TITLE: Haussdorff convergence in $\mathbb{R}^n$
QUESTION [1 upvotes]: Let $\omega_n$ be  bounded open  subsets of $\mathbb{R}^n$
such that $(\omega_n)$ converges to $\omega$ in sens of Hausdorff metric. I would like to know  what are the boundary  conditions, if there exist, that we must consider so that the closure of the sequence $(\omega_n)$, denoted $(\overline{\omega_n})$,  converges to $\overline{\omega}$ in the same sens. \
Thank you for your propositions and please could you provide me a good reference where I should find these kind of convergences.

REPLY [0 votes]: Consider the set $B(X)$ of all nonempty subsets of a metric space $X$. It is well known that $d_H(A,B)=0$ if and only if $\overline{A}=\overline{B}$. And thus the topology $\tau_H$ generated by the extended pseudometric $d_H$ has the following property:

Two points in $\big(B(X), \tau_H\big)$ are topologically indistinguishable if and only if they have equal closure.

Topologically indistinguishable points can be defined in several equivalent ways:

$x,y$ are topologically indistinguishable
every open subset $U$ contains either both $x,y$ or none of them
a net or filter converges to $x$ if and only if it converges to $y$

This should be enough to prove the following lemma:

Let $X$ be a topological space and let $(a_n)$ be a sequence (net) convergent to $a$. Assume that $(b_n)$ is a sequence (net) such that each $a_n$ is topologically indistinguishable from $b_n$. Then $(b_n)$ is convergent to $a$.

So all in all the answer is simple: none condition is needed: $\overline{\omega_n}$ always converges to $\overline{\omega}$.
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