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A subset in R^n is bounded if and only if it is totally bounded.

  1. Jan 7, 2014 #1
    1. The problem statement, all variables and given/known data
    Prove that a subset in R^n, where n is a finite number, is bounded if and only if it is totally bounded.


    2. Relevant equations
    If A is the subset, A is bounded if there is a point b in R^n such that d(x,b)<= K, for a every x in A.

    A is totally bounded if for every e> 0, there is a finite number of ball with radius e, that covers A.


    3. The attempt at a solution

    totally bounded -> bounded

    Chose a number e, there is a finite number P of open balls that cover A, each with radius e.
    Chose an element b in A. s=sup{d(b,l): l is the center of the balls} must exist since we only have finite amount of balls. Then the max distance from b to another point in A is s+e, and hence A is bounded.
    Is this proof correct for this implication?

    bounded -> totally bounded
    There is an element in R^n such that for any element x in A d(x,b) <= K, for a real number K.
    I do not see how to proceed here?
    Since Q^n is dense I guess that if we make balls with radius epsilon around every element in R^n that is rational we have a countable number of balls that cover A?, then the problem is to reduce it to a finite number of balls?
     
    Last edited: Jan 7, 2014
  2. jcsd
  3. Jan 7, 2014 #2
    I like your thinking, simply because it shows you're actually thinking about this, but I don't think that will work.

    Since ##A## is bounded, can we make ##A\subseteq B_K##, where ##B_K=\{x\in\mathbb{R}^n:d(b,x)\leq K\}##?
     
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