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How do you prove supermum?

  1. Nov 27, 2004 #1
    ok, so supermum is the least upper bound, but when you write down on paer, what am I trying to show...does that make sense? I mean, what am I trying to get to that will show something is the supermum.

    so for example: prove sup {1 - 1/n} = 1 for all n in N.

    so I start by saying that I know that 1 is an upper bound.....

    then, for all epsilon > 0, there exist n st n > epsilon, or 1/n < epsilon.

    ok, now what....how else do I need to go to show that 1 is supermum...

    how do I know that I have shown that 1 is the supermum??
     
  2. jcsd
  3. Nov 27, 2004 #2

    shmoe

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    To show 1 is a supremum you need to show 2 things,
    a) that 1 is an upper bound for your set
    b) that you can find points in your set arbitrarily close to 1, That is for every [tex]\epsilon>0[/tex] there is an [tex]x[/tex] in your set where [tex]1-\epsilon<x[/tex].

    That's all. It appears you've done both of these steps, but your organization seems unclear. Do you understand how what you've done relates to the above?


    An alternate to b) above, you can instead prove
    b*)if A is any other upper bound for your set, then 1<=A
     
  4. Nov 27, 2004 #3
    Show that the number is an upperbound AND any number smaller is NOT an upperbound.
     
  5. Nov 27, 2004 #4
    I dotn know what I've done to solve the problem....can you tell me what I did? lol...how haev I already solved the problem?

    btw, is 1 - e < x the same as 1/e < x?
     
  6. Nov 27, 2004 #5

    shmoe

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    This is condition a), though you should should give some justification as to why 1 is an upper bound.

    This is b), but it needs some explanation. To prove b), let [tex]\epsilon >0[/tex] and we want to find something in our set that's larger than [tex]1-\epsilon[/tex]. You've shown that there is an [tex]n\in\Bbb{N}[/tex] where [tex]1/n<\epsilon[/tex]. Now you know that [tex]1-1/n[/tex] is in your set and the last inequality tells us that [tex]1-1/n>1-\epsilon[/tex], and we've found an element in our set and condition b) is satisfied.


    No. Try e=1/2, x=1.
     
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