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

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?

• Wait, why the negative response? This seems like a reasonable question. Commented Apr 28, 2016 at 20:29
• @NoahSchweber See the discussion on meta.math.stackexchange.com/questions/9959/…. This is very important to keep a decent level of discussions on the site (but apparently the OP has got away with this quite a few times already). Commented Apr 28, 2016 at 20:35
• @NoahSchweber This is a good question, but I think it could phrased a little more clearly. For instance, the "distance function" the OP is referring to is unbeknownst to me (it would be nice if it were included). It would be helpful too if the OP mentioned what they've tried and their thoughts on the problem. Commented Apr 28, 2016 at 20:36
• @garabedgulbenkian To be sure (and perhaps suggest a more precise phrasing): are you asking if the set of compact subsets of $\Bbb R^n$, endowed with the Hausdorff-Pompieu distance, is a separable metric space?
– user228113
Commented Apr 28, 2016 at 20:53
• Yes. That is what I am asking. Commented Apr 29, 2016 at 18:23

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.

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.

• Similarly, the Gromov-Hausdorff metric on the set of compact metric spaces modulo isometry is separable, via finite metric spaces with rational distances. Commented Apr 28, 2016 at 20:34
• The first term that I learned for a set like $N$ is $\epsilon$-net. Commented Apr 28, 2016 at 20:35
• @BrianM.Scott Makes sense, I've added it. (I actually haven't heard it before.) Commented Apr 28, 2016 at 20:35
• The topology the Hausdorff metric generates is actually also known as the Vietoris topology, and for this it's also quite easy to see the separability. But this comes at a cost: prove that the metric topology and it coincide. The benefit is that in the Vietoris topology is often easier to some properties like connectedness etc. Commented Apr 28, 2016 at 21:00
• @HennoBrandsma Good point (and for the OP and anyone interested, see en.wikipedia.org/wiki/Hypertopology and heldermann-verlag.de/jca/jca01/jca01013.pdf). Commented Apr 28, 2016 at 21:02