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An example of a set

  1. Aug 12, 2005 #1
    I want to know some examples of an infinite set S with a least upper bound that is not an accumulation point of S. Is this an example... (-oo, 10]?
     
  2. jcsd
  3. Aug 12, 2005 #2
    No, it's not. Every point in that set is an accumulation point of it.

    An example is
    [tex] \left[0 , 1 \right] \cup \{ 2 \} [/tex].

    edit: Another example is Z- ={-1,-2,-3,-4,-5...}, the set of negative integers. It's clearly infinite, bounded above, and does not have any accumulation points.
     
    Last edited by a moderator: Aug 12, 2005
  4. Aug 13, 2005 #3
    Hmmm, thank you for the help...

    By the way, suppose I have Λ as my least upper bound of a set S but Λ is not in S. I want to know how this Λ is an accumulation point....

    My friend told me that for any ε > 0, he can show that there is a point s belonging to S such that Λ - ε < s < Λ. To end up the proof, he used the definition of accumulation point... how do I prove this...? :D
     
  5. Aug 13, 2005 #4

    HallsofIvy

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    Suppose &Lambda; is a least upper bound of a set, A, but not in the set itself. Since &Lamba; is an upper bound for A, there are no members of A larger than &lambda;. Given &epsilon;> 0 suppose there were no members of A between &Lambda;-&epsilon; and &Lambda;. Then there would be no members of A larger than &Lambda;-&epsilon;. That means that &Lamba;-&epsilon; is an upper bound for A, contradicting the fact that &Lambda; is the least upper bound.
     
  6. Aug 13, 2005 #5
    So the concept of accumulation point comes in here "Given ε> 0 suppose there were no members of A between Λ-ε and Λ. ...".
     
  7. Aug 14, 2005 #6
    May I clarify something.... is this the proof for ( I have restated my problem) "Assume that Λ is the least upper bound of a set S but Λ is not in S. Show that Λ is an accumulation point of S" ?
     
  8. Aug 14, 2005 #7
    Yes, you're right.
     
  9. Aug 14, 2005 #8
    Ah.. ok.. I asked that because I thought the proof only shows that /\ is the least upper bound... and not that /\ is an accumulation point.

    From the proof HallsofIvy stated, how did /\ turn out in the end to be an accumulation point (I apologize for being "slow")
     
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