Remark

If in particular we have $f(x) < g(x)$ for all $x \in X$ (or only for all points of a set of the filter $\overline{\mathbf{R}}$) we can infer, by Theorem 1, that $\lim_{\mathfrak{F}} f \leq \lim_{\mathfrak{F}} g$; but we cannot infer the strong inequality $\lim_{\mathfrak{F}} f < \lim_{\mathfrak{F}} g$. For example, if we take $X$ to be the set $\mathbf{N}$ of natural numbers, filtered by the Fréchet filter, and if $f(n) = 0$ and $g(n) = 1/n$, then $f(n) < g(n)$ for all $n$, but $\lim_{n \to \infty} f(n) = \lim_{n \to \infty} f(n) = 0$. Thus we lose strictness when we pass to the limit in a strict inequality.