proof that sup(A U B) <= max{sup(a),sup(b)}

[wewfy]

I need to prove the existance, and the supremum of: intersection and union, of A,B.

Right now I don't know where to start...

 

[pka]

I need to prove the existance, and the supremum of: intersection and union, of A,B.

This is a poorly written question. We must assume that both sets of real numbers are bounded.
If \(\alpha=\sup(A)~\&~\beta=\sup(B)\) then without loss of generality let's assume \(\alpha\ge \beta\).
1) Now if \(\varepsilon>0\) then \((\exists a_1\in A)[\alpha-\varepsilon<a_1\le \alpha]\). Why is this the case?
2) But from 1) this means that \(\alpha=\sup(A\cup B)\). Why is this the case?
Please post your answers.

 

[wewfy]

A and B are both bounded sets of reals.
If [MATH]α=sup(A) [/MATH] then 1) inmediate (2nd condition of supreme)

2) If you assumed that [MATH]α≥β[/MATH], then [MATH]α> b[/MATH], for all [MATH]b[/MATH]'s that belong to B, then [MATH]α=sup(A∪B)[/MATH]
Im very sorry, don't speak english trying to do my best.

Say i have to do the same with the intersection of A and B, provided it's not empty.

 

[pka]

A and B are both bounded sets of reals.
If [MATH]α=sup(A) [/MATH] then 1) inmediate (2nd condition of supreme)

2) If you assumed that [MATH]α≥β[/MATH], then [MATH]α> b[/MATH], for all [MATH]b[/MATH]'s that belong to B, then [MATH]α=sup(A∪B)[/MATH]
Im very sorry, don't speak english trying to do my best.

Say i have to do the same with the intersection of A and B, provided it's not empty.

Well done. Just one correction \((\forall b\in B)[b\le \alpha]\).