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Prove supS <= infT

  1. Mar 28, 2010 #1
    1. The problem statement, all variables and given/known data
    Let S and T be non-empty subsets of R, and suppose that for all s [tex]\in[/tex] S and t [tex]\in[/tex] T, we have s [tex]\leq[/tex] t.

    Prove that supS [tex]\leq[/tex] infT.

    2. Relevant equations

    3. The attempt at a solution

    Since s [tex]\in[/tex] S [tex]\Rightarrow[/tex] s [tex]\in[/tex] T, supT is an upper bound for S.
    Since supS is the least upper bound, supS [tex]\leq[/tex] supT.

    How does this look to you? Feedback is appreciated.
  2. jcsd
  3. Mar 28, 2010 #2


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    Staff Emeritus
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    [tex]supS \leq supT[/tex] doesn't seem that useful as a start to be honest (supT and infT aren't very close to each other in general). To show that [tex] supS \leq infT[/tex], can you show that infT is an upper bound of S?
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