see also algebraic topology, functional analysis and homotopy theory
Basic concepts
topological space (see also locale)
fiber space, space attachment
Extra stuff, structure, properties
Kolmogorov space, Hausdorff space, regular space, normal space
sequentially compact, countably compact, locally compact, sigma-compact, paracompact, countably paracompact, strongly compact
Examples
Basic statements
subsets are closed in a closed subspace precisely if they are closed in the ambient space
closed subspaces of compact Hausdorff spaces are equivalently compact subspaces
open subspaces of compact Hausdorff spaces are locally compact
compact spaces equivalently have converging subnet of every net
paracompact Hausdorff spaces equivalently admit subordinate partitions of unity
injective proper maps to locally compact spaces are equivalently the closed embeddings
locally compact and second-countable spaces are sigma-compact
Theorems
Basic homotopy theory
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 $T_0$ and hence $T_3$ if regular, while every Hausdorff space is $T_3$ if compact) since regularity is easier to formulate and handle than Hausdorffness in locale theory.
There are many ways to say that a space $X$ is compact. The first is perhaps the most common:
Let $(X,\tau)$ be a topological space. Then an open cover an is 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 $X$
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. $\underset{i \in J \subset I}{\cup} U_i = X$.
(compact space)
A topological space is called compact if every open cover has a finite subcover (def. 1).
(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 $n$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).
If excluded middle is assumed, then def. 2 the following reformulations #in terms of closed subsets:
(compactness in terms of closed subsets)
Let $(X,\tau)$ be a topological space. Assuming excluded middle, then the following are equivalent:
$(X,\tau)$ 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: $\underset{i \in J \subset i}{\cap} C_i = \emptyset$.
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, $\underset{i \in I}{\cap} C_i \neq \emptyset$.
For every topological space $(Y,\tau_Y)$ the projection map out of the product topological space $\pi_Y \;\colon\; (X \times Y, \tau_{X \times Y}) \to (Y, \tau_Y)$ is a closed map.
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 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:
$X$ 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 .
$X$ 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:
$X$ 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 (2&1) are for topological spaces, the convergence definitions (3–5) 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
$X$ is compact iff given any directed collection of opens whose union is $X$ (a directed open cover), $X$ belongs to the collection.
As the union is the coproduct in the category of open subsets $Op(X)$, we can also say
$X$ is compact iff it is a compact object in $Op(X)$.
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):
(closed-projection characterization of compactness)
$X$ is compact iff for any space $Y$, the projection map $X \times Y \to Y$ 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:
$X$ is compact iff for any space $Y$ and any open subset $U$ of $X \times Y$, the subset
is open in $Y$.
To remove it from dependence on points, we can also write the definition like this:
$X$ 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:
for every open $V$ in $Y$.
A dual condition is satisfied by an overt space.
(unions and intersection of compact spaces)
Let $(X,\tau)$ be a topological space and let
be a set of compact subspaces.
If $I$ is a finite set, then the union $\underset{i \in I}{\cup} K_i \subset X$ is itself a compact subspace;
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.
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.
(complements of compact with open subspaces is compact)
Let $X$ be a topological space. Let
$U \subset X$ be an open subset.
Then the complement
is itself a compact subspace.
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.
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.
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.
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:
sequentially compact metric spaces are equivalently compact metric spaces
countably compact metric spaces are equivalently compact metric spaces
compact spaces equivalently have converging subnet of every net
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.
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
(closed intervals are compact)
For any $a \lt b \in \mathbb{R}$ the closed interval
regarded with its subspace topology of Euclidean space with its metric topology is a compact topological space.
Since all the closed intervals are homeomorphic it is sufficient to show the statement for $[0,1]$. Hence let $\{U_i \subset [0,1]\}_{i \in I}$ be an open cover (def. 1). We need to show that it has an open subcover.
Say that an element $x \in [0,1]$ is admissible if the closed sub-interval $[0,x]$ is covered by finitely many of the $U_i$. In this terminology, what we need to show is that $1$ is admissible.
Observe from the definition that
0 is admissible,
if $y \lt x \in [0,1]$ and $x$ is admissible, then also $y$ is admissible.
This means that the set of admissible $x$ forms either
an open interval $[0,g)$
or a closed interval $[0,g]$,
for some $g \in [0,1]$. We need to show that the latter is true, and for $g = 1$. We do so by observing that the alternatives lead to contradictions:
Assume that the set of admissible values were an open interval $[0,g)$. Pick an $i_0 \in I$ such that $g \in U_{i_0}$ (this exists because of the covering property). Since such $U_{i_0}$ is an open neighbourhood of $g$, there is a positive real number $\epsilon$ such that the open ball $B^\circ_g(\epsilon) \subset U_{i_0}$ is still contained in the patch. It follows that there is an element $x \in B^\circ_g(\epsilon) \cap [0,g) \subset U_{i_0} \cap [0,g)$ and such that there is a finite subset $J \subset I$ with $\{U_i \subset [0,1]\}_{i \in J \subset I}$ a finite open cover of $[0,x)$. It follows that $\{U_i \subset [0,1]\}_{i \in J \subset I} \sqcup \{U_{i_0}\}$ were a finite open cover of $[0,g]$, hence that $g$ itself were still admissible, in contradiction to the assumption.
Assume that the set of admissible values were a closed interval $[0,g]$ for $g \lt 1$. By assumption there would then be a finite set $J \subset I$ such that $\{U_i \subset [0,1]\}_{i \in J \subset I}$ were a finite cover of $[0,g]$. Hence there would be an index $i_g \in J$ such that $g \in U_{i_g}$. But then by the nature of open subsets in the Euclidean space $\mathbb{R}$, this $U_{i_g}$ would also contain an open ball $B^\circ_g(\epsilon) = (g-\epsilon, g + \epsilon)$. This would mean that the set of admissible values includes the open interval $[0,g+ \epsilon)$, contradicting the assumption.
This gives a proof by contradiction.
In contrast:
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 4 below), but for illustration it is instructive to consider the direct proof:
Pick any $\epsilon \in (0,1/2)$. Consider the open cover of $\mathbb{R}^n$ given by
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.
By the Tychonoff theorem, every product topological spaces of compact spaces is again compact.
This implies the Heine-Borel theorem, saying that:
In particular
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.
The Mandelbrot set, regarded as a subspace of the plane, is a compact space (see this prop.et#MndlbrtIsCompact))
A famous example of a space that is compact, but not sequentially compact is the product space
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
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
This is a convergent sub-net.
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.
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).
sequentially compact topological space, countably compact topological space, paracompact topological space, locally compact topological space, strongly compact topological space, compactly generated topological space
Examples of compact spaces that are not sequentially compact are in
On compact metric spaces:
For proper base change theorem e.g.