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**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.