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

  • #1

Main Question or Discussion Point

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.)
 

Answers and Replies

  • #2
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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.
 
  • #3


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

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