Is the space of subsets of $\mathbb R^n$ with the Hausdorff metric separable?

Question

Let M be the Metric Space whose "points" are the Closed and Bounded subsets of a finite dimensional Euclidean Space and whose "distance function" is the Metric defined by Hausdorff for such point sets. Is M separable?

Answer 1

Yes, it is separable - in fact, it has a countable dense subset consisting entirely of finite sets!

Exercise. Let $C$ be a closed bounded set, and $N\subseteq C$ such that for each $c\in C$, there is an $n\in N$ such that $d(n, c)<\epsilon$. Then the Hausdorff distance between $N$ and $C$ is $<\epsilon$.

Such an $N$ is sometimes called an "$\epsilon$-dense subset" of $C$, or "$\epsilon$-net."

Exercise. Using the previous, show that the set $$\mathcal{F}=\{N: N\subseteq\mathbb{Q}^n\mbox{ is finite }\}$$ is dense in the space of closed bounded subsets of $\mathbb{R}^n$.

Note that it's not quite one line - plenty of closed bounded subsets of $\mathbb{R}^n$ don't contain any rational points at all! So it takes a bit of thought.

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EDIT: in response to a comment below, let me add an even simpler proof sketch:

For a given $\epsilon>0$, divide the plane into $\epsilon$-squares: let $$S_\epsilon(a, b)=[a\epsilon, (a+1)\epsilon)\times[b\epsilon, (b+1)\epsilon)$$ for $a, b$ integers. Let $q^\epsilon_{a, b}$ be a rational point in $S_\epsilon(a, b)$. Now for $X\subset\mathbb{R}^2$ bounded, let $$A_\epsilon(X)=\{q^\epsilon_{a, b}: X\cap S_\epsilon(a, b)\not=\emptyset\},$$ that is, we "mark" each box which $X$ hits.

Since $X$ is bounded, the set $A_\epsilon(X)$ is finite; and it's not hard to bound the Hausdorff distance between $A_\epsilon(X)$ and $X$, and show that it goes to zero as $\epsilon\rightarrow0$. So the set of finite sets of rational points is a dense set in this metric, and is clearly countable.