How Do Supremum and Infimum Relate When s < t for All s in S and t in T?

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SUMMARY

The relationship between supremum and infimum is established as sup S ≤ inf T when subsets S and T of R satisfy the condition s < t for all s in S and t in T. The proof begins by defining sup(S) as 'a' and inf(T) as 'b', leading to the conclusion that a > s and b < t for all s and t. A proof by contradiction is effective, starting with the assumption that inf(T) < sup(S) and demonstrating that this leads to a contradiction, confirming the original statement.

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  • Understanding of supremum and infimum definitions in real analysis
  • Familiarity with proof techniques, particularly proof by contradiction
  • Knowledge of upper and lower bounds in set theory
  • Basic concepts of subsets in real numbers
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wang jia le
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Let S and T be subsets of R such that s < t for each s ∈ S and each t ∈ T. Prove carefully that sup S ≤ inf T.

Attempt:

I start by using the definition for supremum and infinum, and let sup(S)= a and inf(T)= b

i know that a> s and b< t for all s and t. How do i continue? , do i prove it directly starting from s< t or will it be easier to use proof by contradiction?
 
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wang jia le said:
Let S and T be subsets of R such that s < t for each s ∈ S and each t ∈ T. Prove carefully that sup S ≤ inf T.

Attempt:

I start by using the definition for supremum and infinum, and let sup(S)= a and inf(T)= b

i know that a> s and b< t for all s and t. How do i continue? , do i prove it directly starting from s< t or will it be easier to use proof by contradiction?

Try contradiction.
 
Usually the definition of upper/lower bound would only imply [itex]s \leq \sup(S)[/itex] for all [itex]s \in S[/itex] and [itex]\inf(T) \leq t[/itex] for all [itex]t \in T[/itex]. In other words, the upper and lower bounds can be in the set themselves. The stated result should hold regardless though.

Just start with [itex]\inf(T) \lt \sup(S)[/itex] and go from there. There must be an [itex]s \in S[/itex] such that [itex]\inf(T) \lt s[/itex] ( otherwise [itex]\inf(T)[/itex] would be an upper bound of [itex]S[/itex] that's less than [itex]\sup(S)[/itex] ). But then, for similar reasons, there must be a [itex]t \in T[/itex] such that [itex]t \lt s[/itex] ( fill in the details ).
 
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