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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 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 $X,Y$ 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 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 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.

Choose $b_n\in (n,n+1/2)\setminus A$, possible because each interval is uncountable. Every tail $K_n=\{b_k:k\ge n\}$ is closed and disjoint from $A$. Therefore $U_n=\mathbb{R}\setminus K_n$ is an open saturated set for collapsing a subset to a point, so its image in the quotient topology is open. These increasing images cover the quotient, but no finite selection covers it, because the largest selected image still misses an infinite tail of distinct singleton classes. This is an open cover with no finite subcover, proving noncompactness. The argument does not require the countable subset to be closed or nondense. For the empty subset, read the quotient as the original line.

A subset $B\subseteq \mathbb{C}$, with its subspace topology, is compact when every open cover of $B$ has a finite subcover. The Heine-Borel theorem in the complex plane, identified with ${\mathbb{R}}^2$, saysClosedness is in the ambient complex plane. For a continuous real-valued function on a nonempty compact set, the supremum and infimum are finite and attained.

The empty set is bounded. Otherwise choose $p$ in the compact metric space $X$. The open balls $B(p,n)$, $n=1,2,\dots$, form an open cover. A finite subcover exists by compactness; because the balls are nested, the largest one already covers $X$. Hence $X\subseteq B(p,N)$ for some $N$.

For $x,y\in X$, the triangle inequality gives $d(x,y)\le d(x,p)+d(p,y)<2N$. ThusThe proof uses only compactness and the metric, not a coordinate system.

Let $(x_n)$ be any sequence in the compact metric space $X$. There must be an $x\in X$ such that every open ball about $x$ contains $x_n$ for infinitely many indices $n$. Otherwise, for each $x\in X$ choose a ball $B(x,r_x)$ containing only finitely many terms, counted by indices. These balls are an open cover. A finite subcover would imply that only finitely many indices occur anywhere in $X$, a contradiction.

Now choose successively $n_k>n_{k-1}$ with $x_{n_k}\in B(x,1/k)$; infinitely many available indices make this possible. Then $d(x_{n_k},x)<1/k\to 0$, so $(x_{n_k})$ is a convergent subsequence. ThusThe proof counts indices rather than distinct points, so it also handles a value repeated infinitely often.

A metric space $X$ is a compact space if every open cover $X={\bigcup }_{\alpha \in I}{U}_{\alpha }$, with all $U_{\alpha }$ open in $X$, has a finite subcover: finitely many indices ${\alpha }_1,\dots ,{\alpha }_r$ cover $X$.

It is a sequentially compact space if every sequence $(x_n)$ in $X$ has a convergent subsequence $(x_{n_k})$ whose limit of a sequence belongs to $X$. In formulas,A finite subcover is the defining test for compactness; a convergent subsequence is the defining test for sequential compactness.

A topology on $X$ contains $\varnothing ,X$, is closed under arbitrary unions and under finite intersections. The cofinite family has these properties: a nonempty union contains one cofinite member and therefore has finite complement; the complement of a finite intersection is a finite union of finite sets. The empty union and the sets $X,\varnothing$ are included explicitly.

Any two nonempty open sets intersect, because their intersection has finite complement in the infinite set $X$. Thus no two disjoint nonempty open sets can separate $X$, proving connectedness. For compactness, take any open cover and choose a nonempty member $U$. Only finitely many points lie outside $U$. Choose one covering member for each of those points. Together with $U$ these give a finite subcover. Thus $X$ is compact.

A subset is compact if every cover of it by open sets of the ambient space has a finite subcover; equivalently use open sets of its subspace topology. A cover of $A\cup B$ separately covers $A$ and $B$. Compactness of each supplies two finite collections, whose union is finite and covers $A\cup B$. If $A$ is closed in a compact space, add the open complement of $A$ to any open cover of $A$. Compactness supplies a finite subcover of the whole space, and discarding the complement leaves a finite cover of $A$.

For the enlarged space, every old open set is open in $C$ and has the same trace on $X$. The trace on $X$ of a new open set $V\cup \{\infty \}$ is $V$, which is open in $X$ because its complement is closed. Thus the induced topology on $X$ is exactly the original one.

Given an open cover of $C$, choose a member containing infinity. It has the form $V\cup \{\infty \}$ with $K=X\setminus V$ compact. The remaining cover restricted to $K$ has a finite subcover. Together with the chosen infinity-neighborhood, these finitely many sets cover $C$. Hence $C$ is compact, with no Hausdorff assumption needed.

Finally put $a=f(\infty )$. Continuity makes $U=f^{-1}((a-ϵ,a+ϵ))$ open and containing infinity. Therefore $U=V\cup \{\infty \}$ with $W=X\setminus V$ compact and closed in $X$. For every $x\in X\setminus W=V$ we haveThis expresses convergence to the value at infinity outside compact closed subsets.

A topological space is compact if every open cover has a finite subcover:There is no Hausdorff requirement in this definition.

No countable subset works. For any countable set $A\subseteq \mathbb{R}$, choose $b_n\in (n,n+1/2)\setminus A$: each nonempty interval is uncountable. The sequence tends to infinity, so every tailis closed in $\mathbb{R}$ and disjoint from $A$. By part (i), $p(\mathbb{R}\setminus K_n)$ is open in $X_A$. These sets increase and cover $X_A$, since no point belongs to every tail. But any finite selection is contained in its largest member, which still misses every quotient point $p(b_k)$ in that member's remaining tail. The $b_k$ represent distinct singleton classes.

This is an open cover without a finite subcover, provingThe argument does not assume that $A$ is closed, bounded, or discrete. It is the noncompactness after collapsing a countable real subset obstruction.

Let $F$ be a closed subset of the compact space $X$. Express an arbitrary open cover of $F$ in its subspace topology as $\{F\cap U_i:i\in I\}$, where the $U_i$ are open sets of $X$. Since $X\setminus F$ is open, the familyis an open cover of $X$. Take a finite subcover and discard $X\setminus F$ if it occurs. The remaining finitely many $U_i$ still cover $F$, since the discarded set contains no point of $F$. Their traces on $F$ give the required finite subcover. Hence every closed subset of a compact space is compact, with no Hausdorff assumption.

Assume $K$ is compact, and let the closed subsets $F_i$ of $K$ have the finite intersection property. If their total intersection were empty, their complements $K\setminus F_i$ would form an open cover of $K$. A finite subcover would givecontradicting the finite intersection property. Therefore the total intersection is nonempty.

Conversely, assume the stated nonempty-intersection property for every family of closed subsets of $K$. Given an open cover $\{V_i:i\in I\}$ of $K$ in its subspace topology, set $F_i=K\setminus V_i$. Their total intersection is empty. By the contrapositive of the assumed property, this family fails the finite intersection property; hence some finite $J\subseteq I$ satisfiesThis is a finite subcover, proving compactness. Both directions are instances of De Morgan's laws. Thus compactness is equivalent to the finite intersection characterization of compactness.

For completeness, the intersection of the empty family of subsets of $K$ is $K$. If $K$ is empty, no family has the finite intersection property when the empty finite subfamily is included, so the implication remains valid.

A subset $K$ is compact if every open cover has a finite subcover. Explicitly, if $U_i$ are open sets of $X$ and $K\subseteq {\bigcup }_{i\in I}{U}_i$, there are finitely many indices $i_1,\dots ,i_m$ such thatEquivalently, the subspace topology makes $K$ a compact space. An open set in this subspace topology is $K\cap U$ for an ambient open set $U$, which proves equivalence of the two open cover formulations. The empty set is compact: the empty selection is a finite subcover.

For every $k\in K$, the Hausdorff condition gives disjoint open neighbourhoods $U_k$ of $k$ and $V_k$ of $x$. The $U_k$ form an open cover of $K$, so choose a finite subcover $U_{k_1},\dots ,U_{k_m}$. PutThen $U,V$ are open sets, $K\subseteq U$ and $x\in V$. If a point belonged to $U\cap V$, it would belong to some $U_{k_j}$ and to $V_{k_j}$, contrary to their choice. ThusThe finiteness matters: a finite intersection of open sets is open. This is separation of a point and a compact set. If $K=\varnothing$, use $U=\varnothing$, $V=X$.

Every $x\in X\setminus K$ consequently has an open neighbourhood $V$ contained in $X\setminus K$. The complement is open, so a compact subset of a Hausdorff space is closed.

The conclusion can fail without the Hausdorff condition. Take the two-point set $X=\{0,1\}$ with the indiscrete topology $\{\varnothing ,X\}$. The singleton $K=\{0\}$ is compact, since one member of any open cover suffices. It is not closed, because its complement $\{1\}$ is not open. The two points cannot have disjoint open neighbourhoods, so this example is indeed not Hausdorff.