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Total boundedness in R^n

  1. Dec 13, 2008 #1
    I'm trying to prove that boundedness implies total boundedness in R^n. I'm stuck mostly because I can't convert my geometric intuition into maths. I am failing at the pivotal step where I use the finite-dimensionality. Any help would be appreciated.
  2. jcsd
  3. Dec 14, 2008 #2
    I'm looking for an elementary proof. I can think of a non-elementary proof that relies heavily on other theorems but it's not very satisfying.
  4. Dec 14, 2008 #3
    This depends on the Heine-Borel theorem, but does it satisfy you? If a subset of Rn is bounded, take its closure to get a compact set, and by compactness find a finite covering by ε-balls for any ε > 0.
    Last edited: Dec 14, 2008
  5. Dec 14, 2008 #4
    That's what I mean. I can use Heine-Borel theorem but that's not quite what I want. I have this geometric intuition that I would like to know if I can use rigorously:

    Let X be a subset of Rn and assume that it's not totally bounded. Then you can construct an infinite set of points x_n such that d(x_i, x_j)>e for some e>0. The geometric intuition is that if you have infinite number of these points, an infinite number of them must fall on some straight line (because of the finite dimensionality) and then you can use that to show that the diameter of X is not finite thereby completing the proof. The key step that I can't show is that there are an infinity of these points on one straight line.

    I'm not sure if this works, but I'd rather not invoke Heine-Borel...
  6. Dec 14, 2008 #5
    Even better: if a subset X of Rn is bounded, it is contained in some ball (or perhaps a cube might be easier) B. Cover all of Rn with ε-balls centered on points on some sort of lattice; argue that finitely many of these intersect B. Then B is totally bounded, so X, being a subset of B, is as well.
  7. Dec 15, 2008 #6
    LOL. I was about to make a new thread about this topic.

    I have an exam on Wed and I was praticing couple of things. Here is my question,
    Suppose X € R^n is compact. Prove that X is bounded.

    does it make sense? did i do it right?
  8. Dec 15, 2008 #7
    Your proof is invalid. By saying "let α = sup X", there are two things wrong there:
    (1) You're already assuming that X is bounded above, and you're trying to prove that X is bounded.
    (2) sup X (and bounded above) only makes sense if X is a subset of [itex]\mathbb{R}[/itex] (or some other ordered set with the least upper bound property)!
    Additionally, limit points have little to do with boundedness. (What does "α is a limit point of X which contain all limit points" even mean?)

    Remember the definition of compactness. The simplest way to solve that problem is to use the definition of compactness directly; the idea is to find an open cover of X such that each set in the cover is bounded. By compactness, finitely many of these sets cover X, and you have to argue that the union of finitely many bounded sets is bounded.

    As a specific example, I denote [itex]B(x; 1)[/itex] to mean the ball of radius [itex]1[/itex] centered at [itex]x[/itex] (that is; [itex]B(x; 1) = \{y \in \mathbb{R}^n \mid \lvert x - y \rvert < 1[/itex]); these balls are open sets (by the definition of the topology induced by a metric space). My open cover will be [itex]\mathcal{O} = \{B(x; 1) \mid x \in X\}[/itex]; it's easy to verify that [itex]\mathcal{O}[/itex] covers [itex]X[/itex]. By compactness, there is a finite subcover [itex]\mathcal{F} \subseteq \mathcal{O}[/itex]. Since I picked each set in [itex]\mathcal{O}[/itex] to be bounded, [itex]X \subseteq \bigcup \mathcal{F}[/itex] is bounded.
    Last edited: Dec 15, 2008
  9. Dec 15, 2008 #8
    wow its wrong? huh what world am i living in?

    ps: thats alpha.
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