Jamin2112 said:
So if I start with the assumption (rewritten the correct way)
"Suppose for all integers n, an≤b for some integers a and b."
This is also not right. Perhaps it's easier to see if you write it in symbols. Abbreviate the set of positive reals as \mathbb{R}_{>0}.
Your first attempt at the hypothesis was: \forall a \in \mathbb{R}_{>0}.\, \forall b \in \mathbb{R}_{>0}. \, \forall n \in \mathbb{N}.\, an \leq b.
Your second attempt translates to: \forall n \in \mathbb{N}. \, \exists a \in \mathbb{R}_{>0}. \, \exists b \in \mathbb{R}_{>0}. \, an \leq b. You can tell this is not the right hypothesis, because it's
true: for any natural number n > 0, one may choose a = \frac1{2n}, b = 1, and then an = \frac12 < 1 = b. (For n = 0 just choose a = b = 1.)
The correct hypothesis is the negation of the archimedean property. The archimedean property is \forall a \in \mathbb{R}_{>0}. \, \forall b \in \mathbb{R}_{>0}. \, \exists n \in \mathbb{N}. \, an > b, so its negation is \exists a \in \mathbb{R}_{>0}. \, \exists b \in \mathbb{R}_{>0}. \, \forall n \in \mathbb{N}. \, an \leq b. If you don't know how to negate statements with strings of quantifiers like this, you need to learn before you go much further in analysis.
Jamin2112 said:
Looking at the definitions, I see that any M≥b is an upper bound for the set S={na for all integers n and some positive real number a}. The least upper bound, of course, would be b.
Right?
No. Just because b is the smallest upper bound you see in front of your face does not mean it is actually the least upper bound. You must
prove that any
other upper bound is at least b, and you can't do that from the hypotheses at hand.