Proof that given the info below, sup S <= inf T

  • #1

Homework Statement


Let S and T be nonempty subsets of [tex]\mathbb{R} \backepsilon s \leq t \forall s \in S \wedge t \in T.[/tex] A) Observe that S is bounded above and that T is bounded below. B) Prove that [tex]sup S = inf T.[/tex]


Homework Equations





The Attempt at a Solution


Let [tex] s_0 = sup S. \text{ Then } s \leq s_o \forall s \in S \wedge s \leq t \forall s \in S \Rightarrow s_0 \leq t. [/tex]
Let [tex] t_0 = inf S. \text{ Then } t_0 \leq t \forall t \in T \wedge s \leq t \forall t \in T \Rightarrow s \leq t_0. [/tex]

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
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I am probbably missing something, but it doesn't seem that it is even true. COnsider
S=(0,1) and T=(2,3). Then for all s in S s=<t for all t in T, but clearly supS=/=infT. Are you sure the problem is not asking you to show : supS=<infT??? If this is what the problem is asking, then try a proof by contradiction. Namely, assume that infT<supS, and derive a contradiction.
 
Last edited:
  • #3
HallsofIvy
Science Advisor
Homework Helper
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Yes. The title says "sup S[itex]\le[/itex] inf T" but in your post you say "sup S< sup T" which does not follow from your hypotheses. They might be equal. If sup S> sup T, then let [itex]\delta[/itex]= Sup S- sup T. There must be some member of S closer to S than that and so must be some member of S larger than sup T.
 

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