Thread: 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 = 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 is an upper bound of A and < . By the Archimedian property of the rational numbers, there is an x= p/q Q such that



Since we can assume both p and q are positive integers we see that x < 25 implies that p < 25q. Therefore x A and x > . Therefore is not an upper bound. Therefore, if is an upper bound of A, then . Thus = 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 is an upper bound to A, it is by definition greater than all x in A?



A very confused mathematician.

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

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 whilst the condition that is an upper bound, so cannot exist?

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 whilst the condition that is an upper bound, so cannot exist?

Yes, that is what I said.