# Thread: Proving a supremum exists.

1. ## Proving a supremum exists.

The question is:

Let A = {p/q : p,q are positive integers and p < 25q }
Prove, using the definition of the supremum, that supA = 25.

I have proved that $\alpha$ = 25, which is an upper bound. What remains to prove is that it is the least upper bound and this is done on the answer sheet by reductio ad absurdum.

The answer sheet reads "Suppose that $\gamma$ is an upper bound of A and $\gamma$ < $\alpha$. By the Archimedian property of the rational numbers, there is an x= p/q $\in$ Q such that

$\gamma < x < 25 = \alpha$

Since we can assume both p and q are positive integers we see that x < 25 implies that p < 25q. Therefore x $\in$ A and x > $\gamma$. Therefore $\gamma$ is not an upper bound. Therefore, if $\gamma$ is an upper bound of A, then $\gamma \ge \alpha$. Thus $\alpha$ = 25 is the supremum."

I draw your attention to the inequality on a line on its own. (I couldn't seem to make the symbols bold.) Surely if $\gamma$ is an upper bound to A, it is by definition greater than all x in A?

A very confused mathematician.

2. That proof is correct.
Here is what it shows. $\alpha$ is an upper bound and no number less than $\alpha$ is an upper bound, therefore $\alpha$ is the least upper bound, supremum.

3. Ah, thank you for clarifying things. I think I was approaching it from the wrong angle. Is it saying that no x can exist that will fulfil the condition $\gamma < x < 25 = \alpha$ whilst the condition that $\gamma$ is an upper bound, so $\gamma$ cannot exist?

4. Originally Posted by iamahorse
Ah, thank you for clarifying things. I think I was approaching it from the wrong angle. Is it saying that no x can exist that will fulfil the condition $\gamma < x < 25 = \alpha$ whilst the condition that $\gamma$ is an upper bound, so $\gamma$ cannot exist?
Yes, that is what I said.