (X,d) $ will be a metric space and $B(x,r) = \{ y \in X : d(x,y)<r \}$ 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) <r $ 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) <r $ then $ \exists \, 0<s<r$ 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. $d(z,y)<s$ and then $ y \in A_r $ since $s<r$. Then I have found $y \in (D \cap A_r)$ s.t. $d(z,y) <s$ i.e. $A \subset (D \cap A_r)_s$.

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


1 Answer 1


I wasn't able to do so: it is surely bounded and nonempty but why is it closed?

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 $t > r$.

  • $\begingroup$ 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? $\endgroup$
    – Bremen000
    Commented Sep 7, 2018 at 16:19
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Sep 7, 2018 at 17:36
  • $\begingroup$ Yes, you are right! Thank you very much for your help! $\endgroup$
    – Bremen000
    Commented Sep 7, 2018 at 17:38

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