Prove supS ≤ infT - Math Homework

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SUMMARY

The discussion focuses on proving the inequality supS ≤ infT for non-empty subsets S and T of real numbers R, given that every element s in S is less than or equal to every element t in T. The proof begins by establishing that supT serves as an upper bound for S, leading to the conclusion that supS is less than or equal to supT. However, the key insight is to demonstrate that infT is also an upper bound for S, which directly supports the required inequality supS ≤ infT.

PREREQUISITES
  • Understanding of supremum and infimum in real analysis
  • Familiarity with the properties of upper and lower bounds
  • Knowledge of set theory and subsets
  • Basic mathematical proof techniques
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  • Study the definitions and properties of supremum and infimum in real analysis
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  • Explore examples of proving inequalities involving supremum and infimum
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This discussion is beneficial for students studying real analysis, mathematicians interested in set theory, and anyone looking to deepen their understanding of bounds in mathematical proofs.

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Homework Statement


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.

Homework Equations


N/A


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.


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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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