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A review of supremum infimum.

  1. Oct 30, 2006 #1
    i need to show that when A is a subset of B and B is a subset of R (A B are non empty sets) then: infB<=infA<=supA<=SupB

    here's what i did:
    if infA is in A then infA is in B, and by defintion of inf, infB<=infA.
    if infA isnt in A then for every e>0 we choose, infA+e is in A and so infA is in B, so infA+e>=infB, we can find e'>0 such that infA+e>=infB+e'>infB
    so we have: infA-infB>=e'-e, let e=e'/2 then we have infA-infB>e'/2>0 so we have infA>infB. (is this method correct?).
    obviously supA>=infA by defintion.
    i think that the same goes for supA and supB, with suitable changes.
     
    Last edited: Oct 30, 2006
  2. jcsd
  3. Oct 30, 2006 #2

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    You don't need to use an e>0. Go back to the original definition of inf in terms of just [itex]\leq[/itex].
     
  4. Oct 30, 2006 #3
    how exactly to use it?
    i mean if a is in A then a is in B, infA<=a so for every c<=a infA>=c
    because a is in B also then a>=infB, but because infA is the greatest lower bound then we have infB<=infA, correct?
     
  5. Oct 30, 2006 #4

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    Everything there sounds right except for that thing with c which I didn't follow.
     
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