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A subset is Cb-embedded if every bounded continuous real-valued function on it extends to a bounded continuous real-valued function on the ambient space. A bounded extension can be clipped to preserve the original bound. This is weaker than C-embeddedness: the natural numbers in their Stone-Čech compactification have the bounded extension property, but their unbounded identity function cannot extend to a compact space.

Let $D$ be a closed discrete subset of a compact space $X$. Closedness makes $D$ compact. Discreteness makes its singletons open in its subspace topology, so the singleton cover has a finite subcover. Thus $D$ is finite. Closedness is essential: the discrete set $\{1/n:n\ge 1\}$ inside the compact interval $[0,1]$ is infinite and has the missing accumulation point zero.

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.

The map $x\mapsto (x,f(x))$ is continuous into the product topology. It maps a compact space $X$ onto the graph of a function of a continuous $f$. Hence that graph is compact, whether or not the codomain is a Hausdorff space. Projection to $X$ is its continuous inverse.

A compact metric space is a metric space that is compact. Equivalently, it is complete and totally bounded.

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.

Let $K$ be a nonempty compact space and let $\alpha :K\to \mathbb{R}$ be continuous and nowhere zero. On $C(K)$, define $\Vert f{\Vert }_{\alpha }={\sup }_{x\in K}|\alpha (x)f(x)|$. Multiplication by $\alpha$ is an isometry onto $C(K)$ with its ordinary uniform norm, because division by the continuous nonvanishing $\alpha$ provides the inverse. Thus the weighted space is a Banach space. Equivalently, the extreme value theorem gives $0<m=\min |\alpha |\le \max |\alpha |=M<\infty$, and $m\Vert f{\Vert }_{\infty }\le \Vert f{\Vert }_{\alpha }\le M\Vert f{\Vert }_{\infty }$. A weighted Cauchy sequence is uniformly Cauchy, has a continuous uniform limit, and the upper bound gives convergence also in the weighted norm.

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.

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 continuous map that is also an injective function. A continuous injection need not be a topological embedding: its inverse on its image need not be continuous. If its source is a compact space and its target is a Hausdorff space, however, the compact-to-Hausdorff continuous bijection theorem makes that inverse continuous.

This concerns arbitrary compact spaces, with no Hausdorff restriction. For nonempty sets $A_j$, adjoin a distinguished point ${\infty }_j$ and use the four-open-set topology $\{\varnothing ,A_j,\{{\infty }_j\},A_j\cup \{{\infty }_j\}\}$. Each resulting space is compact without choice. The product has the canonical all-infinity point. The closed coordinate conditions $x_j\in A_j$ have the finite intersection property, since finite choice is provable without the axiom of choice. If the countable product is compact, their intersection is nonempty, and any point in it supplies a choice function.

A real-valued function that is continuous on a nonempty compact space attains both a global minimum and a global maximum. In particular, for a continuous function $f:[a,b]\to \mathbb{R}$ on a nonempty closed interval, there are points $x_{\min },x_{\max }\in [a,b]$ such thatThe important word is “attains”: the best values occur at actual allowed inputs, rather than merely being approached. The interval version is ^[ref].

For a worked example, take $f(x)=(x-1{)}^2$ on $[-1,2]$. Its global minimum is $f(1)=0$, and its global maximum is $f(-1)=4$. Nonnegativity gives the first answer; $|x-1|\le 2$ on the specified closed interval gives the second. The endpoint $x=-1$ is essential: looking only for a zero derivative would find the minimum and miss the maximum.

For comparison, the continuous function $g(x)=x$ on the open interval $(0,1)$ has infimum $0$ and supremum $1$, but neither is a value of $g$. Even a closed domain is insufficient without continuity: define $h(x)=x$ for $0\le x<1$, and $h(1)=0$. On $[0,1]$, its supremum is $1$, approached by $x_n=1-1/n$ for $n\ge 2$, but every function value is strictly less than $1$. This endpoint change makes $h$ discontinuous, preventing attainment. On the unbounded real line, the continuous function $x\mapsto x$ has neither a global minimum nor a global maximum. These examples show what can fail when a hypothesis is dropped; functions outside the hypotheses may still have extrema.

The extreme value theorem is an existence tool for optimization: first establish that a continuous objective function on a nonempty compact feasible set has a best value, then investigate where it occurs. It neither locates the optimizing point nor promises uniqueness or differentiability. For example, the absolute value function $|x|$ on $[-1,2]$ has its global minimum at zero although no derivative exists there; a constant function has every input as a minimizer. This role in optimization is explained in University of Toronto MAT237 notes, section 1.4: compactness and the extreme value theorem.

A short proof outline identifies the mechanism. The continuous image of a compact space is compact; a nonempty compact subset of the real line contains its infimum and supremum. For a closed interval, one can instead choose inputs whose outputs approach the supremum, extract a convergent subsequence by the Bolzano-Weierstrass theorem, and use continuity to make its limiting output equal that supremum. The closed interval retains the limiting input. Repeating the argument for the infimum supplies the minimum.

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.

Let $X$ be a nonempty compact metric space and let $f:X\to X$ satisfy $d(f(x),f(y))<d(x,y)$ whenever $x\ne y$. Then $f$ has exactly one fixed point. The distance condition implies continuity, and $x\mapsto d(f(x),x)$ attains a minimum by compactness. If a minimizer $p$ were not a fixed point, then $d(f^2(p),f(p))<d(f(p),p)$ would contradict minimality. Two distinct fixed points would contradict the same strict distance inequality directly. A uniform Lipschitz constant below $1$ is unnecessary; compactness replaces the quantitative hypothesis of the contraction mapping theorem.

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 homeomorphism is a bijection between topological spaces such that both $f$ and $f^{-1}$ are continuous maps. Equivalently, for every subset $U\subseteq X$, $U$ is an open set if and only if $f(U)$ is an open set in $Y$. A homeomorphism therefore transports the entire topology, including connectedness and compactness. It need not preserve numerical metric values, and a bounded set need not map to a bounded set.

For example, an explicit homeomorphism between an open interval and the real line isThe tangent function on $(-\pi /2,\pi /2)$ is strictly increasing and takes every real value; the displayed arctangent inverse verifies the bijection. Both formulas define continuous maps on their stated domains. Computed values are $f(1/4)=-1$, $f(1/2)=0$, and $f(3/4)=1$. Points approach infinity as $x$ approaches an excluded endpoint. Thus a bounded set can have the same topology as an unbounded one; neither of these topological spaces is compact.

A useful shortcut is the compact-to-Hausdorff continuous bijection theorem: a continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Its images of closed sets are compact sets, hence closed sets in the target, which proves continuity of the inverse. The indiscrete topology example shows why the Hausdorff assumption matters. ^[ref]

Every continuous injective map $g:S^1\to S^1$ is a homeomorphism. After normalizing $g(1)=1$ by a rotation, it is homotopic either to the identity or to complex conjugation, and has degree $1$ or $-1$.

To see this without assuming surjectivity, lift $t\mapsto g(e^{2\pi it})$ on $[0,1]$ through the covering map $u\mapsto e^{2\pi iu}$ to a path $G$ with $G(0)=0$. Injectivity makes $G$ injective on $[0,1)$, hence strictly monotone. Its endpoint $d=G(1)$ is an integer. Strict monotonicity excludes $d=0$; if $|d|>1$, the intermediate value theorem gives an interior integer value, contradicting injectivity of $g$. Thus $d=\pm 1$, and $\exp (2\pi i((1-s)G(t)+sdt))$ descends to a homotopy from $g$ to $z\mapsto z^d$. If $g$ omitted a point, a continuous branch of the complex argument on the punctured circle would make it null-homotopic, contradicting this nonzero degree. Hence it is surjective. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism.

Suppose a family $A\subseteq C(K,\mathbb{R})$ on a compact space is closed under pointwise maximum and minimum. If a target $f$ can be approximated arbitrarily closely at each pair of points by a member of $A$, it can be approximated in the supremum norm. For a fixed first point, take finitely many pairwise approximants whose neighborhoods cover $K$ and take their maximum: it lies above $f-\epsilon$ everywhere and below $f+\epsilon$ near that first point. Take finitely many such first-point neighborhoods covering $K$ and take the minimum of their maxima. The result is within $\epsilon$ of $f$ everywhere. Starting with tolerance $\epsilon /2$ ensures a strict uniform approximation error smaller than $\epsilon$.

A product topology on locally compact factors is locally compact if all but finitely many factors are compact. At a point, choose a compact neighborhood in each of the finitely many exceptional factors and take the full compact space in every other coordinate. Tychonoff's theorem makes their product compact, and the product contains an open basic neighborhood because only finitely many coordinates were restricted. In contrast ${\mathbb{R}}^{\mathbb{N}}$ is not locally compact: a basic neighborhood has a free coordinate projecting onto all of the noncompact real line, so no compact set can contain it.