Proposition 8

Let $f$ be a mapping of a set $X$ into a topological space $Y$; then $y \in Y$ is a cluster point of $f$ with respect to $\mathfrak{F}$ if and only if there is a filter $\mathfrak{G}$ on $X$ which is finer than $\mathfrak{F}$ and such that $y$ is a limit of $f$ with respect to $\mathfrak{G}$.

For if $y$ is a cluster point of $f$ with respect to $\mathfrak{F}$, and if $\mathfrak{B}$ is the neighbourhood filter of $y$, then $\overline{f}^{-1}(\mathfrak{B})$ is a filter base on $X$ since every set of $\overline{f}^{-1}(\mathfrak{B})$ meets every set of $\mathfrak{F}$ (§ 6, no. 6). This remark shows also that there is a filter $\mathfrak{G}$ on $X$ which is finer than both $\mathfrak{F}$ and the filter with base $\overline{f}^{-1}(\mathfrak{B})$ (§ 6, no. 2, Proposition 1, Corollary 2), hence that $y$ is a limit point of $f$ with respect to $\mathfrak{G}$.

Notice finally that if $f$ is a mapping of a set $X$ into a topological space $Y$, the set of cluster points of $f$ with respect to a filter $\mathfrak{F}$ on $X$ is closed in $Y$ (no. 2, Proposition 5) and possibly empty.