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A closed filter base in a compact space has nonempty intersection. If the space is Hausdorff and every base member is connected, the intersection is connected. Otherwise normality separates its two closed components by disjoint open sets. Compactness of the complement supplies finitely many base members whose intersection lies in that union; directedness then gives a single base member meeting both sides, contradicting its connectedness.

A filter base is a nonempty family of nonempty subsets directed under reverse inclusion. Its upward closure is a proper filter. Every finite intersection contains a base member, so it has the finite intersection property.

A proper filter $\mathcal{F}$ on $X$ is a nonempty family of subsets, closed under finite intersection and passage to supersets, with $\varnothing \notin \mathcal{F}$. An ultrafilter is a maximal proper filter, equivalently one which contains, for each $A\subseteq X$, exactly one of $A$ and $X\setminus A$. The pushforward filter isInverse images preserve intersections and complements. Thus it is a proper filter, and the either-a-set-or-its-complement criterion shows it is an ultrafilter whenever $\mathcal{F}$ is one.

Convergence of a filter $\mathcal{F}\to x$ means that every neighborhood of $x$ belongs to $\mathcal{F}$. A mapping is continuous exactly when $\mathcal{F}\to x$ implies $f_{\ast }\mathcal{F}\to f(x)$ for all proper filters and points. The forward implication follows from inverse images of neighborhoods. For the converse apply the criterion to the neighborhood proper filter at $x$: each inverse image of a neighborhood of $f(x)$ must be a neighborhood of $x$, which is continuity at $x$.

A topological space is compact exactly when every ultrafilter converges to at least one point; equivalently every proper filter has a cluster point. For completeness, in a compact space the closed sets $\overline{A}$, $A\in \mathcal{U}$, have the finite intersection property, hence a common point $x$. Every neighborhood $N$ of $x$ belongs to the ultrafilter. Indeed, if $N$ did not, choose an open $V$ with $x\in V\subseteq N$. Then $V$ would not belong either, so $X\setminus V$ would belong. That set is closed and excludes $x$, contradicting the choice of $x$ in every closure. Conversely, if an open cover has no finite subcover, its closed complements generate a proper filter. An ultrafilter extension cannot converge anywhere, because each point has a covering neighborhood whose complement belongs to that ultrafilter. This proves the compactness criterion using the permitted extension principle.

Now suppose $b\in \overline{f[A]}$. The familyis a filter base of nonempty sets: closure guarantees each intersection is nonempty, and finite intersections correspond to intersections of neighborhoods. Extend its generated proper filter to an ultrafilter $\mathcal{H}$. It contains $A$ and every $f^{-1}[V]$, so $f_{\ast }\mathcal{H}\to b$.

Conversely, if such an ultrafilter exists, each neighborhood $V$ of $b$ has $f^{-1}[V]\in \mathcal{H}$. Its intersection with $A\in \mathcal{H}$ is nonempty. Thus $V$ meets $f[A]$, provingThis part includes the unnumbered proper filter definitions and criteria preceding the printed subparts.

A filter base on a set $X$ is a nonempty family $\mathcal{F}$ of nonempty subsets such that for every $F,G\in \mathcal{F}$ there is $H\in \mathcal{F}$ with $H\subseteq F\cap G$. Repeated application gives a member contained in any specified finite intersection. In particular every finite intersection is nonempty, so the family has the finite intersection property. Its upward closure $\{A\subseteq X:\exists F\in \mathcal{F},F\subseteq A\}$ is the corresponding proper filter.