Inequality proof involving Infs and Sups

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SUMMARY

The discussion centers on proving that for bounded functions F and G on a set S, if f(x) ≤ g(x) for all x in S, then it follows that inf{f(x): x ∈ S} ≤ inf{g(x): x ∈ S}. The proof strategy involves defining L0 as inf{f(x): x ∈ S} and L1 as inf{g(x): x ∈ S}, and using a proof by contradiction. The approach suggests assuming a scenario where g(x) is less than L0 and demonstrating the inconsistency of this assumption.

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  • Understanding of bounded functions
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homegrown898
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Homework Statement


Let F and G be bounded functions on S. If f(x) <= g(x) for all x in S prove that inf{f(x):x belongs to S} <= inf{g(x):x belongs to S}


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The Attempt at a Solution


Basically the idea is to let L0 = inf{f(x):x belongs to S} and L1 = inf{g(x):x belongs to S} and show that L0 <=L1
 
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I suggest contradiction. Assume there is some value of x where g(x) is less than the inf of all possible f(x)'s, then see if you can show why that just don't make no sense.
 

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