Corollary

Let $\mathfrak{F}_1$ be a filter on a set $X_1$, and $\mathfrak{F}_2$ a filter on a set $X_2$; let $\mathfrak{F}_1 \times \mathfrak{F}_2$ be the product filter (§ 6, no. 7) on $X = X_1 \times X_2$, and let f be a mapping of X into a regular space Y. Suppose that a) $\lim_{\mathfrak{F}_1 \times \mathfrak{F}_2} f$ exists.

b) $\lim_{x_2, \mathfrak{F}_2} f(x_1, x_2) = g(x_1)$ exists for all $x_1 \in X_1$.

Then $\lim_{x_1, \mathfrak{F}_1} g(x_1)$ exists and is equal to $\lim_{\mathfrak{F}_1 \times \mathfrak{F}_2} f$.

Let $X'_1 = X_1 \cup \{\omega_1\}$ (resp. $X'_2 = X_2 \cup \{\omega_2\}$) be the topological space associated with the filter $\mathfrak{F}_1$ (resp. $\mathfrak{F}_2$) (§ 6, no. 5, Example). In the product space $X' = X'_1 \times X'_2$ let $X''$ be the union of the subspaces $X_1 \times X'_2$ and $\{(\omega_1, \omega_2)\}$. $X$ is clearly a dense subspace of $X''$, and the hypotheses imply that $f(y_1, y_2)$ tends to a limit when $(y_1, y_2)$ tend to any point $(x_1, x_2)$ of $X''$ whilst remaining in $X$. The existence of the extension of $f$ by continuity to $X''$ then follows from Theorem 1. Since also $(\omega_1, \omega_2)$ lies in the closure of $X_1 \times \{\omega_2\}$ relative to $X''$, the result follows immediately (§ 7, no. 5).