Theorem 3

The upper limit of a real-valued function $f$ with respect to a filter $\mathfrak{G}$ is equal to the largest cluster value of $f$ with respect to $\mathfrak{G}$.

Let $b$ be a cluster point of $f$ with respect to $\mathfrak{G}$. For each $M \in \mathfrak{G}$, $b$ lies in the closure of $f(M)$, hence $b \leq \sup_{x \in M} f(x)$, and therefore, by (12), $b \leq \lim \sup_{\mathfrak{G}} f = a.$

On the other hand, let $V$ be any open neighbourhood of $a$ in $\overline{\mathbf{R}}$. Then there exists a set $M_0$ in $\mathfrak{G}$ such that, for each $M \in \mathfrak{G}$ contained in $M_0$, we have $\sup_{x \in M} f(x) \in V$; since $V$ is open it follows that $f(M)$ meets $V$, and therefore $a$ is a cluster point of $f$ with respect to $\mathfrak{G}$, and the proof is complete.