nLab
compact space

Context

Topology

topology (point-set topology)

see also algebraic topology, functional analysis and homotopy theory

Introduction

Basic concepts

Universal constructions

Extra stuff, structure, properties

Examples

Basic statements

Theorems

Basic homotopy theory

Compact spaces

Idea

A topological space (or more generally: a convergence space) is compact if everything converges as much as possible. It is a kind of ultimate topological expression of the general idea of a space being “closed and bounded”: every net must accumulate somewhere in the space; by boundedness it cannot escape, and by closure the point is in the space.

There is also a notion of compactness for locales. Observe that (using classical logic) already every locally compact locale is spatial (this prop.).

It is also common to work with compact subsets of a space. These are those subsets which are compact spaces with the subspace topology.

One often wishes to study compact Hausdorff spaces. For locales, one usually speaks of compact regular locales; these are equivalent (since every locale is T 0T_0 and hence T 3T_3 if regular, while every Hausdorff space is T 3T_3 if compact) since regularity is easier to formulate and handle than Hausdorffness in locale theory.

Definitions

There are many ways to say that a space XX is compact. The first is perhaps the most common:

Definition

(open cover)

Let (X,τ)(X,\tau) be a topological space. Then an open cover an is set {U iX} iI\{U_i \subset X\}_{i \in I} of open subsets (i.e. (U iX)τP(X)(U_i \subset X) \in \tau \subset P(X)) such that their union is all of XX

iIU i=X. \underset{i \in I}{\cup} U_i = X \,.

This is called a finite open cover if II is a (Kuratowski-)finite set.

A subcover of an open cover as above is a subset JIJ\subset I of the given open subsets, such that their union still exhausts XX, i.e. iJIU i=X\underset{i \in J \subset I}{\cup} U_i = X.

Definition

(compact space)

A topological space is called compact if every open cover has a finite subcover (def. 1).

Remark

(differing terminology)

Some authors use “compact” to mean “compact and Hausdorff” (a much nicer sort of space, and forming a much nicer category of spaces, see atz compact Hausdorff space), and use the word “quasicompact” to refer to just “compact” as we are using it here. This custom seems to be prevalent among algebraic geometers, for example, and particularly so within Francophone schools.

But it is far from clear to me (Todd Trimble) that “quasicompact” is very well-established outside such circles (despite some arguments in favor of it), and using simply “compact” for the nicer concept therefore carries some risk of creating misunderstanding among mathematicians at large. My own habit at any rate is to say “compact Hausdorff” for the nicer concept, and I will continue using this on the nnLab until consensus is reached (if that happens).

Another term in usage is ‘compactum’ to mean a compact Hausdorff space (even when ‘compact’ is not used to imply Hausdorffness).

If excluded middle is assumed, then def. 2 the following reformulations #in terms of closed subsets:

Proposition

(compactness in terms of closed subsets)

Let (X,τ)(X,\tau) be a topological space. Assuming excluded middle, then the following are equivalent:

  1. (X,τ)(X,\tau) is compact in the sense of def. 2.

  2. Let {C iX} iI\{C_i \subset X\}_{i \in I} be a set of closed subsets such that their intersection is empty iIC i=\underset{i \in I}{\cap} C_i = \emptyset, then there is a finite subset JIJ \subset I such that the corresponding finite intersection is still empty: iJiC i=\underset{i \in J \subset i}{\cap} C_i = \emptyset.

  3. Let {C iX} iI\{C_i \subset X\}_{i \in I} be a set of closed subsets such that it enjoys the finite intersection property, meaning that for every finite subset JIJ \subset I then the corresponding finite intersection is non-empty iJIC i\underset{i \in J \subset I}{\cap} C_i \neq \emptyset. Then also the total intersection is inhabited, iIC i\underset{i \in I}{\cap} C_i \neq \emptyset.

  4. For every topological space (Y,τ Y)(Y,\tau_Y) the projection map out of the product topological space π Y:(X×Y,τ X×Y)(Y,τ Y)\pi_Y \;\colon\; (X \times Y, \tau_{X \times Y}) \to (Y, \tau_Y) is a closed map.

Proof

The equivalence of the first two statements follows by de Morgan's law (complements interchange unions with intersections), the definition of closed subsets as the complements of open sets, and (using excluded middle) that dually the complements of closed subsets are the open subsets:

Let {U iX} iI\{U_i \subset X\}_{i \in I} be an open cover. Write C iX\U iC_i \coloneqq X \backslash U_i for the corresponding closed complements. By de Morgan's law the condition that iIU i=X\underset{i \in I}{\cup} U_i = X is equivalent to iIC i=\underset{i \in I}{\cap} C_i = \emptyset. The second statement is that there is then a finite subset JIJ \subset I such that also iJIC i=\underset{i \in J \subset I}{\cap} C_i = \emptyset, and under forming complements again this is equivalently the first statement.

Then statement 3 is the contraposition of the second, and contrapositives are equivalent under excluded middle.

The proof of the equivalence of the third is discussed at closed-projection characterization of compactness.

If the ultrafilter theorem (a weak form of the axiom of choice) is assumed, compactness may be characterized in terms of ultrafilter (or ultranet) convergence:

Definition

XX is compact iff every ultrafilter 𝒰\mathcal{U} (or ultranet ν\nu) on XX converges to some point xXx \in X, meaning that 𝒰\mathcal{U} contains the filter of neighborhoods of xx (or that ν\nu is eventually in any neighbourhood of xx).

In any case, compactness can be characterized in terms of proper filter or equivalently (see at eventuality filter) of net convergence .

Definition

XX is compact iff every proper filter/net on XX has a convergent proper refinement/subnet.

This is equivalent to the characterization given in the Idea-section above:

Definition

XX is compact iff every proper filter 𝒰\mathcal{U} (or net ν\nu) on XX has a cluster point xx, meaning that every element of 𝒰\mathcal{U} meets (has inhabited intersection with) every neighbourhood of xx (or ν\nu is frequently in every neighbourhood of xx).

While the usual definitions (2&1) are for topological spaces, the convergence definitions (35) make sense in any convergence space.

The definition (2) also works for locales, since it refers only to the frame of open sets. An equivalent way to phrase it is

Definition

XX is compact iff given any directed collection of opens whose union is XX (a directed open cover), XX belongs to the collection.

As the union is the coproduct in the category of open subsets Op(X)Op(X), we can also say

Definition

XX is compact iff it is a compact object in Op(X)Op(X).

Compactness is equivalent to the condition of being “stably closed” (and it is this condition which suggests the correct notion of proper map in algebraic geometry and elsewhere):

Definition

(closed-projection characterization of compactness)

XX is compact iff for any space YY, the projection map X×YYX \times Y \to Y out of their Cartesian product is closed (see e.g. Milne, section 17).

Contrary to possible appearance, the equivalence of this with definition 2 does not require the axiom of choice; see this MO question and answers, as well as this page. See also the page compactness and stable closure (under construction). This equivalence is also true for locales, by way of proper maps; see below.

Moreover, this characterization of compact spaces may be used to give an attractive proof of the Tychonoff theorem, due to Clementino and Tholen. See here for details.

Closely related to the previous definition, a logical characterisation of compactness is used in Abstract Stone Duality:

Definition

XX is compact iff for any space YY and any open subset UU of X×YX \times Y, the subset

XU={b:Y|a:X,(a,b)U} \forall_X U = \{ b : Y \;|\; \forall\; a: X,\; (a, b) \in U \}

is open in YY.

To remove it from dependence on points, we can also write the definition like this:

Definition

XX is compact iff given any space YY and any open UU in X×YX \times Y, there exists an open XU\forall_X U in YY that satisfies the universal property of universal quantification:

V XUX×VU V \subseteq \forall_X U \;\Leftrightarrow\; X \times V \subseteq U

for every open VV in YY.

A dual condition is satisfied by an overt space.

Properties

Various

Proposition

(unions and intersection of compact spaces)

Let (X,τ)(X,\tau) be a topological space and let

{K iX} iI \{K_i \subset X\}_{i \in I}

be a set of compact subspaces.

  1. If II is a finite set, then the union iIK iX\underset{i \in I}{\cup} K_i \subset X is itself a compact subspace;

  2. If all K iXK_i \subset X are also closed subsets then their intersection iIK iX\underset{i \in I}{\cap} K_i \subset X is itself a compact subspace.

Proof

Regarding the first statement:

Let {U jX} jJ\{U_j \subset X\}_{j \in J} be an open cover of the union. Then this is also an open cover of each of the K iK_i, hence by their compactness there are finite subsets J iJJ_i \subset J such that {U j iX} j iJ i\{U_{j_i} \subset X\}_{j_i \in J_i} is a finite cover of K iK_i. Accordingly then iI{U j iX} j iJ i\underset{i \in I}{\cup} \{ U_{j_i} \subset X \}_{j_i \in J_i} is a finite open cover of the union.

Regarding the second statement:

By the axioms on a topology, the intersection of an arbitrary set of closed subsets is again closed. Hence the intersection of the closed compact subspaces is closed. But since subsets are closed in a closed subspace precisely if they are closed in the ambient space then for each iIi \in I the intersection is a closed subspace of the compact space K iK_i. Since closed subspaces of compact spaces are compact it follows that the intersection is actually compact, too.

Proposition

(complements of compact with open subspaces is compact)

Let XX be a topological space. Let

  1. KXK\subset X be a compact subspace;

  2. UXU \subset X be an open subset.

Then the complement

KUX K \setminus U \subset X

is itself a compact subspace.

Proposition

Assuming the axiom of choice, the category of compact spaces admits all small limits. In any case, the category of compact locales admits all small limits.

By the Tychonoff theorem, every product topological space of compact spaces is itself again compact.

Proposition

The direct image of a compact subspace under a continuous map is compact. A topological space becomes a bornological set by taking the bounded sets to be subsets contained in some compact subspace, and under this bornology, every continuous function is a bounded map.

If the spaces in question are T 1T_1, then the sets with compact closure also constitute a bornology and continuous maps become bounded. In a non-Hausdorff situation these bornologies might differ because the closure of a compact set need not be compact.

Proposition

A compact Hausdorff space must be normal. That is, the separation axioms T 2T_2 through T 4T_4 (when interpreted as an increasing sequence) are equivalent in the presence of compactness.

Proposition

The Heine-Borel theorem asserts that a subspace S nS \subset \mathbb{R}^n of a Cartesian space is compact precisely if it is closed and bounded.

Proposition

A discrete space is compact iff its underlying set is finite. In constructive mathematics, a discrete space is compact iff its underlying set is Kuratowski-finite.

Relation to compact objects in TopTop

One might expect that compact topological spaces are precisely the compact objects in Top in the abstract sense of category theory, but this identification requires care, the naive version fails. See at compact object – Compact objects in Top

Examples

General

Example

(closed intervals are compact)

For any a<ba \lt b \in \mathbb{R} the closed interval

[a,b] [a,b] \subset \mathbb{R}

regarded with its subspace topology of Euclidean space with its metric topology is a compact topological space.

Proof

Since all the closed intervals are homeomorphic it is sufficient to show the statement for [0,1][0,1]. Hence let {U i[0,1]} iI\{U_i \subset [0,1]\}_{i \in I} be an open cover (def. 1). We need to show that it has an open subcover.

Say that an element x[0,1]x \in [0,1] is admissible if the closed sub-interval [0,x][0,x] is covered by finitely many of the U iU_i. In this terminology, what we need to show is that 11 is admissible.

Observe from the definition that

  1. 0 is admissible,

  2. if y<x[0,1]y \lt x \in [0,1] and xx is admissible, then also yy is admissible.

This means that the set of admissible xx forms either

  1. an open interval [0,g)[0,g)

  2. or a closed interval [0,g][0,g],

for some g[0,1]g \in [0,1]. We need to show that the latter is true, and for g=1g = 1. We do so by observing that the alternatives lead to contradictions:

  1. Assume that the set of admissible values were an open interval [0,g)[0,g). Pick an i 0Ii_0 \in I such that gU i 0g \in U_{i_0} (this exists because of the covering property). Since such U i 0U_{i_0} is an open neighbourhood of gg, there is a positive real number ϵ\epsilon such that the open ball B g (ϵ)U i 0B^\circ_g(\epsilon) \subset U_{i_0} is still contained in the patch. It follows that there is an element xB g (ϵ)[0,g)U i 0[0,g)x \in B^\circ_g(\epsilon) \cap [0,g) \subset U_{i_0} \cap [0,g) and such that there is a finite subset JIJ \subset I with {U i[0,1]} iJI\{U_i \subset [0,1]\}_{i \in J \subset I} a finite open cover of [0,x)[0,x). It follows that {U i[0,1]} iJI{U i 0}\{U_i \subset [0,1]\}_{i \in J \subset I} \sqcup \{U_{i_0}\} were a finite open cover of [0,g][0,g], hence that gg itself were still admissible, in contradiction to the assumption.

  2. Assume that the set of admissible values were a closed interval [0,g][0,g] for g<1g \lt 1. By assumption there would then be a finite set JIJ \subset I such that {U i[0,1]} iJI\{U_i \subset [0,1]\}_{i \in J \subset I} were a finite cover of [0,g][0,g]. Hence there would be an index i gJi_g \in J such that gU i gg \in U_{i_g}. But then by the nature of open subsets in the Euclidean space \mathbb{R}, this U i gU_{i_g} would also contain an open ball B g (ϵ)=(gϵ,g+ϵ)B^\circ_g(\epsilon) = (g-\epsilon, g + \epsilon). This would mean that the set of admissible values includes the open interval [0,g+ϵ)[0,g+ \epsilon), contradicting the assumption.

This gives a proof by contradiction.

In contrast:

Nonexample

For all nn \in \mathbb{N}, n>0n \gt 0, the Euclidean space n\mathbb{R}^n, regarded with its metric topology, is not compact.

This is a special case of the Heine-Borel theorem (example 4 below), but for illustration it is instructive to consider the direct proof:

Proof

Pick any ϵ(0,1/2)\epsilon \in (0,1/2). Consider the open cover of n\mathbb{R}^n given by

{U n(nϵ,n+1+ϵ)× n1 n+1} n. \left\{ U_n \coloneqq (n-\epsilon, n+1+\epsilon) \times \mathbb{R}^{n-1} \;\subset\; \mathbb{R}^{n+1} \right\}_{n \in \mathbb{Z}} \,.

This is not a finite cover, and removing any one of its patches U nU_n, it ceases to be a cover, since the points of the form (n+ϵ,x 2,x 3,,x n)(n + \epsilon, x_2, x_3, \cdots, x_n) are contained only in U nU_n and in no other patch.

Example

By the Tychonoff theorem, every product topological spaces of compact spaces is again compact.

This implies the Heine-Borel theorem, saying that:

Example

(Heine-Borel theorem)

Every bounded and closed subspace of a Euclidean space is compact.

In particular

Nonexample

The open intervals (a,b)(a,b) \subset \mathbb{R} and the half-open intervals [a,b)[a,b) \subset \mathbb{R} and (a,b](a,b] \subset \mathbb{R} are not compact.

Example

The Mandelbrot set, regarded as a subspace of the plane, is a compact space (see this prop.et#MndlbrtIsCompact))

Compact spaces which are not sequentially compact

A famous example of a space that is compact, but not sequentially compact is the product space

{0,1} I{0,1} [0,1][0,1]{0,1} \{0,1\}^{I} \coloneqq \{0, 1\}^{[0, 1]} \coloneqq \underset{[0,1]}{\prod} \{0,1\}

with the product topology.

This space is compact by the Tychonoff theorem.

But it is not sequentially compact. We now discuss why. (We essentially follow Steen-Seebach 70, item 105).

Points of {0,1} I\{0,1\}^{I} are functions I{0,1}I \to \{0,1\}, and the product topology is the topology of pointwise convergence.

Define a sequence (a n)(a_n) in I II^{I} with a n(x)a_n(x) being the nth digit in the binary expansion of xx (we make the convention that binary expansions do not end in sequences of 11s). Let a(a n k)a \coloneqq (a_{n_k}) be a subsequence and define p aIp_a \in I by the binary expansion that has a 00 in the n kn_kth position if kk is even and a 11 if kk is odd (and, for definiteness and to ensure that this fits with our convention, define it to be 00 elsewhere). Then the projection of (a n k)(a_{n_k}) at the p ap_ath coordinate is 1,0,1,0,...1, 0, 1,0,... which is not convergent. Hence (a n k)(a_{n_k}) is not convergent.

(Basically that’s the diagonal trick of Cantor's theorem).

However, as {0,1} I\{0,1\}^I is compact, aa has a convergent subnet. An explicit construction of a convergent subset, given a cluster point aa, is as follows. A function a:I{0,1}a \colon I \to \{0,1\} is a cluster point of (a n)(a_n) if, for any p 1,,p nIp_1, \dots, p_n \in I the set

A(p 1,,p n){k:a k(p i)=a(p i)i} A(p_1,\dots,p_n) \coloneqq \{k \in \mathbb{R} : a_k(p_i) = a(p_i) \forall i\}

is infinite. We index our subnet by the family of finite subsets of II and define the subnet function (I)\mathcal{F}(I) \to \mathbb{N} to be

{p 1,,p n}minA(p 1,,p n) \{p_1,\dots,p_n\} \mapsto \min A(p_1,\dots,p_n)

This is a convergent sub-net.

In synthetic topology

In synthetic topology, where ‘space’ means simply ‘set’ (or type, i.e. the basic objects of our foundational system), one natural notion of “compact space” is a covert set, i.e. a set whose discrete topology is covert. This includes the expected examples in various gros toposes.

Compact spaces and proper maps

A space XX is compact if and only if the unique map X1X\to 1 is proper. Thus, properness is a “relativized” version of compactness.

For topological spaces, this is either a definition of “proper map” (closed with compact fibers) or follows from the above characterization of compactness in terms of projections being closed maps (if proper maps are defined to be those that are universally closed). For locales, it follows from the definition of proper map (a closed map such that f *f_* preserves directed joins) and the fact that compact locales are automatically covert (see covert space for a proof).

References

Examples of compact spaces that are not sequentially compact are in

  • Lynn Steen, J. Arthur Seebach, Counterexamples in Topology, Springer-Verlag, New York (1970) 2nd edition, (1978), Reprinted by Dover Publications, New York, 1995

On compact metric spaces:

  • Roland Speicher, Compact metric spaces (pdf)

  • Alan Sokal, Compactness of metric spaces (pdf)

For proper base change theorem e.g.

Revised on May 18, 2017 04:02:03 by Urs Schreiber (131.220.184.222)