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Topology - is my proof correct?

  1. Jun 25, 2007 #1

    quasar987

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    1. The problem statement, all variables and given/known data
    The lemma to prove is that "If [-R,R]^{n-1} is compact, then [-R,R]^n is too.

    To help us, we have two other lemmas already proven:

    L1: "[a,b] is compact."

    L2: "If A is R^n is compact and x_0 is in R^m, then A x {x_0} is compact."

    3. The attempt at a solution I found a proof that is substantially simpler than the one provided by the book. Usually, this is not a good sign!

    I say, let [tex]\{U_i\}_{i\in I}[/tex] be an open cover for [-R,R]^n. For all x in [-R,R], [-R,R]^{n-1} x {x} is compact (by L2), so we can find a finite subcover, say [tex]\{U_{i^{(x)}_k}\}_{1\leq k\leq N_x}[/tex]. Let [tex]O_x=\cup_{k=1}^{N_x}U_{i^{(x)}_k}[/tex]. We have that O_x is open and [-R,R]^{n-1} x {x} [tex]\subset[/tex] O_x. Next, observe that the familly [tex]\{O_x\}_{x\in [-R,R]}[/tex] covers [-R,R]^n. As in the proof of L2,

    By construction, the resulting sub-family, say [tex]\{O_{x_j}\}_{1\leq j\leq M} [/tex] is a finite cover of [-R,R]^n. Therefor, [tex]\{U_{i^{(x_j)}_1},...,U_{i^{(x_j)}_{N_{x_j}}}\}_{1\leq j \leq M}[/tex] is a finite subcover of [-R,R]^n that has been extracted from our original cover [tex]\{U_i\}_{i\in I}[/tex].
     
    Last edited: Jun 25, 2007
  2. jcsd
  3. Jun 25, 2007 #2

    matt grime

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    The cover you wrote down just before quoting me (presumably) is not an open cover of anything at all.

    Also, the supposed pull-back idea you want to use is of no use what-so-ever.
     
  4. Jun 25, 2007 #3

    quasar987

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    How come? We have for every x in [-R,R] that, [-R,R]^{n-1} x {x} [itex]\subset[/itex] O_x, therefor,

    [tex]\bigcup_{x\in [-R,R]}([-R,R]^{n-1}\times \{x\})\subset \bigcup_{x\in [-R,R]}O_x[/tex]

    But unless I'm crazy,

    [tex]\bigcup_{x\in [-R,R]}([-R,R]^{n-1}\times \{x\})=[-R,R]^n[/tex],

    so

    [tex][-R,R]^n\subset \bigcup_{x\in [-R,R]}O_x[/tex].


    You're so hard with me :cry:.

    Is it not so that since {O_x} covers [-R,R] x [-R,R], the projection of the last coordinates of the O_x on the real line covers [-R,R] with open sets? we can then extract a finite subcover of these projections, and after we pullback again, the whole O_x's corresponding to the finite subcover actually covers [-R,R]^n because by construction, each O_x covers [-R,R]^{n-1}.

    How is the situation different here than in my other post?
     
    Last edited: Jun 25, 2007
  5. Jun 25, 2007 #4

    StatusX

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    I think what you're trying to do, starting at the quote, is something like this:

    To every x, define A_x to be the open subset of [-R,R] obtained by projecting the sets in O_x onto the last coordinate. Then finitely many of the A_x cover [-R,R], and taking the union of the corresponding O_x gives a finite subcover of U_i.

    If this is what you meant, it's not right. It seems like you're assuming that if the projection of the O_x is A_x (an open subset of [-R,R] containing x), then O_x covers not only [-R,R]n-1x{x}, but in fact all of [-R,R]n-1 x A_x.

    But A_x gives, in a sense, the "tallest" "cross-section" of the O_x. But O_x will only cover [-R,R]n-1 x B_x where B_x is the shortest cross-section of the O_x. For a general set, there might not be such a B_x, since you may have a set in the cover like {(x,y)| |y|<1/|x|}. But here, using the compactness of [-R,R]n-1, you should be able to find such a set.
     
  6. Jun 26, 2007 #5

    matt grime

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    I think I misread part of your post. I still can't quite decide if you need to add anything to your reasoning - I will ponder some more. But I'm beginning to think it is correct. Indeed, I'd now go so far as to say it is obviously correct and I was being an idiot.
     
    Last edited: Jun 26, 2007
  7. Jun 26, 2007 #6

    quasar987

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    Well, in any case, StatusX's post showed me why what I thought I was doing was not correct. I was having difficulty with this proof and Lemma 2's proof because I obstinately tried to visualize the thing in n dimensions because I am always afraid that if I work with R² or R³, I might do something that is not legal in an arbitrary n.

    In any case, it was easy to see in R² what bit of thinking I was doing wrong, but the fact that [-R,R]^{n-1} is compact assures us, (by defining the right function and taking it's inf) that we can find an epsilon_y>0 such that O_y covers [-R,R]^{n-1} x (y-epsilon_y, y+epsilon_y). And the fact that [-R,R] is compact (see L1) assures usagain that inf{epsilon_y: y in [-R,R]} is stricly greater than zero. From there, it's just a technicality to construct a subcover.

    Overall, my proof is only a semi-success, because it leans on the fact that a function defined on a compact takes its infimum, a result not yet seen. And if one formalizes, the idea hinted at by StatusX and that I described just above, the result is a shower of symbols detailing technical details and in the end, the proof is not any easier to read than the one provided by the book. Though it is more intuitive!
     
    Last edited: Jun 26, 2007
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