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

[middleCmusic]

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

 

[micromass]

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.

 

[middleCmusic]

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.