# A sequence of non-empty, compact, nested sets converges to its intersection.

• Sep 9th 2012, 09:54 AM
goedelite
A sequence of non-empty, compact, nested sets converges to its intersection.
A text* I am reading offers a proof that a sequence of non-empty, compact, nested sets converges to its intersection in the Hausdorff metric. I do not follow the second half of the proof which shows that, in the limit, a member of the sequence is contained in the intersection in the sense that any point of that element is near some point of the intersection. The proof in the text applies the theorem that any covering of a compact set has a finite sub-cover. I do not follow the proof.

I wonder if anyone would discuss this proof with me or offer another.

*"Measure, Topology, and Fractal Geometry", by Gerald A. Edgar
• Sep 9th 2012, 10:13 AM
Plato
Re: A sequence of non-empty, compact, nested sets converges to its intersection.
Quote:

Originally Posted by goedelite
A text* I am reading offers a proof that a sequence of non-empty, compact, nested sets converges to its intersection in the Hausdorff metric. I do not follow the second half of the proof which shows that, in the limit, a member of the sequence is contained in the intersection in the sense that any point of that element is near some point of the intersection. The proof in the text applies the theorem that any covering of a compact set has a finite sub-cover. I do not follow the proof.

It is difficult to know what you don't understand from that description.
Could post the part of the proof you don't follow?
• Sep 9th 2012, 02:18 PM
goedelite
Re: A sequence of non-empty, compact, nested sets converges to its intersection.
Quite so! Easier, if one has access to the text, on p.68. I am not at home, but I shall try to re-state Prof Edgar:

The part I find troubling shows that a member of the sequence, A_n belongs to N_e(A), A, the intersection of all A_n. In words, any pt of A_n is within e of some point of the intersection, A. I take that to be for n sufficiently large. Edgar asserts that the following sets are all open sets and form an infinite cover of A_1, the first and largest set of the sequence: \A_n U N_e(A)

Examining the nth set, I find that in the limit n--> inf , the expression includes the entire metric space, S, because all the sets A_n are descending or nested. Thus, the complement as n --> inf combined with the intersection of all is the entire space, in my view. Yes, the entire space does cover A_1, no question! But the entire metric space is not compact. Yes, A_n is compact by hypothesis; so its complement is open. Yes, N_e(A), all points within e of some point of A constitute an open set. I don't see that a finite number of these unions cover A1. Any finite number of them would leave a "ring" of A1 uncovered, heuristically.

Leaving my heuristic objection aside, I note that Edgar then states that because there is a finite cover, A_n belongs to N_e(A). I don't see that either.

I don't know if my word-picture will make clear my difficulty with the proof. Thanks for your interest.
• Sep 9th 2012, 03:32 PM
Plato
Re: A sequence of non-empty, compact, nested sets converges to its intersection.
Quote:

Originally Posted by goedelite
Quite so! Easier, if one has access to the text, on p.68. I am not at home, but I shall try to re-state Prof Edgar: The part I find troubling shows that a member of the sequence, A_n belongs to N_e(A), A, the intersection of all A_n. In words, any pt of A_n is within e of some point of the intersection, A. I take that to be for n sufficiently large. Edgar asserts that the following sets are all open sets and form an infinite cover of A_1, the first and largest set of the sequence: \A_n U N_e(A)
Examining the nth set, I find that in the limit n--> inf ,with the expression includes the entire metric space, S, because all the sets A_n are descending or nested. Thus, the complement as n --> inf combined with the intersection of all is the entire space, in my view. Yes, the entire space does cover A_1, no question! But the entire metric space is not compact. Yes, A_n is compact by hypothesis; so its complement is open. Yes, N_e(A), all points within e of some point of A constitute an open set. I don't see that a finite number of these unions cover A1. Any finite number of them would leave a "ring" of A1 uncovered, heuristically.
Leaving my heuristic objection aside, I note that Edgar then states that because there is a finite cover, A_n belongs to N_e(A). I don't see that either.

I still do not follow you description.
It seems that this has to do with either the Finite Intersection Property or maybe Cantor's nested intervals.

Also explain the notation. What does N_e(A) mean for example?
• Sep 9th 2012, 03:57 PM
goedelite
Re: A sequence of non-empty, compact, nested sets converges to its intersection.
Sorry for the lack of clarity. By N_e(A) I mean a "neighborhood" of the set A, where A is the intersection of the sequence of sets A_n. I am using _ to indicate a subscript follows. The subscript "e" is usually the small epsilon. This is not a neighborhood of a point but rather a neighborhood in the following sense: Any point of the neighborhood is within e>0 of some point of the intersection A.

No, the finite intersection property is not referred to explicitly nor implied, in my opinion. What is cited is the Heine-Borel theorem as stated for real line or plane, or for a compact metric set as: a countable open covering of a compact metric set has a finite sub-covering. I cannot comment on "Cantor's nested intervals", because I do not know what they are, but the sequence in Edgar's text is a sequence of non-vacuous, compact sets in a metric space. Each member is contained within its successor.

The conclusion is that the sequence converges in the Hausdorff metric to the intersection of the sets.

I don't need an explanation of Edgar's proof. Another proof would do just as well. For that purpose, I restate the proposition to be proved (Proposition 2.4.7 in the text, p.68):

A sequence of non-vacuous, compact, descending sets in a metric space, converges to its intersection in the Hausdorrf sense.

lim D(A_n, A) = 0, where A_n is a set of the sequence and A is their intersection.

I see intuitively that in the limit A_n approaches the intersection, because the sets are descending towards it, but that is not a proof.