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Definition of supremum and infimum using epsilons ?

  1. Sep 24, 2009 #1
    Is this what it is:

    "For every [itex]\epsilon > 0[/itex] there exists [itex]x\in A[/itex] such that [itex]x \leq \inf A + \epsilon[/itex]."

    ...and similarly for the supremum?
     
  2. jcsd
  3. Sep 24, 2009 #2

    LCKurtz

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    No, that's not quite it. Any number y which is greater than all numbers in A would satisfy your definition.
     
  4. Sep 24, 2009 #3
    Well in the first definition the OP said "there exists", so really the concern of upper bounds isn't really important. This is a useful proposition, and sometimes the inequality is strict, though that doesn't matter all that much.
     
  5. Sep 24, 2009 #4

    LCKurtz

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    What "first definition"? Are you talking about something other than:

    "For every [tex]\epsilon > 0[/tex] there exists [tex]x\in A[/tex] such that [tex] x \leq \inf A + \epsilon .[/tex]"?

    While that is true about the infinimum, it won't do for the definition for the reason I gave.
     
  6. Sep 24, 2009 #5
    That's correct. We must also state that inf A is a lower bound.
     
  7. Sep 24, 2009 #6
    Yeah I realized I was thinking of the theorem that states that if L is a lower bound for a set A in R, then L = inf A iff for every epsilon > 0, there is an x in A with x - L < epsilon. My apologies, of course it's not a definition.
     
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