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 $$\sup A\leq \sup B$$

1. amilapsn

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
3. amilapsn

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

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