Compact spaces

# Compact spaces

## Idea

A topological space (or more generally: a convergence space) is compact if all sequences and more generally nets inside it converge as much as possible.

Compactness is a topological notion that was developed to abstract the key property of a subspace of a Euclidean space being “closed and bounded”: every net must accumulate somewhere in the subspace. (Roughly, the reason is that boundedness implies the net cannot escape the subspace, and the point to which it accumulates lies in the subspace by closure. This is the statement of the Heine-Borel theorem, see there for for more.)

Compactness provides an intrinsic way of formulating this property in the context of general topological spaces, without the need to view them as subspaces of an ambient space. Still, it is common to work with compact subsets of a given space. These are those subsets which are compact spaces with the subspace topology.

One often wishes to study compact Hausdorff spaces (“compacta”) since these enjoy particularly useful properties. For instance they all arise as one-point compactifications (of locally compact Hausdorff spaces, see this remark).

As for most concepts related to topological spaces, there is also a concept of compactness for locales. Observe that (using classical logic) every locally compact locale is aready spatial (this prop.). Instead of compact Hausdorff locales one usually considers compact regular locales, since regularity is easier to formulate and handle than Hausdorffness in locale theory (these are equivalent, since every locale is $T_0$ separated and hence $T_3$ if regular, while every Hausdorff space is $T_3$ if compact).

## Definitions/characterizations

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

###### Definition

(open cover)

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

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

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

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

###### Definition

(compact space)

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

For the purposes of exposition, this definition will be taken as the baseline definition. Throughout the remainder of this section we state a number of propositions of type “A space is compact (in the sense of Definition ) iff it satisfies property P”, which can be read as saying that property P can be taken as an alternative definition of compactness (and may in fact be considered a more convenient or preferable definition over , depending on author and context).

###### 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 at 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 $n$Lab 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).

The various reformulations of compactness fall into several families. Some are more or less tautological reformulations based on direct logical or set-theoretic manipulations of the open covering definition; we give these first. Others express compactness in terms of notions of convergence (nets, filters, ultrafilters). A third family expresses compactness of a space in terms of “stable” properties of maps sourced is that space; this is perhaps the most intrinsically categorical of the three families.

### Elementary reformulations

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

###### Proposition

(compactness in terms of closed subsets)

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

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

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

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

###### 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 $\{U_i \subset X\}_{i \in I}$ be an open cover. Write $C_i \coloneqq X \backslash U_i$ for the corresponding closed complements. By de Morgan's law the condition that $\underset{i \in I}{\cup} U_i = X$ is equivalent to $\underset{i \in I}{\cap} C_i = \emptyset$. The second statement is that there is then a finite subset $J \subset I$ such that also $\underset{i \in J \subset I}{\cap} C_i = \emptyset$, 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 closed-subset formulations of compactness appear frequently and are often more convenient. For example, compactness theorems in model theory draw on a connection between the finite intersection property and finite satisfiability of sets of axioms.

### Compactness for locales

In another direction, the definition () also works for locales, since it refers only to the frame of open sets. Here is an equivalent way to phrase it that is often convenient for locale theory.

###### Proposition

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

###### Proof

For the “only if” direction: if $\mathcal{U}= \{U_i\}_{i \in I}$ is a directed open cover, then by the open-cover definition of compactness, $X$ is the union of finitely many $U_i$. But by definition of directedness, any finite subfamily of $\mathcal{U}$ has an upper bound in $\mathcal{U}$; since the only upper bound of $X$ is $X$, it follows that $X$ belongs to $\mathcal{U}$.

For the “if” direction: given an open cover $\mathcal{U}$ of $X$, let $\mathcal{U}'$ be the family of open sets that are unions of finite subfamilies of $\mathcal{U}$. This $\mathcal{U}$ is clearly directed, and an open cover of $X$ since $\mathcal{U}$ is. By hypothesis, $X$ belongs to $\mathcal{U}'$, so $X$ is a union of finitely many elements of $\mathcal{U}$. This shows $X$ is compact according to Definition .

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

###### Proposition

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

###### Proof

To say $X$ is a compact object means that $\hom(X, -): Op(X) \to Set$ preserves filtered colimits, or colimits of filtered diagrams. Here we may as well replace $Set$ by the full subcategory consisting of $0, 1$ (a $0$-element set or a $1$-element set) since the hom-sets for posets are of size $0$ or $1$. Further, in a poset like $Op(X)$, a filtered diagram is the same as a nonempty directed diagram $D$, and $\hom(X, \bigcup_{d \in D} d)$ has size $1$ iff $X = \bigcup_{d \in D} d$. On the other hand, $\bigcup_{d \in D} \hom(X, d)$ has size $1$ iff $X = d$ for some $d$ in $D$. Thus that $\hom(X, -)$ preserves filtered colimits amounts to the same thing as saying $X$ belongs to any directed open cover $D$, which is the same as compactness by Proposition .

### Compactness via convergence

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:

###### Proposition

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

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

###### Proposition

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

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

###### Proposition

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

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

### Compactness via completeness

A uniform space $X$ is compact if and only if it is complete and totally bounded. Moreover, a compact Hausdorff topological space has a unique compatible uniformity, which is complete and totally bounded.

In constructive mathematics, “complete and totally bounded” is sometimes taken as a substitute for open-cover compactness (to which it is no longer equivalent); see Bishop-compact space.

### Compactness via stability properties

Frequently in category theory, for example when we discuss internal logic in toposes, we are interested in properties of maps that are stable under pullback, and it turns out that compactness can be reformulated in terms of stability properties.

A first example concerns the property of a topological space $X$ that the unique map $!: X \to 1$ to the point space (the terminal object in Top) is a closed map. As a statement in ordinary point-set topology, this is plainly a tautology, trivially true for any space $X$. (Side remark: it is not at all a tautology in the more general setting of internal locales in toposes; the word “closed” is reserved for locales $X$ having that property. See also closed morphism.) However, even in ordinary point-set topology, $!: X \to 1$ is usually not stably closed. In more detail: the pullback of $!: X \to 1$ along a (or the) map $Y \to 1$ is the projection map

$X \times Y \to Y$

out of the product topological space and the issue is whether this map is closed. This leads us to the following proposition.

###### Proposition

(closed-projection characterization of compactness)

A topological space $X$ is compact (def. ) precisely if for any topological space $Y$, the projection map $X \times Y \to Y$ out of their product topological space is closed.

Thus, compactness of $X$ is equivalent to $!: X \to 1$ being stably closed. For a proof, see closed-projection characterization of compactness.

###### Remark

Contrary to possible appearance, the equivalence in prop. 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, the characterization of compact spaces via prop. may be used to give an attractive proof of the Tychonoff theorem, due to Clementino and Tholen. See here for details.

Of course, the notion of being stably closed applies to maps $p: X \to Y$ besides $!: X \to 1$. An analysis of this notion (that the pullback $f^\ast p: X \times_Y Z \to Z$ is a closed map for every map $f: Z \to Y$) leads to the correct notion of proper map in algebraic geometry and elsewhere.

Closely related to closed-projection characterization of compactness, a characterisation of compactness in terms of logical quantification is featured in Paul Taylor‘s Abstract Stone Duality:

###### Proposition

$X$ is compact iff for any space $Y$ and any open subset $U$ of $X \times Y$, the subset

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

is open in $Y$.

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

###### Definition

$X$ is compact iff given any space $Y$ and any open $U$ in $X \times Y$, there exists an open $\forall_X U$ in $Y$ that satisfies the universal property of universal quantification:

$V \subseteq \forall_X U \;\Leftrightarrow\; X \times V \subseteq U$

for every open $V$ in $Y$.

A dual condition is satisfied by an overt space.

## Properties

### Various

###### Proposition

(unions and intersection of compact spaces)

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

$\{K_i \subset X\}_{i \in I}$

be a set of compact subspaces.

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

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

###### Proof

Regarding the first statement:

Let $\{U_j \subset X\}_{j \in J}$ be an open cover of the union. Then this is also an open cover of each of the $K_i$, hence by their compactness there are finite subsets $J_i \subset J$ such that $\{U_{j_i} \subset X\}_{j_i \in J_i}$ is a finite cover of $K_i$. Accordingly then $\underset{i \in I}{\cup} \{ U_{j_i} \subset X \}_{j_i \in J_i}$ 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 $i \in I$ the intersection is a closed subspace of the compact space $K_i$. 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 $X$ be a topological space. Let

1. $K\subset X$ be a compact subspace;

2. $U \subset X$ be an open subset.

Then the complement

$K \setminus U \subset Xcov$

is itself a compact subspace.

###### Proof

Let $\{V_i \subset K \setminus U\}$ be an open cover of the complement subspace. We need to show that this admits a finite subcover.

Observe that by definition of the subspace topologies on $K \setminus U$ and on $K$:

1. the $V_i \subset K \setminus U$ are still open when regarded as subsets $V_i \subset K$ of $K$;

2. also the intersection $K \cap U \subset K$ is an open subset of the compact subspace $K$.

This means that

$\{ V_i \subset K \}_{i \in I} \sqcup (U \cap K)$

is an open cover of $K$. Hence by the compactness of $K$, this has a finite subcover. If $U \cap K$ is retained in this subcover then it is of the form

$\{ V_j \subset K\}_{j \subset J} \sqcup (U \cap K)$

otherwise it is of the form

$\{ V_j \subset K\}_{j \subset J}$

with $J \subset I$ a finite subset.

Since either of these is still a cover, also their restriction from $K$ to $K \setminus U$ is a cover of $K \setminus U$. But by assumption the $V_j$ are already restricted to $K \setminus U$, while the restriction of $(U \cap K)$ to $K \setminus U$ is empty. Therefore in either of the above cases we find that

$\{ V_j \subset K \setminus U\}_{j \subset J}$

is a finite subcover of the original open cover of $K \setminus U$. Therefore $K \setminus U$ is compact.

###### Proposition

Assuming the axiom of choice, the category of compact spaces admits all small products. And, even without the axiom of choice, the category of compact locales admits all small products.

The first statement follows from the Tychonoff theorem: every product topological space of compact spaces is itself again compact.

###### Remark

However, the category of compact spaces does not admit equalizers. An explicit example is where one takes two maps $[0, 1] \to \{0, 1\}$ where the codomain is Sierpinski space, where one of the maps $f$ has $f^{-1}(1) = [0, 1/2)$, and the other is the constant map at $1$. If an equalizer in $Comp$ existed, then $\hom(1, -): Comp \to Set$ would have to preserve it, so set-theoretically it would have to be $[0, 1/2)$. The topology on $[0, 1/2)$ would then have to be the same as or finer than the subspace topology in order for the equalizer map to be continuous. But if the subspace topology isn’t compact, then no finer topology would make it compact either. (Here we are invoking the contrapositive of the proposition that if $(X, \tau)$ is compact and $\tau' \subseteq \tau$ is a coarser topology, then $(X, \tau')$ is also compact.)

On the other hand, the category of compact Hausdorff spaces does admit all limits, and this fact does not require the full strength of the axiom of choice, but only a weaker choice principle called the ultrafilter principle.

###### 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 $T_1$, 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 $T_2$ through $T_4$ (when interpreted as an increasing sequence) are equivalent in the presence of compactness.

###### Proposition

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

We have:

Hence:

###### Proposition

continuous images of compact spaces are compact

### Relation to compact objects in $Top$

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

###### 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.

###### Example

(closed intervals are compact)

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

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

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

In contrast:

###### Nonexample

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

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

###### Proof

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

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

This is not a finite cover, and removing any one of its patches $U_n$, it ceases to be a cover, since the points of the form $(n + \epsilon, x_2, x_3, \cdots, x_n)$ are contained only in $U_n$ 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 $(a,b) \subset \mathbb{R}$ and the half-open intervals $[a,b) \subset \mathbb{R}$ and $(a,b] \subset \mathbb{R}$ are not compact.

###### Example

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

###### Example

Any set when equipped with the cofinite topology forms a compact space. This space is also $T_1$ (i.e., singletons are closed), but is not Hausdorff if the underlying set is [infinite set|infinite]].

### Compact spaces which are not sequentially compact

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

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

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

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

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

However, as $\{0,1\}^I$ is compact, $a$ has a convergent subnet. An explicit construction of a convergent subset, given a cluster point $a$, is as follows. A function $a \colon I \to \{0,1\}$ is a cluster point of $(a_n)$ if, for any $p_1, \dots, p_n \in I$ the set

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

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

$\{p_1,\dots,p_n\} \mapsto \min A(p_1,\dots,p_n)$

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 $X$ is compact if and only if the unique map $X\to 1$ 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 $f_*$ preserves directed joins) and the fact that compact locales are automatically covert (see covert space for a proof).

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.