Thread: Revised: A sequence of non-empty, compact, nested sets converges to its intersection

Proposition 2.4.7 * is a metric space with metric . is a sequence of descending, non-empty, compact sets.

Then for in the Hausdorff sense.

In that sense, one must show that

(1) and (2)

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

. is an open set, he states, which is straightforward.

He then states that the family

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

for all , where M is some positive integer is a finite sub-cover. The author concludes with

Therefore, , 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

You don't seem to have stated what it is you want to prove!

Well, for starters, how about the last and next to last sentences of my post? Could you help with those?

Now that I have restated more clearly my question with LaTex, where is Mr "Plato" ?

My own solution is that the one offered by Prof Edgar does not make sense. Maybe his graduate assistant screwed up and got away with it. Maybe I should not have mentioned Edgar's name in my posting. That mention might have discouraged a critical review . In any case, I believe I have proved Proposition 2.4.7 by recalling that in taking the limit of the sequence, the index goes to "infinity". With that in mind, I skipped all the stuff about the Heine-Borel theorem and simply looked at the limit in the Hausdorff metric. So, this posting's question has been answered - by the poster! Thanks for all the help.

here is how i look at it:

since A₁ is compact, and we have exhibited an open cover of A₁, there is some finite sub-cover. note that we can take N_ε(A) to be one member of the subcover, so we only need the complements of finitely many of the A_j to "meet up with N_ε(A)" (the rest of the infinite family lies inside the set N_ε(A)).

if the book really DOES say: n ≥ M, instead of n ≤ M, then that is a typo.

It really, really does read as I reported; also, the new edition of 2008 reads the same. Thus, it is not just a typo but carelessness, if in the author's mind he meant less than or equal. Furthermore, the sub-cover, if you are correct about the "typo", omits a "ring" of points in sets of index greater than M+1 and without the intersection of all - unless M is unbounded - because the indexed sets are complements of the given sequence's sets.

My view is far simpler: lim A_n = Intersection of all A_n because the members are descending. The second part of the Hausdorff requirement is thus satisfied. The limit has to be taken in any case to show that the Hausdorff "distance" is zero.