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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..

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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 $s_2\leq s_1$

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