Claim: sup(A)=3\sup(A)=3sup(A)=3
Proof:
Clearly, by definition for all x∈Ax\in Ax∈A , x<sup(A)=3x<\sup(A)=3x<sup(A)=3
Now, for every ϵ:=3−x2>0\epsilon:=3-\frac{x}{2}>0ϵ:=3−2x>0 , 3−ϵ=x2<x3-\epsilon=\frac{x}{2}<x3−ϵ=2x<x <3<3<3, 3−ϵ3-\epsilon3−ϵ is not an upper bound of AAA. Hence done.
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