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amilapsn

  • one year ago

Let \(A\), \(B\) be two non empty bounded subsets of real number set(\(\mathbb{R}\)). Prove that if \(A\subseteq B\) then \(\sup A\leq \sup B\)

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  1. amilapsn
    • one year ago
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    I understand the question. All I want is a rigorous proof. Here's how I understood the problem: Since \(A\) and \(B\) are non empty subsets of \(\mathbb{R}\) and bounded, by the completeness property we can say that there exist \(\sup A\) and \(\sup B\).

  2. ganeshie8
    • one year ago
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    see https://proofwiki.org/wiki/Supremum_of_Subset

  3. amilapsn
    • one year ago
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    As \(A\subseteq B\) there can be two cases. There may be \(x\in A^{'}\cap B\) such that \(x>a\ \ \forall a\in A\). Or there may not. From the case 1(\(x>a\)) we can prove \(\sup A\leq \sup B\) and from the second we can prove \(\sup A=\sup B\), From both cases we can come to the conclusion \(\sup A\leq \sup B\).. But I don't think this is a rigorous proof...

  4. amilapsn
    • one year ago
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    That's easy and elegant. Thanks @ganeshie8 !! So here's the proof: \[ \text{Since }A,B \neq \{\}\text{ and bdd, }\sup A \text{ and } \sup B \text{ exist.} \\ \text{Let }x\in A \\ \ \ x\in B\because A\subseteq B \\ \ \ \sup B\geq x \\ \forall x\in A x\leq \sup B \\ \therefore \sup B \text{is an upper bound of $A$} \\ \therefore \sup A\leq \sup B \]

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