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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 $A \subseteq B$?

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  • $\begingroup$ 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? $\endgroup$
    – GEdgar
    Mar 22 at 14:09
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    $\begingroup$ 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? $\endgroup$
    – joriki
    Mar 22 at 14:09
  • $\begingroup$ @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. $\endgroup$ Mar 22 at 14:23
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    $\begingroup$ It is true if you consider only closed subsets, it is false if you do not. What are your own thoughts about the question? $\endgroup$ Mar 22 at 14:37
  • $\begingroup$ Sorry, regarding my previous comment, we do require the sets to be bounded and non-empty. $\endgroup$ Mar 22 at 15:27

2 Answers 2

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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.

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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.

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