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?


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.


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