Heres a question from a Foundations of Calculus past paper and i was wondering whether the answer given is really needed...

Q: Prove That Sup A = 10, where $\displaystyle A = (-\infty ,10)$

A: Since A is bounded, $\displaystyle \alpha$ = sup A exists by the least-upper bound axiom.

Since $\displaystyle \gamma = 10$ is an upper bound for A, $\displaystyle \alpha \leq 10$ by property 2 (Properties at end of question).

We want to show that $\displaystyle \alpha$ cannot be less than 10, which will imply $\displaystyle \alpha = 10$

Assume that $\displaystyle \alpha < 10$.

Let $\displaystyle x = \frac{1}{2}(10 + \alpha)$.

Then $\displaystyle x < \frac{1}{2}(10 + 10) = 10$ so $\displaystyle x \in A$.

On the other hand $\displaystyle \alpha = \frac{1}{2}(\alpha + \alpha) < \frac{1}{2}(10 + \alpha) = x$.

Therefore, $\displaystyle \alpha$ is not an upper bound for A, which contradicts property 1. Absurd.

Therefore, $\displaystyle \alpha \geq 10$ so $\displaystyle \alpha = 10$.

Properties: A real number $\displaystyle \alpha$ is the supremum of a non-empty set of real numbers A if and only if...

1; $\displaystyle x \leq \alpha$ for all $\displaystyle x \in A$; and

2; if $\displaystyle x \leq \gamma$ for all $\displaystyle x \in A$, then $\displaystyle \alpha \leq \gamma$

And heres my thought... Cant you just use the Archimedian Property of Rational Numbers to say that...

Since A is bounded, $\displaystyle \alpha$ = sup A exists by the least-upper bound axiom.

Since $\displaystyle \gamma = 10$ is an upper bound for A, $\displaystyle \alpha \leq 10$ by property 2.

We want to show that $\displaystyle \alpha$ cannot be less than 10, which will imply $\displaystyle \alpha = 10$

Assume that $\displaystyle \alpha < 10$.

By the A.P.O.R.N there exists an x s.t. $\displaystyle \alpha < x < \gamma$. Therefore, $\displaystyle \alpha$ is not an upper bound. Absurd... etc...