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

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

Since is an upper bound for A, by property 2 (Properties at end of question).

We want to show that cannot be less than 10, which will imply

Assume that .

Let .

Then so .

On the other hand .

Therefore, is not an upper bound for A, which contradicts property 1. Absurd.

Therefore, so .

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

1; for all ; and

2; if for all , then

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

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

Since is an upper bound for A, by property 2.

We want to show that cannot be less than 10, which will imply

Assume that .

By the A.P.O.R.N there exists an x s.t. . Therefore, is not an upper bound. Absurd... etc...