1. Hausdorff metric

Let (X,d) be a metric space. Let {An} be a nested family of non empty compact subsets of X. Let A=Intersection of all An.
We have that A is non empty and compact.

We show An converges to A in the Hausdorff metric (D).

I know D(A,B)= Inf {t>or eq.0 : A C B_t and B C A_t} Where A_t is the t-parallel body of A meaning A_t={x in X: d(x,A) < or eq.t}.

But I am not sure how to proceed.

2. Originally Posted by math8
Let (X,d) be a metric space. Let {An} be a nested family of non empty compact subsets of X. Let A=Intersection of all An.
We have that A is non empty and compact.

We show An converges to A in the Hausdorff metric (D).

I know D(A,B)= Inf {t>or eq.0 : A C B_t and B C A_t} Where A_t is the t-parallel body of A meaning A_t={x in X: d(x,A) < or eq.t}.

But I am not sure how to proceed.
Clearly $\displaystyle A\subseteq (A_n)_t$ for all t>0 (because in fact $\displaystyle A\subseteq A_n$). So we need to show that given t>0 there exists n such that $\displaystyle A_n\subseteq A_t$. This seems to be quite tricky to prove. One way that works (although it's not very elegant) is to suppose that the result is false, and get a contradiction.

So suppose that there exists t>0 such that for all n there is an element $\displaystyle x_n\in A_n$ with $\displaystyle d(x,A)\geqslant t$. Use compactness to conclude that there is a subsequence of $\displaystyle (x_n)$ that converges, to $\displaystyle x_0$ say. Then get a contradiction by showing that $\displaystyle x_0\in A$ but also $\displaystyle d(x_0,A)\geqslant t$.