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:

$\displaystyle (i) : (\forall x \in A+B) x \le (q+p) $ - this says that (q+p) is an upper bound

$\displaystyle (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:

$\displaystyle

(\forall x \in A) p \ge x

$ ( by definition )

$\displaystyle

(\forall y \in B) q \ge y

$ ( by definition )

$\displaystyle

\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)

$\displaystyle

(\forall t1 < p)(\exists x \in A) x > t1

$ (By definition)

$\displaystyle

(\forall t2 < q)(\exists y \in B) y > t2

$ (By definition)

$\displaystyle

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

$ ( note: t1 < p t2<q )

$\displaystyle

\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.