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An affine open cover of a scheme is an open cover by affine open subsets. Every scheme has such a cover by definition. A quasi-compact scheme has a finite affine open cover. Whether the intersections are affine is a further condition, supplied in particular by separatedness over an affine base.

An affine open subscheme is an open subset of a scheme with its restricted structure sheaf, which is itself an affine scheme. Every scheme has an open cover by such subschemes. On an affine scheme, the principal open subschemes give an open basis.

On a Stein manifold $X$, every coherent analytic sheaf $\mathcal{F}$ has $H^q(X,\mathcal{F})=0$ for $q>0$. In particular this applies to the sheaf of holomorphic functions. If every nonempty finite intersection in an open cover is Stein, these vanishing results and the acyclic cover theorem compute the corresponding sheaf cohomology from its Čech cochain complex.

A closed cover of a topological space is a family of closed sets whose union is the whole space. A finite closed cover can be used in separation and antipodal arguments: in a metric space, vanishing distance to a closed member means actual membership. Unlike an open cover, a closed cover does not automatically have a partition of unity subordinate to its members.

Every closed subset $F$ of a compact space $X$ is compact in its subspace topology. An open cover of $F$ by ambient open sets, together with $X\setminus F$, covers $X$. A finite subcover for $X$ still covers $F$ after the extra member is discarded. The result does not require $X$ to be Hausdorff.

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 $f:X\to Y$ be a continuous map that is a closed map and whose fibres are compact. Then $f^{-1}(K)$ is compact for every compact set $K\subseteq Y$, without a Hausdorff hypothesis. Given an open cover, cover each fibre by a finite union $V_y$ of its members. The set $W_y=Y\setminus f(X\setminus V_y)$ is open, contains $y$, and has $f^{-1}(W_y)\subseteq V_y$. A finite subcover of $K$ by the $W_y$ uses only finitely many original cover members and covers $f^{-1}(K)$. Empty fibres cause no exception.

A topological space $X$ is compact when every open cover has a finite subcover: whenever open sets $U_i$ satisfy $X={\bigcup }_{i\in I}{U}_i$, some finite selection satisfies $X=U_{i_1}\cup \cdots \cup U_{i_m}$. A subset $K$ of an ambient topological space is compact when its subspace topology is compact; equivalently, every open cover of $K$ by ambient open sets has a finite subcover. The word every matters: finding one finite subcover of one open cover is not enough to establish compactness. ^[ref]

The closed interval $[0,1]$ is compact by the Heine-Borel theorem. For a concrete open cover, take the three open intervals $(-0.1,0.45)$, $(0.35,0.8)$, and $(0.7,1.1)$. Their union is $(-0.1,1.1)$, so they cover every point of $[0,1]$, including both endpoints. The overlaps have lengths $0.1$ and $0.1$, leaving no gaps. Each member is necessary for this particular open cover: removing them respectively leaves $0$, $0.5$, or $1$ uncovered.

Above, hollow markers denote the excluded endpoints of the covering open intervals, while the solid endpoints belong to the covered closed interval. Below, each displayed open interval is one member of an infinite open cover of $(0,1)$; finitely many rows always leave points near zero uncovered. The rows are vertically separated for readability and all refer to the same real line.

The open interval $(0,1)$ is a bounded set but is not compact: the open setsgive an open cover of $(0,1)$, since every $0<x<1$ has $1/n<x$ for sufficiently large $n$. These are also open sets for the subspace topology on $(0,1)$. Any nonempty finite selection has a largest index $N$, so its union is just $(1/N,1)$ and misses the point $1/(N+1)$, which still belongs to $(0,1)$. Thus no finite subcover exists, however many members we select. The same family is not an open cover of $[0,1]$, since it misses both endpoints.

In a metric space, compactness is equivalent to sequential compactness: every sequence in the subset has a convergent subsequence whose limit of a sequence belongs to that subset. This equivalence does not extend to all topological spaces. For example, $x_n=1/n$ for $n\ge 2$ has limit of a sequence zero in $[0,1]$. The same sequence lies in $(0,1)$, but every subsequence still has limit of a sequence zero on the real line, so no subsequence converges to a point of $(0,1)$. Merely approaching a boundary is harmless when the required point remains available. ^[ref]

Both panels show the sequence $x_n=1/n$ for $n\ge 2$ approaching zero. In the closed interval, zero is an allowed limit of a sequence; in the open interval, the hollow marker identifies the missing point. This single sequence disproves sequential compactness of $(0,1)$; compactness of $[0,1]$ still requires control of every open cover or every sequence, supplied by the Heine-Borel theorem.

An infinite set with the discrete topology similarly fails compactness, using the open cover by all its singletons. Every finite set is compact in any topology: select one covering open set for each point. The empty set is compact because the empty selection is a finite subcover. This open cover definition does not require the topological space to be Hausdorff; in algebraic geometry it is often called quasi-compactness.

The beauty of compactness is that information valid around each point can often be controlled using finitely many regions. A finite subcover lets us combine finitely many local bounds, or choose the smallest of finitely many positive tolerances, to obtain control across the whole compact space. This is why compactness repeatedly turns approximation into existence and why it is so useful in optimization through the extreme value theorem.

Every closed set in a compact space is compact, and a continuous image of a compact space is compact. The latter proof pulls an open cover back by inverse images, selects a finite subcover, and maps forward. This makes compactness useful for proving existence: the extreme value theorem guarantees that a continuous function from a nonempty compact space to the real line attains both a minimum and a maximum. For example, $q(x)=x(1-x)=1/4-(x-1/2{)}^2$ on $[0,1]$ has maximum $1/4$ at $1/2$ and minimum zero at its endpoints. On $(0,1)$ its infimum is still zero but is never attained. ^[ref]

Let $K\subseteq X$, where the ambient topological space $X$ is Hausdorff and $K$ is compact in its subspace topology. Then $K$ is a closed set in $X$: the complement $X\setminus K$ is an open set in $X$. The ambient space need not be compact. For example, $[0,1]$ is a compact closed interval in the non-compact Hausdorff real line.^[ref]

The ambient Hausdorff hypothesis matters even when $K$ itself is Hausdorff. In the Sierpiński space $S=\{0,1\}$, the open sets are $\varnothing ,\{1\},S$, so the closed sets are $\varnothing ,\{0\},S$. Thus $K=\{1\}$ is compact and Hausdorff in its one-point subspace topology, but is not closed in $S$: its complement $\{0\}$ is not an open set. Every open neighborhood of $0$ contains $1$, so separation from $K$ fails. In contrast, the discrete topology on the same underlying set is Hausdorff and makes both singletons closed.

The diagram lists every open set and closed set, so the distinction can be checked directly. All subsets in both panels are compact: for any open cover of a finite set, choose one covering open set per point, obtaining a finite subcover. Compactness alone therefore cannot distinguish the two panels.

Here is the finite-selection mechanism in the general proof. For $x\in X\setminus K$ and each $k\in K$, the Hausdorff property gives disjoint open neighborhoods $U_k$ of $k$ and $V_k$ of $x$. The $U_k$ give an ambient open cover of $K$. Select a finite subcover $U_{k_1},\dots ,U_{k_m}$ and setThen $K\subseteq U$, $x\in V$, and $U\cap V=\varnothing$. The finite intersection makes $V$ an open set; an infinite intersection would not guarantee this. Thus every point of $X\setminus K$ has an open neighborhood contained in that complement, proving that $K$ is closed. The empty set is already closed.

This theorem combines with the continuous image of a compact space theorem to make every continuous map from a compact space to a Hausdorff space a closed map: a closed subset of its domain is compact, and its image is compact and hence closed in the target. This is the reason the inverse in the compact-to-Hausdorff continuous bijection theorem is continuous. The converse implication, from closed to compact, needs additional hypotheses: the real line is closed in itself but not compact; in a compact space, every closed subset is compact.

Only finite subsets are compact in the cocountable topology. An infinite subset contains distinct $c_1,c_2,\dots$. The open sets obtained by removing the tails $\{c_j:j\ge n\}$ form an increasing open cover, but every finite selection misses a tail point. Finite sets are compact in every topology.

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.

A cover of a set $E$ is a family of sets whose union contains $E$. The members may overlap and may include points outside $E$; intersecting each member with $E$ gives a cover by its subsets. An open cover requires its members to be open sets. In Hausdorff measure, countable covers with a bound on each member's diameter determine geometric size. Constructing one inexpensive cover supplies an upper bound, while a lower bound must control all allowed covers.

Every subset $A$ of a cofinite topological space is a compact subset. To see the finite reduction directly, take an open cover of a nonempty $A$ by open sets of $X$ and choose one member $U$ containing a point of $A$. Since $U$ has finite complement, the set $A\setminus U$ is finite. Choose a covering member for each of those remaining points. Those members together with $U$ are a finite subcover. The empty set is already compact.

Equivalently, the subspace topology on $A$ is itself cofinite: for finite $F\subseteq X$, the trace of $X\setminus F$ is $A\setminus (F\cap A)$. The argument applies again inside $A$. A concrete open cover $\{X\setminus \{k\}:k\in X\}$ of $X={\mathbb{N}}_{>0}$ needs only the two members $X\setminus \{1\}$ and $X\setminus \{2\}$ when these labels are distinct: each fills the other's single gap.

On $X={\mathbb{N}}_{>0}$, the subset of even integers is compact but not closed, because it is infinite and proper. A compact subset of a Hausdorff space is guaranteed to be closed, whereas compactness without the Hausdorff space assumption does not suffice. Changing the topology changes compactness: the same infinite set with its discrete topology has the singleton open cover, which has no finite subcover. The finite-exception argument is special to finite complements; in the cocountable topology, compact subsets of a cocountable space must instead be finite.

A sheaf of modules over the sheaf of smooth functions is fine when every locally finite open cover admits a subordinate family of sheaf endomorphisms summing to the identity. Fine sheaves on a paracompact space are acyclic, so their positive-degree sheaf cohomology vanishes.

A topological space is compact exactly when every family of closed sets with the finite intersection property has nonempty intersection. Take complements and apply De Morgan's laws: an empty total intersection corresponds to an open cover, while an empty finite intersection corresponds to a finite subcover. This works without any Hausdorff condition. With the convention that an empty-family intersection is the whole ambient set, the finite intersection property includes nonemptiness of that ambient set, and the criterion correctly treats the empty compact space.

A finite subcover is a subcover with finitely many members, including possibly none when $K$ is the empty set. Requiring a finite subcover for every open cover is the definition of compactness.

If $X={\bigcup }_i{U}_i$ is an open cover by open subschemes, and morphisms of schemes $f_i:U_i\to Y$ agree on every overlap, there is a unique morphism of schemes $f:X\to Y$ restricting to them. The underlying continuous maps glue because their restrictions agree. The maps of structure sheaves glue using the sheaf gluing axiom, and the required local condition on stalks can be checked on the original cover. No affineness or separation assumption on $X$, $Y$, or the overlaps is needed. This is the local construction behind affine test schemes determine scheme morphisms.

For every finite $d\ge 1$, a subset $K\subseteq {\mathbb{R}}^d$ of Euclidean space is compact if and only if it is a closed set and a bounded set. Here closed means closed in the ambient Euclidean space, not merely in $K$ itself, and the bounded set condition uses the Euclidean norm. The empty set is included: it is a closed set, a bounded set, and compact. ^[ref]

This makes an apparently infinite test practical. To verify compactness of a closed interval $[a,b]$ with finite endpoints, we need not examine every open cover: it is closed in the real line and a bounded set. By contrast, the open interval $(0,1)$ is a bounded set but is not closed in the real line, as the missing-endpoint figure in compact space illustrates. The whole real line is closed but is not a bounded set, so it too fails compactness. These examples show why both hypotheses are needed.

The same criterion holds in any real or complex finite-dimensional vector space equipped with a norm: all norms there are equivalent norms, so they give the same open sets, closed sets, and bounded sets. Since compactness depends only on the topology, choosing a different norm does not change the conclusion. Without an appropriate topology, a bare vector space alone has no notion of compactness.

The restriction to a finite-dimensional vector space is essential. In the l2 sequence space, an infinite-dimensional vector space that is also a Hilbert space, let $e_n$ be the vectors of its standard orthonormal basis. All lie in the closed unit ball $B=\{x:\Vert x\Vert \le 1\}$, which is a closed set and a bounded set in the norm topology. For distinct indices, their inner product is zero, givingThus distinct terms remain distance $\sqrt{2}$ apart. No subsequence is a Cauchy sequence, hence none is a convergent sequence. The equivalence of compactness and sequential compactness in a metric space proves that this closed unit ball is not compact. The conclusion concerns the norm topology.

A short proof outline in Euclidean space explains the role of a finite-dimensional vector space. For a sequence in a bounded set, apply the Bolzano-Weierstrass theorem to each of the Cartesian coordinates in turn, retaining a subsequence at each of the finitely many steps. The final subsequence is a convergent sequence in all Cartesian coordinates, and a closed set contains its limit of a sequence. This yields sequential compactness, hence compactness. Conversely, a compact subset of a metric space is a closed set and a bounded set. ^[ref]

A Lindelöf space is a topological space in which every open cover has a countable subcover. A closed discrete subset of a Lindelöf space must be countable: cover each of its points by an open set meeting it at only that point and add the open complement. In a metrizable space, the Lindelöf property is equivalent to topological separability.

For a quasi-compact separated scheme and a global regular function $f$, global sections on $X_f$ are the localization of global sections by $f$. Choose a finite affine open cover. Its intersections are affine, so the sheaf gluing axiom describes global sections as the kernel of the difference map between two finite products of affine coordinate rings. Exactness of localization preserves this kernel, and localization commutes with those finite products. Localizing the same cover gives the claimed isomorphism. This applies before knowing whether $X$ itself is affine.