Compactness

Let $(X, \mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

• An open cover of $(X,\mathcal{T})$ is a family $\mathcal{U} \subseteq \mathcal{T}$ of open sets with $X = \bigcup \mathcal{U}$, where $\bigcup \mathcal{U} = \{\, x \in X : x \in U \text{ for some } U \in \mathcal{U} \,\}$.
• A subcover of $\mathcal{U}$ is a subfamily $\mathcal{V} \subseteq \mathcal{U}$ that is itself an open cover.
• A family $\mathcal{V}$ of sets is finite when $\mathcal{V} = \varnothing$ or there are $n \in \mathbb{N}$ and sets $V_0, \dots, V_n$ with $\mathcal{V} = \{V_0, \dots, V_n\}$; repetitions in the list are allowed and harmless.
• $(X,\mathcal{T})$ is compact when every open cover of it has a finite subcover: for every open cover $\mathcal{U}$, either $X = \varnothing$ and the empty subfamily covers it, or there are $n \in \mathbb{N}$ and $U_0, \dots, U_n \in \mathcal{U}$ with $X = U_0 \cup \dots \cup U_n .$
• A subset $A \subseteq X$ is a compact subset of $X$ when the subspace $(A, \mathcal{T}_A)$ is a compact topological space, $\mathcal{T}_A$ being the subspace topology (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

Compactness of a subset is defined intrinsically, and only intrinsically. The last clause speaks about the subspace $(A, \mathcal{T}_A)$ and its own open sets, not about families of open subsets of the ambient $X$. The two readings do agree, but that is a theorem and not a convention: it is A subspace is compact exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it, and no item of this library may use the ambient reading without citing it. Taking the intrinsic reading makes compactness a property of the topological space $(A, \mathcal{T}_A)$ alone. Hence it is preserved when $A$ is embedded homeomorphically as a subspace, or when another ambient space induces the same topology on $A$; it need not be preserved if the induced topology changes. This is exactly the convention already fixed for metric spaces by Open cover, subcover, compact metric space, and compact subset of a metric space, and the agreement of that definition with this one is For a metric space with its metric topology, compactness in the topological sense is compactness in the metric sense, and the two notions of compact subset coincide.

The empty space is compact, since the empty subfamily of any family covers it; this is the reason the clause above is written with the two cases. Every space listed as $\{x_0, \dots, x_n\}$ is compact too: given a cover, each $x_i$ lies in some member, and finitely many members named in this way already cover. So every finite space is compact, whatever its topology, and in particular the discrete topology on a finite set is compact while the discrete topology on an infinite set is not (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

The finiteness convention. "Finite" above is the listing form. It agrees with the definition of finiteness by equinumerosity with a natural number (Finite, countably infinite, countable, uncountable), in both directions, and the agreement is the one already discharged in Open cover, subcover, compact metric space, and compact subset of a metric space: a nonempty set equinumerous with $m \ge 1$ is listable, and a set listed as $\{a_0, \dots, a_n\}$ injects into $\sigma(n)$ by sending $x$ to the least $i \le n$ with $a_i = x$. Neither direction uses a choice principle; the second selects nothing, taking a least index instead.

Quasicompact is not used here. Some authors, following Bourbaki, reserve compact for a space that is both compact in the above sense and Hausdorff, and call the open-cover condition alone quasicompact. This library follows the more widely adopted convention: compact means the open-cover condition and nothing more, and a Hausdorff hypothesis is always written out. The fork is recorded in The quasicompact convention, why compactness of a subset is read intrinsically here, and what each result on this page costs in choice.

Why open covers rather than covers by arbitrary sets. Nothing in the definition would break if $\mathcal{U}$ were allowed to consist of arbitrary subsets of $X$, but the resulting notion would be uninteresting: every space is covered by its singletons, and only a finite space would survive. Openness of the members is what makes the condition a genuine restriction, and it is what A subspace is compact exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it has to keep track of when the ambient space changes.

A warning about the word "cover". A family may cover $A \subseteq X$ without being a family of subsets of $A$: the members are open subsets of $X$ and their union merely contains $A$. That is the ambient reading, and it is a different statement from "$\mathcal{U}$ is an open cover of the space $(A, \mathcal{T}_A)$", whose members are open subsets of $A$. Which of the two is meant is written out everywhere on this page.

Let $(X, \mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), let $A \subseteq X$ and let $(A, \mathcal{T}_A)$ be the subspace (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace). Then:

1. Compactness read in the ambient space. $A$ is a compact subset of $X$ (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right), that is $(A, \mathcal{T}_A)$ is a compact space, if and only if for every family $\mathcal{U} \subseteq \mathcal{T}$ with $A \subseteq \bigcup \mathcal{U}$ there are $n \in \mathbb{N}$ and $U_0, \dots, U_n \in \mathcal{U}$ with $A \subseteq U_0 \cup \dots \cup U_n$, or else $A = \varnothing$.
2. The same in indexed form. $A$ is a compact subset of $X$ if and only if for every set $I$ and every family $(U_i)_{i \in I}$ of open subsets of $X$ with $A \subseteq \bigcup_{i \in I} U_i$ there are $n \in \mathbb{N}$ and indices $i_0, \dots, i_n \in I$ with $A \subseteq U_{i_0} \cup \dots \cup U_{i_n}$, or else $A = \varnothing$.

Claim 2 is the form used by almost every later proof on this page, because a cover is usually produced by a rule that attaches an open set to each point or to each index, and a set of open sets forgets that rule. No choice principle is used anywhere below; the one place a selection is made is over a finite index set, and Every natural-number-indexed list of nonempty sets has a choice function on its family of values is a theorem of ZF.

Why the ambient reading needed a proof at all. A subset $A$ of $X$ carries two candidate notions of open cover: families of sets open in $(A, \mathcal{T}_A)$, and families of sets open in $X$ whose union contains $A$. The trace description of the subspace topology is what turns one into the other, and it shows that compactness can be checked using ambient open sets for this fixed induced topology. Another ambient is guaranteed to give the same answer when it induces the same topology on $A$; if the induced topology changes, the answer may change. Every later item on this page that covers a subset by ambient open sets is using claim 1 or claim 2, and says so.

The traces do not remember their sources. A single relatively open $V$ is usually the trace of many different ambient open sets, and that is exactly why step 3.1 has to recover indices at all. Recovering infinitely many at once would be a choice principle; recovering finitely many is not, and the proof is arranged so that only finitely many are ever needed.

The metric statement of the same fact is A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it, whose claims 2 and 3 are claims 1 and 2 above with the open subsets of a metric space in place of the members of an abstract topology. Its proof carries an extra first claim, that relative openness in a metric subspace is a trace, which here is the definition of the subspace topology and so needs no argument. Neither statement is used in the proof of the other; that the two agree is For a metric space with its metric topology, compactness in the topological sense is compactness in the metric sense, and the two notions of compact subset coincide.

Let $(X,d)$ be a metric space (Metric space: $d(x,y) = 0$ iff $x = y$, symmetry, and the triangle inequality; pseudometric and ultrametric) and let $\mathcal{T}_d$ be its metric topology (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement), so that $(X, \mathcal{T}_d)$ is a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison) and is metrizable (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not). Then:

1. $(X,d)$ is a compact metric space (Open cover, subcover, compact metric space, and compact subset of a metric space) if and only if $(X, \mathcal{T}_d)$ is a compact topological space (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right).
2. For every $A \subseteq X$: $A$ is a compact subset of the metric space $(X,d)$ if and only if $A$ is a compact subset of the topological space $(X, \mathcal{T}_d)$, the two readings of "compact subset" being the metric subspace $(A, d_A)$ (Isometry, isometric embedding, and the subspace metric on a subset) and the topological subspace $(A, (\mathcal{T}_d)_A)$ (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

Nothing here is a coincidence and nothing is transported. The open-cover condition of Open cover, subcover, compact metric space, and compact subset of a metric space quantifies over families of subsets open in $(X,d)$, and by The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement those are exactly the members of $\mathcal{T}_d$; so the two conditions are not merely equivalent, they are the same condition written twice. No choice principle is used.

What this theorem buys, and why it is stated so early on the page. Every theorem proved on compactness-in-metric-spaces about compact metric spaces and their compact subsets is, by this theorem, a theorem about metrizable topological spaces and their compact subsets, once a metric inducing the topology has been named. Heine-Borel in $\mathbb{R}^n$ (Heine-Borel in $\mathbb{R}^n$: with the Euclidean metric a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line), the extreme value theorem (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value) and the equivalence of the four compactness conditions for metric spaces (For a metric space, compact, countably compact, limit point compact, sequentially compact, and complete together with totally bounded are all equivalent, given countable choice and dependent choice) are all available in that form and are used below without being reproved.

It also forbids a second notion. Since the metric development already fixed the intrinsic reading of "compact subset" (Open cover, subcover, compact metric space, and compact subset of a metric space) and this page fixes the same reading (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right), claim 2 says that the phrase means one thing throughout the library, whichever of the two developments a reader arrives from. Had either page taken the ambient reading as its definition the phrase would have meant two things, and the agreement would have had to be proved through A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it and A subspace is compact exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it rather than directly.

Let $(X, \mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison). For a family $\mathcal{A}$ of subsets of $X$ write

$\bigcap \mathcal{A} \;:=\; \{\, x \in X : x \in A \text{ for every } A \in \mathcal{A} \,\},$

so that $\bigcap \varnothing = X$, matching the convention for the empty finite intersection in Finite intersection property. Then:

1. $(X, \mathcal{T})$ is compact (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right) if and only if every family $\mathcal{A}$ of closed subsets of $X$ with the finite intersection property (Finite intersection property) satisfies $\bigcap \mathcal{A} \ne \varnothing$.
2. Equivalently: $(X, \mathcal{T})$ is compact if and only if every family of closed subsets of $X$ that is contained in some filter on $X$ (Filter on a set) has nonempty intersection, a family of subsets of $X$ lying in a filter exactly when it has the finite intersection property (A family lies in a filter exactly when it has the finite intersection property).

No choice principle is used in either direction: complementation is a canonical bijection, so no member of a family ever has to be selected.

What the condition says, and why it is the useful form. Compactness in the open-cover form is a statement about families that already cover; the closed-set form is a statement about families that already have all their finite intersections nonempty. In practice the second is easier to apply, because a nested family of nonempty closed sets has the finite intersection property for free, and the theorem then produces a point lying in all of them at once. That is how it is used below in Assuming dependent choice, every locally compact Hausdorff space is a Baire space, whose step 7.1 turns a decreasing sequence of nonempty closed sets into a point common to all of them. In a compact Hausdorff space every quasicomponent is connected, so quasicomponents and components coincide uses the theorem in the opposite direction: from a family of closed sets whose intersection is empty it extracts a finite subfamily whose intersection is already empty.

The finite intersection property is not a topological notion. Finite intersection property is a condition on an arbitrary family of subsets of a set, and A family lies in a filter exactly when it has the finite intersection property shows it is exactly the condition for the family to sit inside a filter. The topology enters this theorem only through the word "closed"; the theorem is that compactness of the topology is what makes that combinatorial condition detect a common point.

The metric special case is A metric space is compact if and only if every family of closed subsets with the finite intersection property has nonempty intersection, stated there for a metric space and its closed sets. It is not used above, and it is not needed: by For a metric space with its metric topology, compactness in the topological sense is compactness in the metric sense, and the two notions of compact subset coincide the metric statement is the present one applied to a metric topology.

Let $(X, \mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), with subspaces as in Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace and compactness as in Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right. Then:

1. Closed in compact is compact. If $(X, \mathcal{T})$ is compact and $F \subseteq X$ is closed in $X$, then $F$ is a compact subset of $X$.
2. Finite unions. If $n \in \mathbb{N}$ and $K_0, \dots, K_n$ are compact subsets of $X$, then $K_0 \cup \dots \cup K_n$ is a compact subset of $X$. The union of the empty list is $\varnothing$, which is a compact subset of every space.

Claim 1 needs $X$ to be compact and claim 2 does not; no hypothesis of any kind is placed on $X$ in claim 2. No choice principle is used: claim 1 selects nothing, taking a least index where a selection would be natural, and claim 2 makes finitely many selections through Every natural-number-indexed list of nonempty sets has a choice function on its family of values, a theorem of ZF.

Claim 1 is where the two hypotheses do different work. Compactness of $X$ supplies a finite subcover of $X$; closedness of $F$ is what makes $X \setminus F$ available as one more open set, so that a cover of $F$ can be enlarged to a cover of $X$ by adding a single member. Neither hypothesis can be dropped: an open subspace of a compact space need not be compact, and without compactness of $X$ there is nothing to thin.

The converse of claim 1 fails, and that is the subject of the next item. A compact subset of an arbitrary space need not be closed; it is closed as soon as the ambient space is Hausdorff (In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones), and FALSE: a compact subset of a topological space is closed records the failure without that hypothesis.

The metric special case is A closed subset of a compact metric space is compact. It is stated there for a closed subset of a compact metric space and is not used above; by For a metric space with its metric topology, compactness in the topological sense is compactness in the metric sense, and the two notions of compact subset coincide it is claim 1 applied to a metric topology. The general theorem is proved from the general definitions and borrows nothing from the metric development, which is why the metric statement does not appear among its dependencies.

Let $(X, \mathcal{T})$ be a Hausdorff topological space (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), with compact subsets as in Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right. Then:

1. A point and a disjoint compact set are separated. If $K \subseteq X$ is compact and $x \in X \setminus K$, there are $U, V \in \mathcal{T}$ with $x \in U, \qquad K \subseteq V, \qquad U \cap V = \varnothing .$
2. Two disjoint compact sets are separated. If $K, L \subseteq X$ are compact and $K \cap L = \varnothing$, there are $U, V \in \mathcal{T}$ with $L \subseteq U, \qquad K \subseteq V, \qquad U \cap V = \varnothing .$
3. Compact implies closed. Every compact subset of $X$ is closed in $X$.
4. In a compact Hausdorff space the two classes coincide. If in addition $(X, \mathcal{T})$ is compact, then a subset of $X$ is compact if and only if it is closed.

The proof is written choice-free, and that is not a stylistic preference. The textbook argument says "for each $y \in K$ choose disjoint open $U_y, V_y$", which is a selection over an arbitrary index set and therefore an appeal to the full Axiom of Choice. What is done below instead is to take the family of all open $V$ that admit some open $U \ni x$ disjoint from them — a family cut out by a formula, with nothing selected — extract a finite subcover from it, and only then make finitely many selections, which Every natural-number-indexed list of nonempty sets has a choice function on its family of values supplies as a theorem of ZF.

Where each hypothesis is spent. The Hausdorff condition is used exactly once, at step 2.1, to know that the family $\mathcal{V}$ covers $K$; compactness of $K$ is used exactly once, at step 3.1, to cut that cover down to finitely many members. Claim 2 then reuses claim 1 in the same shape, with the roles of point and compact set played by a point of $L$ and the compact set $K$.

Why the family is defined and not chosen. For each $y \in K$ the Hausdorff condition asserts that some pair $(U, V)$ exists; it provides no rule for naming one. A proof that writes $U_y$ and $V_y$ has selected a pair for every $y \in K$ at once, and for an arbitrary compact $K$ that is the Axiom of Choice. Collecting instead every $V$ that works for some $U$ replaces the selection by a formula, and the only selection left is over the finite index set $\sigma(n)$, where Every natural-number-indexed list of nonempty sets has a choice function on its family of values applies.

Claim 3 fails without the Hausdorff hypothesis, and FALSE: a compact subset of a topological space is closed records the failure with a witness. Claim 4 is the converse pairing: closedness is enough for compactness only when the ambient space is compact (A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact), and compactness is enough for closedness only when it is Hausdorff.

Let $(X, \mathcal{T}_X)$ and $(Y, \mathcal{T}_Y)$ be topological spaces (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), and let $\mathbb{R}$ carry its usual topology, the metric topology of $d_{\mathbb{R}}(s,t) = |s-t|$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not). Then:

1. Continuous images. If $f : X \to Y$ is continuous (Continuity of a map of topological spaces at a point and globally) and $(X, \mathcal{T}_X)$ is compact (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right), then $f[X]$ is a compact subset of $Y$. More generally, if $K \subseteq X$ is a compact subset of $X$ then $f[K]$ is a compact subset of $Y$.
2. Extreme values. If $(X, \mathcal{T}_X)$ is compact and nonempty and $g : X \to \mathbb{R}$ is continuous, then $g[X]$ has a maximum and a minimum (Maximum and minimum of a set): there are $x_{\max}, x_{\min} \in X$ with $g(x_{\min}) \;\le\; g(x) \;\le\; g(x_{\max}) \qquad \text{for every } x \in X .$
3. Compact to Hausdorff. If $(X, \mathcal{T}_X)$ is compact, $(Y, \mathcal{T}_Y)$ is Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not) and $f : X \to Y$ is a continuous bijection, then $f$ is a homeomorphism (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

Nonemptiness in claim 2 is a hypothesis and not an oversight: for $X = \varnothing$ the image is empty and has neither a maximum nor a minimum. No choice principle is used: the one selection made below is over a finite index set, where Every natural-number-indexed list of nonempty sets has a choice function on its family of values is a theorem of ZF.