Thread: Least upper bounds of subsets

1. Least upper bounds of subsets

I'm supposed to prove that if A is a subset of B then sup(A) <or= to sup(B), provided that these least upper bounds exist.

So I started by using the definition of a least upper bound. I let x = sup(A) therefore by definition there exists a x in B such that for ever a in A, a <or= to x.
also there exists a c in B such that for every a in A, a <or= c and c >or= x.

My problem is though I'm not completely sure though whether c is necessarily in B. because then I let y=sup(B) and say because c in B, by definition y >or= c.

Then by transitivity x <or= y.

Also if there are any ideas of how this proof could be done better I'd appreciate it because this doesnt seem solid enough.

2. actually I could shorten it even more and say because x in B then by definition of y=sup(B) for every x in B y >or= x.

But then again I'm not sure I can say x is in B.

3. Originally Posted by nataliemarie
I'm supposed to prove that if A is a subset of B then sup(A) <or= to sup(B), provided that these least upper bounds exist.

So I started by using the definition of a least upper bound. I let x = sup(A) therefore by definition there exists a x in B such that for ever a in A, a <or= to x.
also there exists a c in B such that for every a in A, a <or= c and c >or= x.

My problem is though I'm not completely sure though whether c is necessarily in B. because then I let y=sup(B) and say because c in B, by definition y >or= c.

Then by transitivity x <or= y.

Also if there are any ideas of how this proof could be done better I'd appreciate it because this doesnt seem solid enough.
Problem: Let $\displaystyle A\subset B\subset\mathbb{R}$. Prove that $\displaystyle \sup \text{ }A\le\sup\text{ }B$

Proof: By definition $\displaystyle \sup\text{ }B$ is an upper bound for $\displaystyle B$, therefore $\displaystyle \forall x\in B\quad x\le \sup\text{ }B$. Also, by definition $\displaystyle A\subset B\Longleftrightarrow \bigg[\left(x\in A\right)\implies \left(x\in B\right)\bigg]$. Therefore $\displaystyle \sup\text{ }B$ is an upper bound for $\displaystyle A$. Thus, by definition, $\displaystyle \sup\text{ }A\le\sup\text{ }B$