Theorem 3

Every non-empty subset of the real line which is bounded above (resp. bounded below) has a least upper bound (resp. greatest lower bound).

Let $A$ be a non-empty subset of $\mathbf{R}$, bounded above, and let $b$ be an upper bound of $A$, so that $A \subset ]\leftarrow, b]$. For each $x \in A$ consider the set $A_x$ of numbers $\geqslant x$ which belong to $A$; the sets $A_x$ form a filter base $\mathfrak{B}$ on $\mathbf{R}$, since $A_y \subset A_x$ if $y \geqslant x$. Let $a$ be a point of $A$. For each $x \geqslant a$ belonging to $A$, $A_x$ is contained in the compact interval $[a, b]$ and thus the filter base $\mathfrak{B}$ has a cluster point $c$. Since the intervals $[x, \rightarrow[$ are closed, $c$ belongs to their intersection and therefore $c$ is an upper bound of $A$. But, on the other hand, every upper bound $z$ of $A$ is $\geqslant c$, otherwise the neighbourhood $]z, \rightarrow[$ of $c$ would not contain any point of $A$. Hence $c$ is the least upper bound of $A$.

We can argue similarly for a non-empty set $B$ bounded below, or else simply remark that $-B$ is non-empty and bounded above, and that if $c$ is the least upper bound of $-B$, then $-c$ is the greatest lower bound of $B$.

The least upper bound $c$ of $A$ can be characterized by the following two properties: (i) For each $x \in A, x \leqslant c_*$. (ii) For each $a < c$, there exists $x \in A$ such that $a < x \leqslant c$.

The least upper bound of a closed set (non-empty and bounded above) belongs to the set and is its greatest element; and the least upper bound of any non-empty subset $A$ of $\mathbf{R}$ which is bounded above may be defined as the largest real number in the closure of $A$.