Compactness in Metric Spaces

Let $(X,d)$ be a metric space (Metric space: $d(x,y)=0$ iff $x=y$, symmetry, and the triangle inequality; pseudometric and ultrametric), with open sets as in 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 and balls as in Open ball, closed ball and sphere in a metric space.

• An open cover of $(X,d)$ is a family $\mathcal{U}$ of open subsets of $X$ 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 $\mathcal{V}=\mathrm{\varnothing }$ 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.
• $(X,d)$ is compact when every open cover of it has a finite subcover: for every open cover $\mathcal{U}$, either $X=\mathrm{\varnothing }$ and the empty subfamily covers it, or there are $n\in \mathbb{N}$ and $U_0,\dots ,U_n\in \mathcal{U}$ with $X=U_0\cup \cdots \cup U_n\mathrm{.}$
• A subset $A\subseteq X$ is a compact subset of $X$ when the metric subspace $(A,d_A)$ is a compact metric space, $d_A$ being the restriction of $d$ to $A\times A$ (Isometry, isometric embedding, and the subspace metric on a subset).

Compactness of a subset is defined intrinsically, and only intrinsically. The last clause speaks about the subspace $(A,d_A)$ 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 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 no item of this library may use the ambient reading without citing it. Taking the intrinsic reading as the definition is what makes "compact" a property of the metric space $(A,d_A)$ alone, so that a set compact in one ambient space is compact in every other one containing it isometrically.

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. The one-point space is compact too, and so is every space listed as $\{x_0,\dots ,x_n\}$: given a cover, each $x_i$ lies in some member, and finitely many members chosen in this way already cover.

The finiteness convention, and how it is used both ways. "Finite" above is the listing form, matching the finite lists of Finite intersection property. It agrees with the definition of finiteness by equinumerosity with a natural number (Finite, countably infinite, countable, uncountable), and both directions of the agreement are available and are used below:

• A nonempty finite set $F$ in the sense of Finite, countably infinite, countable, uncountable satisfies $F\approx m$ for some $m\ge 1$, and a bijection $m\to F$ is exactly a listing $F=\{a_0,\dots ,a_{m-1}\}$.
• Conversely a set listed as $A=\{a_0,\dots ,a_n\}$, that is the image of a function $a$ with domain $\sigma (n)$, is finite in the sense of Finite, countably infinite, countable, uncountable: the map sending $x\in A$ to the least $i\le n$ with $a_i=x$ is an injection of $A$ into $\sigma (n)$, so $A$ is equinumerous with a subset of $\mathbb{N}$ bounded above, and such a subset is finite (Every subset of an at most countable set is at most countable).

Neither direction uses a choice principle: the second selects nothing, taking a least index instead.

Why open covers rather than closed ones. Nothing in the definition would break if $\mathcal{U}$ were allowed to consist of arbitrary sets, 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 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 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 metric space $(A,d_A)$", whose members are open subsets of $A$. Which of the two is meant is written out everywhere on this page.

Let $(X,d)$ be a metric space (Metric space: $d(x,y)=0$ iff $x=y$, symmetry, and the triangle inequality; pseudometric and ultrametric), let $A\subseteq X$ and let $(A,d_A)$ be the metric subspace (Isometry, isometric embedding, and the subspace metric on a subset). Then:

1. Relative openness is a trace. A set $V\subseteq A$ is open in $(A,d_A)$ (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) if and only if $V=U\cap A$ for some $U$ open in $(X,d)$.
2. Compactness read in the ambient space. $A$ is a compact subset of $X$ (Open cover, subcover, compact metric space, and compact subset of a metric space), that is $(A,d_A)$ is a compact metric space, if and only if for every family $\mathcal{U}$ of open subsets of $X$ with $A\subseteq \bigcup \mathcal{U}$ there are $n\in \mathbb{N}$ and $U_0,\dots ,U_n\in \mathcal{U}$ with $A\subseteq U_0\cup \cdots \cup U_n$, or else $A=\mathrm{\varnothing }$.
3. 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 $A\subseteq U_{i_0}\cup \cdots \cup U_{i_n}$, or else $A=\mathrm{\varnothing }$.

Claim 3 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 $(A,d_A)$, and families of sets open in $X$ whose union contains $A$. Claim 1 is what turns one into the other, and it is the reason compactness of $A$ does not depend on which space $A$ is regarded as sitting inside. Every later item on this page that covers a subset by ambient balls is using claim 2 or claim 3, 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 6.1 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.

Let $(X,d)$ be a metric space (Metric space: $d(x,y)=0$ iff $x=y$, symmetry, and the triangle inequality; pseudometric and ultrametric), with open sets as in 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 and open covers, subcovers, finiteness and compactness as in Open cover, subcover, compact metric space, and compact subset of a metric space.

• $(X,d)$ is countably compact when every open cover $\mathcal{U}$ of $X$ that is at most countable (Finite, countably infinite, countable, uncountable) has a finite subcover.
• $(X,d)$ is sequentially compact when every sequence $(x_k)$ in $X$, that is every function $x:\mathbb{N}\to X$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), has a subsequence $(x_{n_j})$ converging to a point of $X$ (Convergence of a sequence in a metric space: $x_k\to x$ iff $d(x_k,x)\to 0$ in $\mathbb{R}$), the index map $n:\mathbb{N}\to \mathbb{N}$ being strictly increasing (A strictly increasing index map satisfies $n_k\ge k$).
• $(X,d)$ is limit point compact when every infinite subset $A\subseteq X$ has a limit point in $X$, that is a point $p\in X$ with $B(p,r)\cap (A\setminus \{p\})\mathrm{\ne }\mathrm{\varnothing }$ for every real $r>0$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space). Here infinite means not finite in the sense of Finite, countably infinite, countable, uncountable, equivalently not listable as $\{a_0,\dots ,a_n\}$ and not empty (Open cover, subcover, compact metric space, and compact subset of a metric space).

A subset $A\subseteq X$ is called countably compact, sequentially compact or limit point compact when the metric subspace $(A,d_A)$ is (Isometry, isometric embedding, and the subspace metric on a subset), exactly as for compactness.

The countable covers may be listed. A nonempty at most countable family $\mathcal{U}$ admits a surjection $\mathbb{N}\to \mathcal{U}$ (A nonempty set is at most countable iff it is a surjective image of $\mathbb{N}$), so countable compactness says: for every sequence $(U_n{)}_{n\in \mathbb{N}}$ of open sets with $X={\bigcup }_{n\in \mathbb{N}}{U}_n$ there are finitely many indices whose sets already cover $X$. That surjection is produced from the countability assumption alone and no choice principle is involved; the empty family covers only the empty space, which is compact anyway.

Limit points are computed where the set lives. For $S\subseteq A\subseteq X$ and $a\in A$, the identity $B_A(a,r)=B_X(a,r)\cap A$ (Isometry, isometric embedding, and the subspace metric on a subset) shows that $a$ is a limit point of $S$ in the subspace $(A,d_A)$ exactly when $a$ is a limit point of $S$ in $X$ and lies in $A$. So "$A$ is limit point compact" says that every infinite $S\subseteq A$ has a limit point belonging to $A$; a limit point outside $A$ does not count, and that is what distinguishes the property from a statement about $X$.

Three conditions, and none of them is compactness by definition. Each of the three weakens or replaces the open-cover condition of Open cover, subcover, compact metric space, and compact subset of a metric space: countable compactness restricts the covers tested, sequential compactness speaks about sequences instead of covers, and limit point compactness speaks about subsets. That the four conditions are not equivalent for topological spaces in general is standard and is quoted from the references, not proved here. For metric spaces they do coincide, but the coincidence is a theorem with a choice cost that varies from implication to implication, and it is proved on this page one arrow at a time (In any metric space compactness implies countable compactness and limit point compactness, and each of countable compactness and limit point compactness implies sequential compactness; every implication here is proved without a choice principle, 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, What each implication between the compactness properties of a metric space costs: which are theorems of ZF, which use countable choice, and which use dependent choice).

Indexing starts at $0$. A sequence here is a function on $\mathbb{N}$ and $\mathbb{N}$ contains $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), so a subsequence is $(x_{n_j}{)}_{j\in \mathbb{N}}$ with $n_0<n_1<\cdots$ and $n_j\ge j$ (A strictly increasing index map satisfies $n_k\ge k$). Every recursive construction of a subsequence on this page produces $n_0$ first and then $n_{j+1}>n_j$, and every radius written $1\mathrm{/}(j+1)$ is written that way because $1\mathrm{/}j$ is undefined at $j=0$.

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 $\epsilon$ be a real with $\epsilon >0$.

• A finite $\epsilon$-net for $(X,d)$ is a finite subset $F\subseteq X$ with $X={\bigcup }_{y\in F}B(y,\epsilon ),$ the balls being those of $(X,d)$ (Open ball, closed ball and sphere in a metric space). Finite is the listing form fixed in Open cover, subcover, compact metric space, and compact subset of a metric space: $F=\mathrm{\varnothing }$, or $F=\{y_0,\dots ,y_m\}$ for some $m\in \mathbb{N}$ and points $y_0,\dots ,y_m\in X$.
• $(X,d)$ is totally bounded when it has a finite $\epsilon$-net for every real $\epsilon >0$.
• A subset $A\subseteq X$ is totally bounded when the metric subspace $(A,d_A)$ is (Isometry, isometric embedding, and the subspace metric on a subset); its nets are then finite subsets of $A$ and its balls are the balls $B_A(a,\epsilon )$ of the subspace.

The empty space is totally bounded, the empty net serving for every $\epsilon$, since a union over no indices is empty. Every space listed as $\{x_0,\dots ,x_n\}$ is totally bounded too, itself being an $\epsilon$-net for every $\epsilon$.

The centres are required to lie in the space. Writing the condition with centres in $X$ and balls of $X$ is what makes total boundedness a property of the metric space $(X,d)$ alone, matching the treatment of compactness in Open cover, subcover, compact metric space, and compact subset of a metric space. For a subset $A$ this matters: the nets of $(A,d_A)$ consist of points of $A$, not of nearby points of the ambient space.

Total boundedness is stronger than boundedness and is not the same thing. A totally bounded space is bounded in the sense of Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space — that is claim 1 of A totally bounded metric space is bounded, every subspace of a totally bounded space is totally bounded, and the closure of a totally bounded subset is totally bounded — and the converse fails, as FALSE: a bounded metric space is totally bounded records. Boundedness asks for one ball containing the space; total boundedness asks for finitely many balls of every prescribed radius, and it is the second condition that controls how spread out the space is at small scales.

Why $\epsilon$ ranges over the reals here. Convergence and the Cauchy condition are tested against rational $\epsilon$ in this library (Convergence of a sequence in a metric space: $x_k\to x$ iff $d(x_k,x)\to 0$ in $\mathbb{R}$, Cauchy sequence in a metric space), because that is how Limits and Cauchy sequences of reals is written; total boundedness is not a limit condition and is stated for real $\epsilon >0$ directly. Nothing turns on the difference: a net for a rational ${\epsilon }^{\mathrm{\prime }}\le \epsilon$ is a net for $\epsilon$, since $B(y,{\epsilon }^{\mathrm{\prime }})\subseteq B(y,\epsilon )$.

A net is not unique and is not part of the data. Total boundedness asserts that nets exist; it names none. Producing one net for each $\epsilon$ simultaneously, as a function of $\epsilon$, is a further act of selection, and where a proof needs that function it says so and pays for it — see A complete, totally bounded metric space is compact, proved from countable choice used exactly once and A compact metric space has a countable dense subset, by countable choice, each of which spends the Axiom of Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)) exactly once and at exactly that point.

Let $(X,d)$ be a metric space (Metric space: $d(x,y)=0$ iff $x=y$, symmetry, and the triangle inequality; pseudometric and ultrametric), with total boundedness as in Finite $\epsilon$-net and totally bounded metric space and boundedness as in Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space. Then:

1. If $(X,d)$ is totally bounded, it is bounded.
2. If $(X,d)$ is totally bounded and $A\subseteq X$, then the metric subspace $(A,d_A)$ is totally bounded (Isometry, isometric embedding, and the subspace metric on a subset).
3. If $A\subseteq X$ is totally bounded, so is its closure $\bar{A}$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space).

No choice principle is used. The one selection made is over a finite index set, which Every natural-number-indexed list of nonempty sets has a choice function on its family of values supplies in ZF.

Where the halving is needed. In claim 2 the net of the subspace has to consist of points of $A$, and a point $y_j$ of a net for $X$ need not lie in $A$; moving from $y_j$ to a point of $A$ within $\epsilon \mathrm{/}2$ of it costs the other half of $\epsilon$. The same halving appears in claim 3, where the point being approximated lies in the closure rather than in $A$.

Claim 1 does not reverse. A bounded metric space need not be totally bounded; FALSE: a bounded metric space is totally bounded states the false converse and $\mathbb{N}$ with the discrete metric is bounded and is not totally bounded ↗ exhibits the witness.

Let $(X,d)$ be a metric space (Metric space: $d(x,y)=0$ iff $x=y$, symmetry, and the triangle inequality; pseudometric and ultrametric), with closed sets as in 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 and the finite intersection property as in Finite intersection property. For a family $\mathcal{A}$ of subsets of $X$ write

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

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

Then $(X,d)$ is compact (Open cover, subcover, compact metric space, and compact subset of a metric space) if and only if every family $\mathcal{A}$ of closed subsets of $X$ with the finite intersection property satisfies $\bigcap \mathcal{A}\mathrm{\ne }\mathrm{\varnothing }$.

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

The empty family and the empty space. The conventions are not decoration. $\bigcap \mathrm{\varnothing }=X$ makes the finite intersection property fail outright for every family of subsets of the empty space, so the right-hand condition is vacuously true there; and the empty space is compact, the empty subfamily covering it. The equivalence therefore holds at $X=\mathrm{\varnothing }$ as well, with both sides true.

Why no choice is spent. The natural-looking step "the finite subcover consists of sets $X\setminus F_i$, so pick $F_i\in \mathcal{A}$" would be a selection if a member of the family could be the complement of several different members. It cannot: complementation inside $X$ is injective, so $F_i$ is recovered from $U_i$ by a formula. That is the whole reason this characterisation, and the completeness half of A compact metric space is complete and totally bounded, and neither implication uses any choice principle that runs through it, cost nothing.

Let $(X,d)$ be a compact metric space (Open cover, subcover, compact metric space, and compact subset of a metric space, Metric space: $d(x,y)=0$ iff $x=y$, symmetry, and the triangle inequality; pseudometric and ultrametric) and let $F\subseteq X$ be closed in $X$ (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). Then $F$ is a compact subset of $X$: the metric subspace $(F,d_F)$ is a compact metric space (Isometry, isometric embedding, and the subspace metric on a subset).

No choice principle is used.

The hypothesis that $X$ is compact cannot be dropped, and neither can closedness. A closed subset of a non-compact space need not be compact: the whole space is closed in itself. And a non-closed subset of a compact space need not be compact, since a compact subset of any metric space is closed (A compact subset of a metric space is closed and bounded).

Why the augmented family is the whole trick. The set $F$ is covered by the $U_i$, but $X$ need not be; adjoining the single open set $X\setminus F$ repairs that at no cost, and it is the only member of the resulting finite subcover that has to be discarded again at the end.

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 $K\subseteq X$ be a compact subset (Open cover, subcover, compact metric space, and compact subset of a metric space). Then $K$ is closed in $X$ (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) and bounded (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

No choice principle is used: both covers below are given by a rule, and the indexed form of 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 returns indices rather than sets.

The converse is false in general. A closed and bounded subset of an arbitrary metric space need not be compact (FALSE: a closed and bounded subset of a metric space is compact); it is exactly in ${\mathbb{R}}^n$ that the converse holds (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).

Both conclusions use compactness through the same characterisation. The first cover is by concentric balls of every positive radius, which is what boundedness is about; the second is by balls small enough to keep a fixed outside point away, which is what closedness is about. In each case what compactness returns is a finite list of indices, and a maximum or a minimum of finitely many positive reals then does the rest.

Hausdorffness is what makes the second argument work, and every metric space has it (Distinct points of a metric space have disjoint balls around them). The statement is false for topological spaces without that separation property, which is why the proof cites the separation lemma rather than the metric axioms directly.

Let $(X,d)$ be a compact metric space (Open cover, subcover, compact metric space, and compact subset of a metric space, Metric space: $d(x,y)=0$ iff $x=y$, symmetry, and the triangle inequality; pseudometric and ultrametric). Then $(X,d)$ is totally bounded (Finite $\epsilon$-net and totally bounded metric space) and complete (Complete metric space: every Cauchy sequence converges in the space).

Both implications are theorems of ZF. Completeness is obtained here from the finite intersection characterisation (A metric space is compact if and only if every family of closed subsets with the finite intersection property has nonempty intersection) applied to the closures of the tails of a Cauchy sequence, and not from the extraction of a convergent subsequence, which would route the argument through sequential compactness. What matters for the ledger is that the route taken below selects nothing at all; the first remark below says why the other route was not taken.