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Limit and Continuity Question

  • Thread starter roam
  • Start date
  • #1
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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:

Answers and Replies

  • #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)?
 
  • #3
1,265
11
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.
 
  • #4
Dick
Science Advisor
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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?
 
  • #5
HallsofIvy
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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

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

l-ε < x < l+ε

|f(x)-l| < ε

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

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