Question

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

Answer 1

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 \(\sup B\).

Answer 2

see
https://proofwiki.org/wiki/Supremum_of_Subset

Answer 3

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...

Answer 4

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