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    $\begingroup$ I'd be quite surprised if you could prove that that family you define doesn't cover the whole of [0,1] without using something very like the nested-intervals version of the diagonal argument. (That is, I'd like to know more about your proof that [0,1] has outer measure 1.) $\endgroup$ Commented Nov 25, 2010 at 22:15
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    $\begingroup$ I edited my post with a sketch of a proof that $[0,1]$ has outer measure $1$, modulo compactness of closed intervals. I do not see that I'm using a diagonal/nested intervals argument, but I could be totally wrong. The usual proof of Heine-Borel also doesn't appear to me to use a diagonal argument (see e.g. math.utah.edu/~bobby/3210/heine-borel.pdf). At no point in the proof there is an enumeration of an infinite set. $\endgroup$ Commented Nov 26, 2010 at 0:38
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    $\begingroup$ Maybe you've convinced me. One can take the family you describe and define x to be the supremum over all reals x such that [0,x] can be covered by finitely many of its members. Such an x can't be covered itself (since the sets are open). This is just the proof of Heine-Borel in this special case. Also, your lemma gives us that x can't be 1. So one ends up using the least upper bound axiom instead of the nested-intervals property, which is perhaps enough to qualify the argument as genuinely different. $\endgroup$ Commented Nov 27, 2010 at 18:06
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    $\begingroup$ this proof appears in hardy's pure mathematics, about 100 years ago. $\endgroup$ Commented Dec 5, 2010 at 5:36
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    $\begingroup$ I am sorry I have the book in hard, but can't find where the proof is. Could you tell me where that is? $\endgroup$ Commented Dec 22, 2010 at 16:38