1. ## Supremum

Let A,B be subsets of an ordered field F
Let A+B denote the set {(x+y)|x is in A, y is in B)}
Let q=sup A, p=sup B
Show that supA+B = q+p

If this were a test question (which it isn't!), and I turned the following in, what grade do you think would I receive and why?.

Here is what I need to show:
$(i) : (\forall x \in A+B) x \le (q+p)$ - this says that (q+p) is an upper bound
$(ii) : (\forall t < (q+p))(\exists x \in A+B) x>t$ - this says that nothing less than (q+p) is an upper bound

Here is the proof:
$
(\forall x \in A) p \ge x
$
( by definition )
$
(\forall y \in B) q \ge y
$
( by definition )

$
\implies (\forall x \in A, y \in B) p+q \ge x+y \implies (\forall (x+y) \in A+B) p+q \ge (x+y)
$

(so (i) is true)

$
(\forall t1 < p)(\exists x \in A) x > t1
$
(By definition)
$
(\forall t2 < q)(\exists y \in B) y > t2
$
(By definition)
$
\implies (\forall t=t1+t2 < p+q) (\exists x \in A, y\in B) x+y > t1+t2=t
$
( note: t1 < p t2<q )
$
\implies (\exists (x+y) \in A+B) (x+y) > t
$

So (ii) is true
Therefore sup A+B = q+p

Thanks :-)
I am trying to go thru a book on my own before school starts, but I am always worried because there is no answer key.

2. Originally Posted by billa
Let A,B be subsets of an ordered field F
Let A+B denote the set {(x+y)|x is in A, y is in B)}
Let q=sup A, p=sup B
Show that supA+B = q+p
It should be clear to you that $\sup \left( {A + B} \right) \leqslant q + p$.

So suppose that $\sup \left( {A + B} \right) < q + p$.
Then let $\varepsilon = \frac{{q + p - \sup \left( {A + B} \right)}}{2}$.
Then
$\left( {\exists a' \in A} \right)\left[ {q - \varepsilon < a' \leqslant q} \right]\;\& \,\left( {\exists b' \in B} \right)\left[ {p - \varepsilon < b' \leqslant p} \right]$

Now there is a contradiction lurking in all that.
Can you find it?

3. That was a cool and short way of doing it

When you add the two inequalities you get sup A+B < sup A+B

BUT, since you demonstrated that you clearly understand the problem, can you read my proof and tell me if it is right. I had no doubt that what i was trying to prove was actually true, but I am not so sure that my proof is right.

Thanks

4. Originally Posted by billa
BUT, since you demonstrated that you clearly understand the problem, can you read my proof and tell me if it is right. I had no doubt that what i was trying to prove was actually true, but I am not so sure that my proof is right.
Actually no. I fear that I connot follow what you have done.
Sorry.

5. ## question about the field

Don't you need some sort of assumption on the completeness of the field to be sure this exists? Or at least that it is a linear continuum or something?

I think your proof looks okay provided your field is nice enough. you have the inequalities backwards in those two lines under (so (i) is true). Also I think you need to be a little more careful with your proof writing. I am assuming $t1\in A$ and $t2\in B$. When you say let $t=t1+t2< p+q$ you are only sure that at least one of t1 or t2 is less than p or q respectively. It could be the case that t1=p and t2<q. But the main idea of the proof looks okay to me.

I am just slightly concerned about the lack of assumptions on the field itself, like how do you know for sure that if you have $t_1 for $t_1,p \in F$ that there exists some $f\in F$ such that $t_1 < f < p$?

6. Thanks for the reply :-)
I see what you mean about being more clear on t1 and t2, and I cannot believe i put the inequalities backwards - I read over that post many times lol

Originally Posted by Gamma
Don't you need some sort of assumption on the completeness of the field to be sure this exists? Or at least that it is a linear continuum or something?

I am just slightly concerned about the lack of assumptions on the field itself, like how do you know for sure that if you have $t_1 for $t_1,p \in F$ that there exists some $f\in F$ such that $t_1 < f < p$?
I don't think so, here is why (warning:this could be quite wrong because I am very tired)
First, $t_1 < f \le p$ and $f \in A$
If $f$ didn't exist, then t1 would be an upper bound on A, and p wouldn't be the least upper bound

7. p could be in the set A too though, so $t_1$ is not an upper bound either. for instance $sup[0,1]=1\in [0,1]$.

I am just kind of worried about like what if there were only two elements in A or something, you might not be able to find an element between the sup and your t1.

I am not sure exactly if there are ordered fields out there like that, all finite fields in the algebraic sense are of the form $\mathbb{Z}_{p^n}$ for a prime p in terms of isomorphism class, but I have no clue what kinds of order topologies could be put on them. I am not sure how one would put a proper ordering on fields like that where something like that could happen.

I am just thinking the field needs to be a linear continuum at least for these two proofs that have been supplied to go through. I am pretty sure the statement has to still be correct; however I am not sure how to prove it without sort of constructing a new element that I cannot be sure necessarily exists.

8. Originally Posted by Gamma
p could be in the set A too though, so $t_1$ is not an upper bound either. for instance $sup[0,1]=1\in [0,1]$.

I am just kind of worried about like what if there were only two elements in A or something, you might not be able to find an element between the sup and your t1.
If $p \in A$, then can't $f = p$ ?

I think that if you can't find an $x \in A$ between t1 and $sup A$ then $sup A$ isn't the least upper bound.

In the set {0,1} in an ordered Field, I think you can always pick 1 for x, and no matter what your value of $p>t1 \implies 1=x>t1$

I am not sure exactly if there are ordered fields out there like that, all finite fields in the algebraic sense are of the form for a prime p in terms of isomorphism class, but I have no clue what kinds of order topologies could be put on them. I am not sure how one would put a proper ordering on fields like that where something like that could happen.
I don't know what any of these things are

9. ## proof

I think it may be best to do this one by contradiction. Try this on for size.

Say suppose there exists $s=s_1+s_2 \in A+B$ which is an upper bound and $s=s_1 + s_20$ which means at least one of the things in parentheses is positive so without loss of generality suppose $0 or in other words, $s_1 and $s_1 \in A$. But then by your definition of LUB, there exists $s_1'\in A$ such that $s_1.

Now we consider the point $s_1' + s_2 \in A +B$ and see that $s_1'+s_2>s_1+s_2=s$ contradicting the fact that $s$ was in fact an upper bound for A+B.