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

Proof. This is immediate since if









Next, we prove that does not satisfy the least-upper-bound axiom.
Proof. Let
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