Convergence in the Hausdorff metric

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SUMMARY

The discussion focuses on the convergence of a nested family of non-empty compact subsets {An} in a metric space (X, d) to their intersection A in the Hausdorff metric (D). It establishes that A is non-empty and compact, and defines the Hausdorff distance D(A, B) using the concept of t-parallel bodies. The participants explore the proof strategy involving an epsilon argument to demonstrate that for a given epsilon > 0, there exists an N such that A_N is contained within A_epsilon, leading to a contradiction if this is not the case.

PREREQUISITES
  • Understanding of metric spaces and compactness
  • Familiarity with the Hausdorff metric and its properties
  • Knowledge of convergence concepts in topology
  • Ability to work with sequences and subsequences in mathematical analysis
NEXT STEPS
  • Study the properties of compact sets in metric spaces
  • Learn about the Hausdorff metric and its applications in topology
  • Explore epsilon-delta definitions of convergence in analysis
  • Investigate the concept of t-parallel bodies and their significance in geometric analysis
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Mathematicians, students of topology, and researchers interested in convergence properties within metric spaces will benefit from this discussion.

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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.
 
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Haven't thought this through completely, but some thoughts:

Let epsilon > 0.

Show there exists N such that A_N\subset A_{\epsilon}.

Suppose not. Get sequence of x_n in A_n but not in A_epsilon, then thin to convergent subsequence.

Obtain contradiction.
 

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