Given a partially ordered set $S$, the maximum element of $S$, if it exists, is some $M \in S$ such that for all $n \in S$, $n \leq M$.

For example, the maximum element of the set $S_1 = \{0, e, \pi, 4\}$ of real numbers is $4$, since it is larger than every other element of the set.

Every finite subset of a totally ordered set such as the reals has a maximum. However, many infinite sets do not. The integers, $\mathbb Z$ have no maximum, since for any $n \in \mathbb Z$ we can find $m \in \mathbb Z$ such that $m > n$. (Taking $m = n + 1$ works nicely.)

A more subtle example of this phenomenon is the set $K = \left\{0, \frac 12, \frac 23, \frac 34, \frac 45, \ldots\right\} = \left\{1 - \frac 1n \mid n \in \mathbb{Z}_{> 0}\right\}$. While this set has a least upper bound 1, it has no maximum.

The previous example suggests the following formulation: if $S$ is a set contained in some larger ordered set $R$ with the least upper bound property, then $S$ has a maximum if and only if the least upper bound of $S$ is a member of $S$.

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