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Spivak's proof of A closed bounded subset of R^n is compact

  1. Sep 20, 2012 #1
    Spivak's proof of "A closed bounded subset of R^n is compact"

    Hi guys,

    I'm currently taking a differential geometry course and decided I would read Spivak's Calculus on Manifolds, and then move on to his Differential Geometry series. There's a proof in here that feels unjustified to me, so I'm hoping you guys can point out what I'm missing. It's on p. 10 and it reads as follows:

    1-7 Corollary. A closed bounded subset of ℝn is compact. (The converse is also true (Problem 1-20).)

    Proof. If A[itex]\subset[/itex][itex]ℝ^{n}[/itex] is closed and bounded, then A[itex]\subset[/itex]B for some closed rectangle B. If [itex]\wp[/itex] is an open cover of A, then [itex]\wp[/itex] together with [itex]ℝ^{n}-A[/itex] is an open cover of B. Hence a finite number of [itex]U_1, ..., U_n[/itex] of sets in [itex]\wp[/itex], together with [itex]ℝ^{n}-A[/itex] perhaps, cover B. Then [itex]U_1, ..., U_n[/itex] cover A.

    The part in red is the part that I don't understand. How can we jump to saying that a finite number of open sets cover B? Isn't that sort of assuming the result?

    (I ask these questions not because I doubt the veracity of Spivak's proof, but because I don't understand it.)
     
  2. jcsd
  3. Sep 20, 2012 #2

    micromass

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    Re: Spivak's proof of "A closed bounded subset of R^n is compact"

    In Corollary 1-6, Spivak has already proved that closed rectangles are compact. So he applies the result to B. So since from that result follows that B is compact, it also follows that it has finite subcover.
     
  4. Sep 20, 2012 #3
    Re: Spivak's proof of "A closed bounded subset of R^n is compact"

    Ah, thank you. I had a feeling I was missing something obvious.
     
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