show that sup(XUY)=max{supX supY}
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Answer:
Set x=supAx=supA, y=supBy=supB. Given a∈A,b∈Ba∈A,b∈B we have a≤x,b≤ya≤x,b≤y so a+b≤x+ya+b≤x+y so it's an upper bound. Now take ε>0ε>0 and find a,ba,b such that a>x−ε/2,b>y−ε/2a>x−ε/2,b>y−ε/2 and you have a+b>x+y−εa+b>x+y−ε.
That suffices (since it means that every potential "smaller upper bound" x+y−εx+y−ε is not really an upper bound)
Step-by-step explanation:
I USE A and B except X and Y.
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Answer:
Let A,B not empty,bounded subsets of R.Show that
sup(A∪B)=max{supA,supB}.
That's what I have done so far: Let x∈A∪B⇒x∈A or x∈B⇒x≤sup(A) or x≤sup(B)⇒x≤max{sup(A),sup(B)}. But,how can I continue?
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Feb 7 '14 at 14:16
evinda
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You're nearly there. You've shown that every x∈A∪B satisfies x≤max{sup(A),sup(B)}. Therefore, by definition (or an elementary property, depending on your definition) of sup, sup(A∪B)≤max{sup(A),sup(B)}.
Conversely, every element of A is also in A∪B, so for exactly the same reason, supA≤sup(A∪B). Symmetrically, supB≤sup(A∪B). By definition of max, max{sup(A),sup(B))≤sup(A∪B).4
Down vote
s is an upper bound for A∪B iff it is an upper bound for A and for B, so s≥supA and s≥supB.
The least upper bound, supA∪B is then the least such s - which is clearly max{supA,supB}.
Step-by-step explanation:
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