Proposition 5

Let $f$ be a real-valued function defined on a set $X$. On the set $\mathfrak{F}(X)$ of all finite subsets of $X$, directed with respect to the relation $\subset$, the real-valued function $H \to \sup_{x \in H} f(x)$ is increasing, the real-valued function $H \to \inf_{x \in H} f(x)$ is decreasing, and we have

Let $\varphi(H) = \sup_{x \in H} f(x)$. Clearly $\varphi$ is increasing, and therefore has a limit $a$ (no. 2, Theorem 2); and since $\varphi(H) \leq \sup_{x \in A} f(x)$ for all $H$, we have $a \leq \sup_{x \in A} f(x)$ (no. 2, Theorem 1). If we had $a < \sup_{x \in A} f(x)$, then there would exist $x_0 \in X$ such that $a < f(x_0)$; but this is a contradiction since $\varphi(H) \geq f(x_0)$ whenever $x_0 \in H$.

In particular, by (1), if $A$ is any non-empty subset of $\overline{\mathbf{R}}$, we have