In: Advanced Math
Suppose that a subset S of an ordered field F is not bounded above in F. Let T be a subset of F satisfying the property that, for each x ∈ S, there exists y ∈ T such that x ≤ y. Prove that T is not bounded above in F.
We will prove by contradiction that is not bounded above in .
Suppose is bounded above in then there exist such that ,
for all .
Let , be arbitrary .
By the given condition , as .
Since and are subset of order field and ,
Since be arbitrary so for all
Hence is bounded above , a contradiction to is not bounded above .
Hence is not bounded above in .
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