- #1
MathematicalPhysicist
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let A,B be nonempty sets of real numbers, prove that:
if A,B are bounded and they are disjoint, then supA doesn't equal supB.
here's my proof:
assume that supA=supB=c
then for every a in A a<=c and for every b in B b<=c.
bacuse A.B are bounded then: for every e>0 there exists x in A such that
c-e<x<=c and there exists y in B such that c-e<y<=c so we have two elements that are both in A and B, but this is a contradiction.
is this proof valid?
i feel that i should show that y=x, but i think bacuse e is as we choose, we have to find elements which are both in A and B.
if A,B are bounded and they are disjoint, then supA doesn't equal supB.
here's my proof:
assume that supA=supB=c
then for every a in A a<=c and for every b in B b<=c.
bacuse A.B are bounded then: for every e>0 there exists x in A such that
c-e<x<=c and there exists y in B such that c-e<y<=c so we have two elements that are both in A and B, but this is a contradiction.
is this proof valid?
i feel that i should show that y=x, but i think bacuse e is as we choose, we have to find elements which are both in A and B.