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For ultrafilters $\mathcal{U},\mathcal{V}\in \beta \mathbb{N}$, define $\mathcal{U}+\mathcal{V}$ byThis operation is associative, and each right translation $\mathcal{U}\mapsto \mathcal{U}+\mathcal{V}$ is continuous, so $\beta \mathbb{N}$ is a compact Hausdorff space with an associative operation continuous in its first variable for each fixed second variable.

A Banach algebra is an associative algebra over $\mathbb{R}$ or $\mathbb{C}$ whose norm makes it a Banach space and satisfies submultiplicativity:An identity is an additional assumption; when present, its normalization $\Vert 1\Vert =1$ should be stated if needed. Submultiplicativity makes multiplication continuous, and completeness makes norm-controlled infinite algebraic operations possible.^[ref]

Examples include bounded linear operators on a Banach space with composition and the operator norm, and continuous complex-valued functions on a compact Hausdorff space with pointwise multiplication and the supremum norm. The former is generally noncommutative, whereas the latter is commutative. A closed subalgebra of a Banach algebra inherits both the product estimate and completeness.

For $\Vert a\Vert <1$, the powers satisfy ${\sum }_n\Vert a^n\Vert <\infty$, so the Neumann series constructs $(1-a{)}^{-1}$. This simple estimate connects analysis to invertibility and leads to the spectrum of an element and holomorphic functional calculus.

A Banach star-algebra is a complex Banach algebra equipped with a continuous algebra involution. Thus the involution is conjugate-linear, reverses products, and squares to the identity. Some conventions additionally require $\Vert a^{\ast }\Vert =\Vert a\Vert$; this is the isometric convention. In the unital case the involution fixes the unit. The C-star identity is a stronger additional condition. The algebra of continuous complex-valued functions on a compact Hausdorff space, with pointwise operations, complex conjugation and the supremum norm, satisfies both the isometric condition and the C-star identity.

Let continuous maps $f,g:X\rightrightarrows Y$ between compact Hausdorff spaces have an underlying split coequalizer $q:Y\to Z$ with set maps $s,t$ satisfying $qs=1$, $ft=1$ and $gt=sq$. Its equivalence relation is the image of the closed subset $\{(x,x^{\prime })\in X^2:g(x)=g(x^{\prime })\}$ under $(f,f)$. One inclusion follows from $qf=qg$; for the other use $x=t(y)$ and $x^{\prime }=t(y^{\prime })$ whenever $q(y)=q(y^{\prime })$. This image is compact and hence closed in the Hausdorff product $Y^2$. The quotient is therefore compact Hausdorff, giving the preserved coequalizer required by the Beck monadicity theorem.

A compact group is a topological group whose underlying topological space is a compact Hausdorff space. The circle group is a compact group: the unit circle is a closed bounded subset of the complex plane, so the Heine-Borel theorem gives compactness. The additive group $(\mathbb{R},+)$ is a locally compact group and a Hausdorff space, but is not a compact group: the open cover $\{(-n,n):n\ge 1\}$ has no finite subcover. A finite group with the discrete topology is another compact group; compactness does not require a smooth manifold structure.

The great advantage of a compact group is that one can average over every element without choosing a preferred origin. Precisely, every locally compact group that is a Hausdorff space admits a nonzero left Haar measure $\mu$, a regular Borel measure finite on compact sets, unique up to a positive scalar. Left invariance means $\mu (gA)=\mu (A)$ for every Borel set $A$ and every $g\in G$. There is also a right Haar measure; in general left and right Haar measures can differ. On a compact group, $0<\mu (G)<\infty$, and normalizing to $\mu (G)=1$ gives a unique regular Borel probability measure invariant under both left and right translation. See Tornier, Haar Measures, Theorem 2.2 for existence and uniqueness, and Ram's notes on left and right Haar measure for the distinction.

For a finite group $G$, normalized Haar measure assigns each element mass $1/|G|$, so integration is simply $|G{|}^{-1}{\sum }_{g\in G}f(g)$. For the circle group, the corresponding integral is uniform angle averaging, computed below. On $(\mathbb{R},+)$, Lebesgue measure is a Haar measure, but its total mass is infinite: no positive scalar multiple makes it a probability measure. Thus the existence of Haar measure does not make every topological group a compact group.

Historically, Alfréd Haar introduced the general invariant-measure construction in Der Massbegriff in der Theorie der kontinuierlichen Gruppen, Annals of Mathematics 34 (1933), pp. 147-169. This bibliographic identification is also recorded in André Weil's archived seminar notes, Mesure de Haar, reference Haa33. The conceptual step is to let group symmetry itself determine the weights used in integration, rather than relying on coordinates in an ambient Euclidean space.

Let $X$ be a compact Hausdorff space and $R\subseteq X\times X$ a closed equivalence relation. The quotient space $X/R$ is compact and Hausdorff. Indeed, the saturation $R[C]$ of any closed $C\subseteq X$ is the projection of the compact set $R\cap (C\times X)$, so is closed. Thus the quotient map is a closed map.

Two different equivalence classes are disjoint compact subsets of $X$ and have disjoint open neighbourhoods $U,V$. To see the latter assertion, separate each pair of points by disjoint neighbourhoods and use compactness twice to extract finite covers. Now $U^{\prime }=X\setminus R[X\setminus U]$ and $V^{\prime }=X\setminus R[X\setminus V]$ are open saturated subsets containing the respective classes, and lie inside $U,V$. Their quotient images are disjoint open neighbourhoods, proving the Hausdorff condition. Compactness follows from the continuous quotient map. This applies to finite gluing along closed boundary pieces when the generated equivalence relation is closed; that last condition must be checked.

The inclusion of compact Hausdorff spaces in topological spaces has a left adjoint. Its unit $X\to RX$ is universal among continuous functions from $X$ to compact Hausdorff spaces and has dense image. Completeness follows from products in a category and closed equalizers. For any map $X\to K$, its closed image has a dense subset of cardinality at most $|X|$ and hence cardinality at most $2^{2^{|X|}}$. Representatives of all such compact Hausdorff spaces and maps establish the solution-set condition, so the general adjoint functor theorem gives the reflection. The unit need not be injective or a topological embedding for an arbitrary $X$.

For $A\subseteq \mathbb{N}$, the basic set $\hat{A}=\{U:A\in U\}$ in the Stone-Čech compactification of the natural numbers is a clopen set, since its complement is $\widehat{\mathbb{N}\setminus A}$. Distinct ultrafilters are separated by complementary basic sets. A basic cover with no finite subcover would give complements with the finite intersection property, whose generated filter on a set extends to an ultrafilter missing the entire cover. Thus the space is a compact Hausdorff space.

Let $K$ be a compact Hausdorff space. A test subspace $H\subset C(K,\mathbb{R})$ containing a strictly positive function is sufficient for uniform convergence of positive linear operators on continuous functions if, for each $x\in K$, the only positive linear functional agreeing with point evaluation on $H$ is that evaluation itself. By the Riesz-Markov-Kakutani representation theorem, this is uniqueness of the positive measure with the prescribed moments. Boundedness from the positive test function and the Banach-Alaoglu theorem prove the assertion by compactness. When $1\in H$, nonnegative tests with a unique zero at each prescribed $x$ give a convenient sufficient condition.

Let $X$ be a compact Hausdorff space and let $K_i$ be compact subsets such that all finite intersections are nonempty and connected. Their total intersection $K$ is nonempty by the finite intersection characterization of compactness, and compact because it is closed in $X$.

If $K$ were not connected, its separation of a topological space would divide it into two nonempty compact subsets. Separation of disjoint compact subsets gives disjoint ambient open sets $U,V$ containing the two pieces. The family consisting of $X\setminus (U\cup V)$ and all $K_i$ has empty total intersection, so some finite subfamily has empty intersection. Hence a finite intersection $K_J$ of the $K_i$ lies in $U\cup V$. Since $K\subseteq K_J$ meets both open sets, they disconnect $K_J$, a contradiction. The hypothesis on all finite intersections is stronger than requiring each individual $K_i$ to be connected.

Let $X,Y$ be topological spaces, let $X$ be compact, and let $f:X\to Y$ be a continuous map. Then its image $f(X)$ is compact with the subspace topology inherited from $Y$. More generally, a continuous map sends every compact subset $K\subseteq X$ to a compact subset $f(K)\subseteq Y$, by applying the same statement to its restriction to $K$. Neither $X$ nor $Y$ needs to be Hausdorff.^[ref]

For example, the polynomial function $f(x)=x^2$ sends the compact closed interval $[0,1]$ onto $[0,1]$. The open intervals $V_1=(-1/10,16/25)$ and $V_2=(9/25,11/10)$ give an open cover of the image. Their inverse images in the domain areBoth are open sets in the subspace topology of $[0,1]$, even though neither is an open set in the ambient real line. They cover the domain, so the corresponding two target open sets cover the image.

Matching colors identify the target open sets and their inverse images. Filled endpoints are included and hollow endpoints excluded. This particular open cover illustrates the mechanism; compactness requires a finite subcover for every open cover.

For the general proof, write an arbitrary open cover of $f(X)$ as $\{f(X)\cap V_i:i\in I\}$ with each $V_i$ an open set in $Y$. By continuity, the inverse images $f^{-1}(V_i)$ form an open cover of $X$. Its compactness provides indices $i_1,\dots ,i_m$ such thatThis gives the required finite subcover of $f(X)$. The argument uses inverse images because continuous maps need not send open sets to open sets. For the empty set, the empty selection suffices.

The image of a compact Hausdorff space under a continuous map need not itself be Hausdorff. Take $D=\{0,1\}$ with the discrete topology and $S=\{0,1\}$ with the Sierpiński space topology $\{\varnothing ,\{1\},S\}$. The identity map $D\to S$ is a continuous bijection, since the inverse image of every target open set is an open set in $D$. Its image is compact, but the two points of $S$ have no disjoint open neighborhoods. If the target $Y$ is Hausdorff, its subspace topology does make $f(X)$ Hausdorff, and the compact subset of a Hausdorff space theorem also makes $f(X)$ closed in $Y$.

The practical payoff is existence: the extreme value theorem uses the compact image of a nonempty compact space under a real-valued continuous function to obtain an attained minimum and maximum. When $f$ is also a bijection onto a Hausdorff target, the compact-to-Hausdorff continuous bijection theorem supplies continuity of its inverse without a separate formula.

Every nonempty compact Hausdorff space carrying a left-topological semigroup structure contains an idempotent element of a semigroup.

Choose an inclusion-minimal nonempty compact subsemigroup $K$, using Zorn's lemma and the finite intersection property. For $a\in K$, the set $aK$ is a nonempty compact subsemigroup, so minimality gives $aK=K$. Hence $F=\{x\in K:ax=a\}$ is nonempty. It is a compact closed subset of $K$ by continuity and the Hausdorff space assumption, and is a subsemigroup because $a(xy)=(ax)y=ay=a$ for $x,y\in F$. Minimality gives $F=K$, so $a^2=a$.^[ref]

Reversing multiplication gives the same conclusion for a right-topological semigroup. Joint continuity of multiplication is not needed.

For complex $C(K)$ on a compact Hausdorff space, the extreme points of its dual unit ball are precisely unimodular multiples of Dirac measures. A measure whose variation measure has mass on two disjoint sets splits as a nontrivial convex combination of normalized restrictions and is not extreme. This description is the key to the Banach–Stone theorem.

The kification $kX$ of a topological space $X$ has the same points and the finest topology making every original continuous map $K\to X$ from a compact Hausdorff space continuous. Equivalently, $A\subseteq kX$ is closed exactly when $f^{-1}(A)$ is closed in $K$ for every such test map. This topology is at least as fine as the original one, so the identity $kX\to X$ is continuous.^[ref]

If $X$ is already a compactly generated space, its topology is recovered unchanged. In particular this holds for metric spaces. The operation is idempotent: $k(kX)=kX$. It lets one form products and mapping spaces within compactly generated spaces: the categorical product is the kification of the ordinary product. Compact test maps retain their original continuity, while the new topology supplies the closed sets they detect.

Let $K$ be a compact Hausdorff space, $U$ a real linear subspace of $C(K,\mathbb{R})$, $f\in C(K,\mathbb{R})$ and $u\in U$. Put $e=f-u$ and $E=\{x:|e(x)|=\Vert e{\Vert }_{\infty }\}$. Then $u$ is a best uniform approximation from $U$ precisely when every $v\in U$ has some $x\in E$ with $e(x)v(x)\le 0$.^[ref]

If a direction $v$ has strictly matching error signs everywhere on $E$, compactness makes $u+tv$ a uniformly better approximant for sufficiently small $t>0$. Conversely, for any candidate $u+v$, an active point with $e(x)v(x)\le 0$ has $|e(x)-v(x)|\ge |e(x)|$, preventing improvement. The linear subspace hypothesis ensures that differences of candidates are admissible directions; for a general nonlinear family, the quantified direction set would have to change. This criterion yields the Chebyshev alternation theorem.

For a commutative complex unital Banach algebra, maximal ideals correspond bijectively to characters of an algebra by taking kernels. Give the maximal ideal space the Gelfand topology through this correspondence. It is a compact Hausdorff space, since the characters form a weak-star closed subset of the dual unit ball, which is compact by the Banach-Alaoglu theorem.

A nonempty topological dynamical system $(Y,T)$ on a compact Hausdorff space is minimal if it has no proper nonempty closed forward-invariant subset. Equivalently, every forward orbit is dense in $Y$, because an orbit closure is closed and forward invariant. Minimality implies $T(Y)=Y$.

A minimal subsystem is a nonempty closed forward-invariant subset on which the restricted dynamical system is a minimal dynamical system. Every continuous map on a nonempty compact Hausdorff space has one: intersections of chains of nonempty closed invariant sets remain nonempty by compactness, so the Zorn lemma gives a minimal member.

For $x$ in a compact Hausdorff space $K$, its quasi-component $Q$ is the intersection of all clopen sets containing $x$. It is closed and therefore compact. Suppose $Q=A\bigsqcup B$ is a separation of a topological space into two nonempty relatively closed sets, with $x\in A$. The sets $A,B$ are compact and hence closed in $K$. Normality of a compact Hausdorff space gives disjoint open neighborhoods $U,V$ of $A,B$.

The compact set $K\setminus (U\cup V)$ is disjoint from $Q$. For each point $y$ of that set choose a clopen $C_y$ containing $x$ but not $y$. Finitely many complements cover $K\setminus (U\cup V)$, so their corresponding intersection $C$ is clopen, contains $Q$, and is contained in $U\cup V$. Now $C\cap U$ is clopen in $C$, hence clopen in $K$, and contains $x$ while missing $B$. This contradicts the definition of $Q$. Thus $Q$ is connected. Any connected set containing $x$ lies in every clopen neighborhood of $x$, so the connected component of $x$ equals $Q$.

For the second assertion, $D$ is closed in the ambient Hausdorff space $X$, and is itself compact Hausdorff. Let $C$ be a connected component of $D$ disjoint from ${\partial }_{X}D$. Choose $p\in C$. By the first assertion $C$ is the quasi-component of $p$ in $D$. For each $y\in {\partial }_{X}D$ choose a clopen subset of $D$ containing $p$ but not $y$. Compactness of ${\partial }_{X}D\subseteq D$ gives a finite intersection $H$ of these sets which contains $C$ and avoids the whole boundary. Thus $H\subseteq {\mathrm{int}}_{X}D$. It is closed in $X$, because it is closed in the closed set $D$, and is open in $X$, because it is relatively open in $D$ and contained in its interior. It is nonempty and proper, contradicting connectedness of $X$. Hence every connected component of $D$ meets its boundary.

Finally suppose a closed subset $A$ of the square has the point-separating clopen property specified in the paper. Every nonempty relatively open subset of the square contains a nontrivial line segment: it contains an interior point of the square and a small ball about that point. If such a segment lay in $A$, a clopen subset of $A$ separating its endpoints would induce a separation of a topological space of the connected segment. This is impossible. Therefore $A$ contains no nonempty relatively open subset of the square. Being closed, it is nowhere dense in the square, and also nowhere dense in ${\mathbb{R}}^2$.

Take the natural numbers as the dense discrete subspace of their Stone-Čech compactification $\beta \mathbb{N}$. The ambient compact Hausdorff space is normal, and the Stone-Čech extension property makes this subspace $C^b$-embedded. It is not $C$-embedded: the continuous function $n\mapsto n$ is unbounded, whereas every continuous real-valued function on a compact space is bounded. The subspace is not closed, so this does not contradict the preceding closed-subspace theorem.