0
$\begingroup$

For every metric space $(X,d)$ we have the Hausdorff metric space $(\mathcal{H}(X),H)$ that assosiates with it, where $\mathcal{H}(X)$ is the space of nonempty compact subsets of $X$ and $H$ is the hausdorff metric.

As part of a fractals seminar I'm working on I want to use that any nonincreasing sequence of subsets in $\mathcal{H}(X)$: $A_1\supseteq A_2 \supseteq...$ is a Cauchy sequence.

Similarly, I want to show somthing similar for nondecreasing subsets, though since this clearly doesn't hold (take $([0,n])_{n\in\mathbb{N}}$) I though to add the condition that all elements of the sequence are contained in some compact subset.

Though I'm pretty sure that those statements should be true, I'm having trouble with proving them. Maybe it's just that I get confused with all the max-min stuff.

Are those 2 statments true or am I missing some condition?

Could you help me prove them? If they are both true, and the proofs are similar, please proof one and leave the other as an exercise.

Thanks

$\endgroup$

1 Answer 1

0
$\begingroup$

I think that for the nonincreasing sequence, you can actually show that the sequence converges toward $\cap A_n$ (suppose it does not and then you get a sequence $x_{k_n}\in A_{k_n}$ where $(k_n)$ is strictly increasing and $d(x_{k_n},\cap A_n)>r_0>0$). You get that $(x_{k_n})$ up to extraction converges in $A_1$ and eventually is in all $A_n$ hence the limit must be in $\cap A_n$ contradicting $d(x_{k_n},\cap A_n)>r_0$ for all $n\in \mathbb{N}$.

For the second statement, you might be right. I think that if $(A_n)$ is a non decreasing sequence of compact included in $K$ compact then you can show that the limit of $(A_n)$ will be :

$$\overline{\bigcup A_n} $$

This is compact because it is a closed subset of $K$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .