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Inf proof question

  1. Nov 27, 2008 #1
    its given that "x" is a low bound of non empty group A.
    prove that x=inf A for e>0 there is "y" which x<=y<=x+e

    i cant understand this question

    i know how to prove that a certain number is inf or sup using only the formula of the group

    but here i dont have the formula of the geoup
    and i have this 'y' i dont know why its for??
  2. jcsd
  3. Nov 27, 2008 #2
    Perhaps you need to rewrite the question. x is a lower bound for nonempty set A? y is an element of A? For all e or a fixed e?
  4. Nov 27, 2008 #3
    its given that "x" is a lower bound for non empty set A.
    prove that x=inf A for e>0 there is "y" y is element of A which x<=y<=x+e
    e is fixed
  5. Nov 27, 2008 #4
    i know that inf is the biggest integer under the lower bound
    and if we add a certain "e" then it will cease to be inf
    and it will become inside the group
  6. Nov 27, 2008 #5
    inf A is the greatest lower bound for the set A.
    So inf A + e will not be a lower bound for the set A for some positive real e.

    What does this statement mean: "for e>0 there is "y" y is element of A which x<=y<=x+e"?
  7. Nov 27, 2008 #6
    that there is a fixed "e" for which x<=y<=x+e

    y is element of A
  8. Nov 27, 2008 #7
    It is stated that A is bounded below, x being a lower bound, and there is a specific e such that all elements of A fall in the interval [x, x + e]. So A is a bounded set. Perhaps you need to reread the question (or your notes that pertain to the question).
  9. Nov 28, 2008 #8
    i think i got it
    A bounded from the bottom
    x is a low bound of A

    i need to prove that x=inf A

    this expression is just the definition of inf

    so i tried to solve it as if i got a formula
    I presume that "x" is not inf
    then our inf is x+e
    now i proove that x+e is not infA

    but here i got stuck because in this step i would do a formula operation.

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