Why is the Archimedean property a necessary condition, together with completeness for the Least Upper Bound property of a totally ordered field?

I am taking an introductory Analysis class in university as part of a Mathematics Minor (my major is Physics). In his course notes, our professor mentions that the supremum property of the totally ordered field ($$\mathbb{R}$$, +, $$\cdot$$, $$\leq$$) has as a consequence that $$\mathbb{R}$$ is complete, that is to say, every Cauchy sequence in $$\mathbb{R}$$ converges to some limit in $$\mathbb{R}$$. But he also mentions that the opposite is not true. The field must be an Archimedean for completeness to imply the supremum property. But I can't seem to figure out why this is the case. What am I missing here?

Because there are other complete ordered fields other than $$\Bbb R$$, which are non-archimedian; you will find an example here. In such a field, the least upper bound property does not hold.
Indeed, suppose that the field is not Archimedean; then there exists $$t$$ such that $$t>n$$, for every positive integer $$n$$ (the rationals can always be uniquely embedded in an ordered fields, so we can identify rationals with elements of the field).
Since the supremum property holds and the set of positive integers is bounded, it has a supremum $$s$$. Clearly $$1 (because $$2\le s$$), so $$s-1>0$$. By definition of supremum there exists a positive integer $$m$$ such that $$m>s-1$$; therefore $$m+1>s$$: a contradiction.