# Is subset relation preserved under limit for Hausdorff metric?

Let $$X$$ be a metric space. I consider elements in $$Y=2^X\setminus \emptyset$$ and use the Hausdorff metric for $$Y$$.

Suppose that $$A_n \subseteq B_n$$ for $$A_n,B_n \in Y$$ and $$A_n \rightarrow A$$ and $$B_n \rightarrow B$$, where $$A,B \in Y$$ as well.

Is it true that $$A \subseteq B$$?

• Do we allow Hausdorff metric to take the value $+\infty$? Or do we restrict to bounded subsets of $X$? What happens when you try to prove this? Mar 22 at 14:09
• The Hausdorff distance is a metric only on the set of non-empty compact subsets of $X$. In particular, $d(A,B)=0$ whenever $A$ and $B$ have the same closure. So I think this won‘t work in general. Are you considering only compact subsets? Mar 22 at 14:09
• @joriki The Hausdorff metric works fine on non-empty closed subsets of a metric space. It even works fine on non-empty general subsets, but as a pseudometric, rather than a metric. Mar 22 at 14:23
• It is true if you consider only closed subsets, it is false if you do not. What are your own thoughts about the question? Mar 22 at 14:37
• Sorry, regarding my previous comment, we do require the sets to be bounded and non-empty. Mar 22 at 15:27

The answer is pretty close to yes. As Joriki points out, there's some issues involving closure, but what we can say is this:

If $$A_n \to A$$ and $$B_n \to B$$ under the Hausdorff (pseudo)metric on $$2^X \setminus \{\emptyset\}$$ and $$A_n \subseteq B_n$$ for all $$n$$, then $$\overline{A} \subseteq \overline{B}$$.

Suppose that $$\overline{A} \not\subseteq \overline{B}$$. Then there exists some $$x \in \overline{A} \setminus \overline{B}$$. Since $$x \in X \setminus \overline{B}$$, which is open, there exists an open ball around $$x$$ exists that is disjoint from $$\overline{B}$$ and hence $$B$$. Moreover, in this ball, there must exist some point in $$A$$. Without loss of generality, by replacing $$x$$ with this new point, possibly reducing the radius, let us suppose that $$x \in A$$, but $$r > 0$$ is such that $$B(x; r) \cap B = \emptyset$$.

Now, since $$A_n \to A$$, we know that $$\sup_{a \in A} d(a, A_n) \to 0$$ as $$n \to \infty$$. So, some $$M$$ exists such that $$n \ge M \implies \sup_{a \in A} d(a, A_n) < \frac{r}{3} \implies d(x, A_n) < \frac{r}{3}.$$ We can then fix $$a_n \in A_n$$ such that $$d(x, a_n) < \frac{r}{3}$$.

We also know, since $$B_n \to B$$, that $$\sup_{b \in B_n} d(b, B) \to 0$$ as $$n \to \infty$$. In particular, there must exist some $$N$$ such that $$n \ge N \implies \sup_{b \in B_n} d(b, B) < \frac{r}{3}.$$ As $$B_n \subseteq A_n$$, we then have $$n \ge N \implies \sup_{b \in A_n} d(b, B) < \frac{r}{3} \implies d(a_n, B) < \frac{r}{3}.$$ Thus, if we fix any $$n \ge N$$, we have $$d(x, B) \le d(x, a_n) + d(a_n, B) < \frac{r}{3} + \frac{r}{3} < r.$$ But this implies there exists some $$b \in B$$ such that $$d(x, b) < r$$, which would place $$b \in B(x; r) \cap B = \emptyset$$. We have a contradiction, and the result is proven.

As stated, the answer to your question is negative. The reason is that the Hausdorff pseudo-distance on $$Y=2^X\setminus \{\emptyset\}$$ defines non-Hausdorff topology on $$Y$$. A simple example is given by $$X=[0,1]$$ with the standard metric. Consider the constant sequences $$B_n=B={\mathbb Q}\cap [0,1]$$ and $$A_n:= B_n$$. Then $$B_n\to B$$, while $$A_n\to X$$ as $$n\to\infty$$.

Such pathological examples explain why one usually restricts the Hausdorff distance only to closed (and bounded) subsets.