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posted by  Laysha on 9/6/2008 3:24:06 PM  |  status: Live  

Real Analysis

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 Axiom C: Completeness Axiom
      Every nonempty set S of real numbers which has an upper bound has a least upper bound.
 
Proposition 1:  Let L and U be nonempty subsets of  with  = L  U and such that for each l in L and each u in U we have l < u.  Then either L has a greatest element or U has a least element.
 
Now let's prove proposition 1 using Axiom C.
 
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posted by WSAN on 9/6/2008 7:58:24 PM  |  status: Live
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Response Details:
suppose Axiom C holds.
we prove proposition 1.
suppose L and U are non empty subsets of R such that LU = R , x is in L and y is in U
 ==> x < y.
in other words every member of L is less than every member of U.
i.e. every member of U is an upper bound of L.
so, L is a non empty subset of real numbers which is bounded above.
so, by axiom C ,  L has the greatest element . let it be α. then x < α ==> x is in L and any element L .
if y > α , then y is in U.
 
SWAN
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