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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.

For every ultrafilter $\mathcal{H}$ on $X$, its image is an ultrafilter on the compact space $Y$, so it has a limit. That limit is unique because $Y$ is Hausdorff: distinct limits would have disjoint neighborhoods both belonging to the same proper filter. Therefore $\Lambda$ is well-defined. Since $f[X]$ is dense, every $b\in Y$ lies in $\overline{f[X]}$. Part (a), with $A=X$, supplies an ultrafilter mapping to $b$, proving surjectivity.

Suppose disjoint subsets have disjoint image closures. If $\mathcal{H}\ne \mathcal{K}$, choose $A\in \mathcal{H}$ not in $\mathcal{K}$; the ultrafilter criterion gives $X\setminus A\in \mathcal{K}$. Part (a) givesThe two closures are disjoint by assumption, so the limits differ. Thus $\Lambda$ is injective.

Conversely, suppose $\Lambda$ is injective and disjoint $A,B\subseteq X$ had a common point $b$ in their image closures. Part (a) would supply ultrafilters $\mathcal{H},\mathcal{K}$ containing $A,B$ respectively, both with image limit $b$. They must be distinct, since a proper filter cannot contain the disjoint sets $A$ and $B$ simultaneously. This contradicts injectivity. Hence

A proper filter is an upward-closed family of subsets closed under finite intersections and excluding the empty set. This excludes the improper family of all subsets. An ultrafilter is a maximal proper filter.

A point $b$ lies in $\overline{f[A]}$ exactly when some ultrafilter containing $A$ has pushforward converging to $b$. For the forward implication, the sets $A\cap f^{-1}[V]$ for neighborhoods $V$ of $b$ form a proper filter topological basis; extend to an ultrafilter. For the reverse implication each such intersection belongs to a proper filter and is nonempty.