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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 \(\sup A\leq \sup B\)
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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amilapsn
 one year ago
Best ResponseYou've already chosen the best response.1I 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 \(\sup B\).

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.1As \(A\subseteq B\) 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...

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.1That'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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