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Show set (which is a subset of R^n) is bounded

  1. Mar 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that [itex]D = { (x,y,z) \in \mathbb{R}^{3} | 7x^2+2y^2 \leq 6, x^3+y \leq z \leq x^2y+5y^3}[/itex] is bounded.

    2. Relevant equations
    Definition of bounded:[itex]D \subseteq \mathbb{R}^{n}[/itex] is called bounded if there exists a M > 0 such that [itex]D \subseteq \{x \in \mathbb{R}^{n} | ||x|| \leq M\}[/itex]


    3. The attempt at a solution
    I have to find a M such that [itex]D \subseteq \{(x,y,z) \in \mathbb{R}^{3} | x^2 + y^2 + z^2 \leq M\}[/itex]. I thought of just picking a very high M, say 999999. But how do I show it works?
     
  2. jcsd
  3. Mar 3, 2012 #2

    Office_Shredder

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    It's often easier to show that each coordinate is bounded, say x,y and z are all smaller than 10 maybe. Then ||(x,y,z)|| < ||(10,10,10)|| = M
     
  4. Mar 3, 2012 #3
    So if I say something like: [itex]7x^2+2y^2 \leq 6 \Rightarrow y^2 \leq 3 \lt 4 \Rightarrow y \lt 2[/itex]
    and [itex]7x^2+2y^2 \leq 6 \Rightarrow x^2 \leq \frac{6}{7} \lt 1 \Rightarrow x \lt 1[/itex]
    and [itex]z \leq x^2y+5y^3 \Rightarrow z \lt (2+5*8)=42[/itex]
    So choose M = 4+1+42^2 = 1769. And this M will do.
     
  5. Mar 3, 2012 #4

    Office_Shredder

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    Careful, M=||(1,2,42)|| is the square root of the number you put up. As long as M is larger than 1 that won't matter, but if the norm happens to be smaller than 1 failing to take the square root can give a value of M that doesn't work
     
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