Show that if a subset of an ordered field has a least upper bound, then any two least upper bounds for it have to coincide Show that if a subset of an ordered field has a least upper bound, then any two least upper bounds for it have to coincide Should I prove by contradiction then turn my proof into a direct proof using contrapositives? That seems unnecessarily complicated..

Hint:

1. Least upper bound is also an upper bound.
2. If $M$ is an upper bounded, then least upper bounded $\leq M$.



Let $s_1,s_2$ are two least upper bounds of $S$, then derive $s_1\leq s_2$ and $s_2\leq s_1$