2
$\begingroup$

Let $\mathbb H$ be family of all non empty compact subsets of $\mathbb R^n$ and $d_H$ be Hausdorff distance on $\mathbb H$. Some called this $(\mathbb H,d_H)$ as Fractal Space. Is boundedness equal to totally boundedness in this metric space?

$\endgroup$
1
$\begingroup$

Yes. This follows from some well-known theorems about the Hausdorff metric but is easy to do by hand.

We just have to show that every closed ball in the space is totally bounded. So let $E$ be a compact set in $\mathbb{R}^n$ and $r>0$. Let the ball be $\mathcal{B} = \{F\in\mathbb{H} : d_H(F,E)\leq r\}$. Fix $\epsilon >0$.

Every set $F$ in $\mathcal{B}$ is contained in the closed $r$-neighborhood of $E$ in $\mathbb{R}^n$. This is a compact set in $\mathbb{R}^n$, so it is contained in finitely many balls $B(x_i, \epsilon/2)$ in $\mathbb{R}^n$, where $i=1, 2, \dots, N$ (and of course $N$ depends on $\epsilon$).

Now take all subsets $\{i_1, \dots, i_k\}$ in $\{1,2,\dots, N\}$. Look at the balls $\mathcal{B}_{i_1, \dots, i_k} = \{F\in\mathbb{H} : d_H(F,\{x_{1_1}, \dots, x_{i_k}\})\leq \epsilon\}$. (Note that each set $\{x_{i_1}, \dots, x_{i_k}\}$ is a point in $\mathbb{H}$).

These cover $\mathcal{B}$, and there are finitely many of them.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.