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Property of Compactness

  1. Jun 9, 2012 #1
    Suppose S[itex]\subset[/itex]ℝn is compact, f: S-->R is continous, and f(x)>0 for every x [itex]\in[/itex]S. Show that there is a number c>0 such that f(x) ≥ c for every x[itex]\in[/itex]S.



    Attempt:
    Since S is contained in Rn is compact, then S is closed and bounded.
    By the extreme value thm there exists values a,b that are an absolute minimum and absolute maximum respectively. Let c = f(a). Therefore by EVT f(c) ≤ f(x) in R.


    Well I have the solution manual and their solution is way different to what I attempted. Is there anything right about this?
     
  2. jcsd
  3. Jun 9, 2012 #2

    micromass

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    Your solution is correct. You might want to explain why c>0 however.
     
  4. Jun 9, 2012 #3

    micromass

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    Also, this should be c≤f(x).
     
  5. Jun 9, 2012 #4
    But isn't c [itex]\in[/itex] S. SO wouldn't the function map c to f(c)?



    Is the fact c having to be positive because if c < 0 ==> f(a) < f(c) ?
     
  6. Jun 9, 2012 #5

    micromass

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    You defined c=f(a). So c is an element of [itex]f(S)\subseteq \mathbb{R}[/itex]. Writing f(c) makes no sense.

    Again, f(c) makes no sense.
     
  7. Jun 9, 2012 #6
    got it now, thanks foe the help.
     
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