Let S be a nonempty bounded subset of R. Prove that sup S is unique.

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mlehuzzah | Student, Graduate | (Level 1) Associate Educator

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The supremum is the least upper bound.  That is, it is an upper bound (so everything in S is less than or equal to it) AND it is the LEAST upper bound (so any other upper bound is larger).


We know that S is bounded, so there is some M such that every element of S is less than or equal to M.  So an upper bound exists.  Now, the sup is the smallest of all the upper bounds.


Suppose we had two sups (by contradiction).  That is suppose M and N were both the sup of S.  One of them is the smaller of the two (if they are the same number, then we don't actually have two sups).  Call the smaller one M and the larger one N.  If both are supremums of S, then they are both upper bounds of S.  But now we have a problem:  N is a sup, but there is an upper bound (namely M) that is smaller than N.  If N were a sup, it can't have any upper bounds that are smaller.  So N can't really be a sup of S, and we have a contradiction.


Thus, we must only have one supremum.  So the supremum of S is unique.