Does the supremum of the set $S=\{x\in\mathbb{Q}: x>2\}\subset \mathbb{R}$ exist?

Question

Does the supremum of the set $S=\{x\in\mathbb{Q}: x>2\}\subset \mathbb{R}$ exist? If not, prove it.

I don't think the supremum exist, but I don't know how to prove it.

Answer 1

A supremum does not exist in this set. (With the standard ordering and operations for $\mathbb{R}$)

Proof:

1. Suppose a Supremum exists, call it M.
2. M+1>M and $M+1\in\mathbb{Q}: x>2$

Therefore M was not the supernum

If an unnatural order relation exists, then you could make it so that there is a supernum as you could define that 42>x for all other x in R (the hitchhiker's guide metric), then since 42>43 in this case the proof is invalid. But in this case you're really just defining the order relation to have a supernum.