Recall, the Archimedean property states that if and is arbitrary, then there exists an integer such that .
Further, recall that the least upper bound axiom states that every nonempty set of real numbers which is bounded above has a supremum.
Now, prove that satisfies the Archimedean property, but not the least-upper-bound axiom.
First, we prove that satisfies the Archimedean property.
Proof. This is immediate since if are arbitrary elements in with , then they are also in since . Thus, by the Archimedean property in we know there is an such that . Hence, the Archimedean property is satisfied in
Next, we prove that does not satisfy the least-upper-bound axiom.
Then is non-empty since . Further, is bounded above by 4 since . (Of course, there are better upper bounds available, but we just need any upper bound.)
Now, we must show that has no supremum in to show that the least-upper-bound property fails in (since this will mean is a nonempty set which is bounded above, but fails to have a least upper bound in ).
Suppose otherwise, say with . We know (I.3.12, Exercise #11). Thus, by the trichotomy law we must have either or .
Case 1: If , then there exists such that (since the rationals are dense in the reals, see I.3.12, Exercise #6). But then, (since ) and implies with , contradicting that is an upper bound for . Hence, we cannot have .
Case 2: If , then there exists such that . But then, , so if we have ; hence, is an upper bound for which is less than . This contradicts that is the least upper bound of . Hence, we also cannot have .
Thus, there can be no such (since by the trichotomy exactly one of must hold, but we have shown these all lead to contradictions).
Hence, does not have the least-upper-bound property