Proposition 2.4.7 * $\displaystyle S $ is a metric space with metric $\displaystyle \rho $. $\displaystyle {A_n} $ is a sequence of descending, non-empty, compact sets.

Then for $\displaystyle \epsilon > 0,\ lim A_n = A = \bigcap A_n $ in the Hausdorff sense.

In that sense, one must show that

(1) $\displaystyle A \subseteq N_\epsilon (A_n) $ and (2) $\displaystyle A_n \subseteq N_\epsilon (A ) $

Prof Edgar shows (1) clearly, but I have difficulty with his demonstration of (2). I shall try to restate it:

By the $\displaystyle \epsilon $ neighborhood of $\displaystyle A $, he means:

$\displaystyle N_\epsilon (A) = \{y: \rho (x,y) < \epsilon \ for \ some \ x\in A \} $. $\displaystyle N_\epsilon (A) $ is an open set, he states, which is straightforward.

He then states that the family $\displaystyle \{N_\epsilon (A) \}\bigcup \{S \backslash A_n : n \in N \} $

is an open cover of $\displaystyle A_1 $ . Herein $\displaystyle S\backslash A_n $ is the complement of $\displaystyle A_n $. It seems to me that the family described covers not only $\displaystyle A_1 $ but the entire metric space $\displaystyle S $ , all points of the intersection and all points "outside" the intersection . So, yes, it does cover $\displaystyle A_1 $ . The author then continues that there is therefore a finite subcover of $\displaystyle A_1 $ , since A_1 is compact by hypothesis. He states that:

$\displaystyle N_\epsilon (A) \bigcup \{S\backslash A_n\} \supseteq A_1 $ for all $\displaystyle n \geq M $ , where M is some positive integer is a finite sub-cover. The author concludes with

Therefore, $\displaystyle A \sqsubseteq N_\epsilon (A_n) $ , and (2), above, is satisfied.

My difficulty: The expression for the finite sub-cover does not seem finite to me because it is a family for all n > M.

Can anyone explain?

*"Measure, Topology, and Fractal Geometry", by Gerald A. Edgar (Springer Verlag, New York, 2000) p.68