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Limit and Continuity Question

  1. Apr 9, 2008 #1
    Let [tex]l \in R[/tex] be the least upper bound of a nonempty set S of real numbers.
    Show that for every [tex]\epsilon < 0[/tex] there is an [tex]x \in S[/tex] such that
    [tex]x > l - \epsilon[/tex]




    I don't understand this question very well, I appreciate it if you could give me some hints.



    l is the l.u.b on S, therefore it is greater than or equal to any [tex]s \in S[/tex]

    By the definition of the limit; |f(x)-l| < ε if 0 < |x-a| < δ

    l-ε < x < l+ε

    |f(x)-l| < ε

    "?"
     
    Last edited: Apr 9, 2008
  2. jcsd
  3. Apr 9, 2008 #2

    Dick

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    Suppose it's NOT true. Then there is an epsilon such that x<=l-epsilon for all x in S. Can l really be lub(S)?
     
  4. Apr 10, 2008 #3
    Thanks, so that's how you derive the contradiction.

    But could you explain it in more detail please(especially the last bit)?

    I appreciate that.
     
  5. Apr 10, 2008 #4

    Dick

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    l is supposed to be the LEAST upper bound. If x<=l-epsilon for all x in S, then l-epsilon is ALSO an upper bound. What the order relation between l and l-epsilon?
     
  6. Apr 10, 2008 #5

    HallsofIvy

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    What limit and what f(x) are you talking about? There is no function nor limit of a function mentioned in the problem

    Dick has given the hint you need: use proof by contradiction.
     
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