How to prove a number is a supremum of a set

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SUMMARY

The discussion centers on proving that a number \( u \) is the supremum of a set \( S \) given two conditions: (i) for all natural numbers \( n \), \( u - (1/n) \) is not an upper bound of \( S \), and (ii) for all natural numbers \( n \), \( u + (1/n) \) is an upper bound of \( S \). The conclusion drawn is that if both conditions hold, then \( u \) must equal \( \sup S \). The proof strategy involves contradiction, exploring scenarios where \( u \) is either greater than or less than \( \sup S \) to demonstrate that both lead to violations of the given conditions.

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  • Understanding of supremum and infimum in real analysis
  • Familiarity with the properties of upper bounds
  • Knowledge of proof techniques, particularly proof by contradiction
  • Basic concepts of natural numbers and their properties
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  • Learn about upper and lower bounds in the context of sets
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  • Investigate examples of supremum in various sets, such as bounded and unbounded sets
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Homework Statement


Prove that if
(i) [tex]\forall n[/tex][tex]\in N[/tex], u - (1/n) is not an upper bound of s
(ii) [tex]\forall n[/tex][tex]\in N[/tex], u + (1/n) is an upper bound of S
then, u = supS

Homework Equations





The Attempt at a Solution


It (i) and (ii) are true, then
[tex]\exists s[/tex][tex]\in S[/tex] s.t. u - (1/n) < s
and u+(1/n)>s for all s.
I'm not sure where to go from here.
 
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How about a proof by contradiction?

Suppose [itex]u > \sup S[/itex]. Then there exists [itex]n \in \mathbb{N}[/itex] such that [itex]\sup S < u - 1/n < u[/itex]. Does this violate one of (i) or (ii)?

Next suppose [itex]u < \sup S[/itex]. Can you rule this out as well?
 

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