Corollary 1

Let $X, Y$ be two topological spaces, $f$ a mapping of $X$ into $Y$ which is continuous at a point $a \in X$; then, for every filter base $\mathfrak{B}$ on $X$ which converges to $a$, the filter base $f(\mathfrak{B})$ converges to $f(a)$. Conversely if, for every ultrafilter $\mathfrak{U}$ on $X$ which converges to $a$, the ultrafilter base $f(\mathfrak{U})$ converges to $f(a)$, then $f$ is continuous at $a$.

The first assertion is an immediate consequence of Proposition 9. To prove the second, suppose that $f$ is not continuous at $a$; then there is a neighbourhood W of $f(a)$ in Y such that $\overline{f}^{-1}(W)$ does not belong to the filter $\mathfrak{B}$ of neighbourhoods of $a$ in X. Hence (§ 6, no. 4, Proposition 7) there is an ultrafilter $\mathfrak{U}$, finer than $\mathfrak{B}$, which does not contain $\overline{f}^{-1}(W)$ and therefore contains its complement $A = X - \overline{f}^{-1}(W)$ (§ 6, no. 4, Proposition 5); since $f(A) \cap W = \varnothing$, $f(\mathfrak{U})$ does not converge to $f(a)$.