Proposition 2

Let $f$ and $g$ be two real-valued functions defined on a set $X$ filtered by a filter $\mathfrak{F}$. If $\lim_{\mathfrak{F}} f$ and $\lim_{\mathfrak{F}} g$ exist, and if $\lim_{\mathfrak{F}} f > \lim_{\mathfrak{F}} g$, then there is a set $A \in \mathfrak{F}$ such that $f(x) > g(x)$ for all $x \in A$.

Let $a = \lim_{\mathfrak{F}} f$, let $b = \lim_{\mathfrak{F}} g$ and let $c$ be such that $b < c < a$. The interval $]c, +\infty]$ of $\overline{\mathbf{R}}$ (resp. $]-\infty, c[$) is a neighbourhood of $a$ (resp. $b$); hence there is a set $M \in \mathfrak{F}$ (resp. a set $N \in \mathfrak{F}$) such that $f(x) > c$ for all $x \in M$ [resp. $g(x) < c$ for all $x \in N$]. The set $A = M \cap N$ belongs to $\mathfrak{F}$, and we have $f(x) > c > g(x)$ for all $x \in A$.

As a particular case of Proposition 1 we have the following theorem: