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

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

Every finite subset of the reals (or any other totally ordered set) has a minimum. However, many infinite subsets do not. The integers, $\mathbb Z$ have no minimum, 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 $K = \left\{1, \frac 12, \frac 13, \frac 14, \frac 15, \ldots\right\} = \left\{\frac 1n \mid n \in \mathbb{Z}_{> 0}\right\}$. While this set has a greatest lower bound 0, it has no minimum.

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

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