Bounds and boundedness
Let S be a nonempty set of real numbers. A number M is an upper bound for S if x ≤ M for every x in S. A number m is a lower bound if m ≤ x for every x in S. A set with an upper bound is bounded above, with a lower bound is bounded below, and with both is a bounded set. Note that bounds need not belong to S, and a set has infinitely many of them: if M works, so does anything larger.
The maximum of S, if it exists, is an upper bound that lies *inside* S. But many sets have no maximum: the open interval (0, 1) has no largest element, because for any x < 1 there is a larger element still below 1. We need a substitute that always exists for bounded sets — and that is the supremum.
The supremum: the least upper bound
The supremum of S, written sup S, is the least upper bound: a number L such that (1) L is an upper bound of S, and (2) no smaller number is an upper bound — if M < L, then M fails to bound S. Dually, the infimum inf S is the greatest lower bound. When sup S happens to lie in S, it equals the maximum; (0, 1) has sup = 1 but no maximum.
Claim: sup of S = { 1 - 1/n : n = 1, 2, 3, ... } equals 1.
Step 1 (upper bound): for every n, 1/n > 0, so 1 - 1/n < 1.
Hence 1 is an upper bound.
Step 2 (least): fix any epsilon > 0. We must find x in S with x > 1 - epsilon.
Choose n large enough that 1/n < epsilon
(possible by the Archimedean property — guide 3).
Then 1 - 1/n > 1 - epsilon, and 1 - 1/n is in S.
By the epsilon characterization, sup S = 1.
Note 1 is NOT in S, so S has no maximum.