Contents

# Contents

## Idea

A compact Hausdorff space or compactum, for short, is a topological space which is both a Hausdorff space as well as a compact space. This is precisely the kind of topological space in which every limit of a sequence or more generally of a net that should exist does exist (this prop.) and does so uniquely (this prop).

One may consider the analogous condition for convergence spaces, or for locales (see also at Hausdorff locale and compact locale). Even though these are all different contexts, the resulting notion of compactum is (at least assuming the axiom of choice) always the same. Interestingly, there is even an algebraic definition, not one that uses only finitary operations, but one which uses a monad.

## Definitions

If you know what a compact space is and what a Hausdorff space is, then you know what a compact Hausdorff space is, so let's be fancy. (Full justifications will be provided in section on compacta as algebras.)

Given a set $S$, let $\beta S$ be the set of ultrafilters on $S$. Note that $\beta$ is an endofunctor on Set; every function $f: S \to T$ induces a function $\beta f: \beta S \to \beta T$ using the usual application of functions to filters. In fact, $\beta$ is a monad; it comes with a natural (in $S$) unit $\eta_S: S \to \beta S$, which maps a point $x$ to the principal ultrafilter that $x$ generates, and multiplication $\mu_S: \beta \beta S \to \beta S$, which maps an ultrafilter $U$ on ultrafilters to the ultrafilter of sets whose principal ultrafilters of ultrafilters belong to $U$. That is,

• $A \in \eta x \;\Leftrightarrow\; x \in A$, so $\eta x = \{ A \subseteq S \;|\; x \in A \}$;
• $A \in \mu U \;\Leftrightarrow\; \{ F \in \beta S \;|\; A \in F \} \in U$.

Then a compactum is simply an algebra for this monad; that is, a set $X$ together with a function $\lim: \beta X \to X$, such that

• each point $x$ is the limit ($\lim$) of the principal ultrafilter $\eta x$, and
• given an ultrafilter $U$ on ultrafilters, the limit of $\mu U$ is the limit of $(\beta \lim) U$.

It is then a theorem that this $\lim$ generates a convergence on $S$ that is compact, Hausdorff, and topological. The converse, that every compact Hausdorff topological convergence is of this form, is equivalent to the ultrafilter principle.

Every compact Hausdorff space is regular and sober and so defines a compact regular locale. Again, the axiom of choice gives us a converse: every compact regular locale is spatial and so comes from a compactum.

Probably this is also equivalent to the ultrafilter principle, but I need to check.

Note that every compact Hausdorff space (topological or localic) is not only regular but also normal. See compact Hausdorff spaces are normal.

## Compacta as algebras

Throughout this section, $CH$ will be used to denote the category of compact Hausdorff spaces (compacta).

### The space of ultrafilters

Let $Bool$ be the category of Boolean algebras. The functor $\hom(-, \mathbf{2}): Bool^{op} \to Set$ has a left adjoint $P: Set \to Bool^{op}$ given by power sets, and we define the ultrafilter monad to be the composite $\beta \coloneqq \hom(P -, \mathbf{2})$.

For a set $S$, topologize $\beta S$ by declaring a basic open set to be one of the form

$\hat{A} \coloneqq \{F \in \beta S: A \in F\}$

for $A$ a subset of $S$. Notice $\hat{\emptyset}$ is empty. Indeed, $\widehat{(-)}$ defines a Boolean algebra map

$P(S) \to P(\beta S)$

so that in particular $\widehat{A \cap B} = \hat{A} \cap \hat{B}$, which immediately implies that the $\hat{A}$ form a basis of a topology.

###### Remark

In fact, $\widehat{(-)}$ is the component at $P(S)$ of the counit of the adjunction $P \dashv \hom(-, \mathbf{2})$, as a morphism in $Bool^{op}$. If $f: S \to T$ is a function, the commutativity of the naturality square

$\array{ P(T) & \stackrel{\widehat{(-)}}{\to} & P(\beta T) \\ \mathllap{P(f)} \downarrow & & \downarrow \mathrlap{P(\beta f)} \\ P(S) & \overset{\widehat{(-)}}{\to} & P(\beta S) }$

implies that if $U = \hat{B} \in P(\beta T)$ is a basic (cl)open, then so is $(\beta f)^{-1}(U) = P(\beta f)(\hat{B}) = \widehat{f^{-1}(B)}$. It then follows that $\beta f$ is continuous.

These results show that the monad $\beta: Set \to Set$ lifts through the forgetful functor $U: Top \to Set$.

The unit of the monad $\beta$ is given componentwise by functions

$prin_X: X \to \beta X$

where $prin_X$ takes $x \in X$ to the principal ultrafilter

$prin_X(x) = \{A \in P(X): x \in A\}.$

It is evident that $prin_X$ is injective.

###### Proposition

The injection $prin_X: X \to \beta X$ exhibits $X$ as a dense subset of $\beta X$.

###### Proof

If $\hat{A}$ is a basic open neighborhood containing an ultrafilter $F$, then $A$ is nonempty and hence contains some $x \in X$, which is to say $A \in prin_X(x)$ or that $prin_X(x) \in \hat{A}$.

### Ultrafilters form a compactum

###### Proposition

$\beta S$ is Hausdorff.

###### Proof

Let $F, G$ be distinct ultrafilters, so there is $A \subseteq S$ with $A \in F$ and $\neg A \in G$. Then $\hat{A}$ and $\widehat{\neg A}$ are disjoint neighborhoods which contain $F$ and $G$ respectively.

###### Proposition

$\beta S$ is compact.

###### Proof

It is enough to show that if $\mathcal{O}$ is a collection of opens such that the union of any finite subcollection is a proper subset, then the union of $\mathcal{O}$ is also proper.

If $\mathcal{O}$ covers $U$, it admits a refinement by basic clopens also covering $U$, and thus we may assume WLOG that $\mathcal{O}$ consists of basic clopens $\hat{A}$. If every finite union of elements of $\mathcal{O}$ is a proper subset of $\beta S$, then every finite intersection $\widehat{\neg A_1} \cap \ldots \cap \widehat{\neg A_n}$ is nonempty, so that the $\neg A$ generate a filter, which is contained in some ultrafilter $F$. This $F$ lies outside the union of all the $\hat{A}$‘s.

### Convergence

###### Definition

Let $X$ be a topological space, and $F$ an ultrafilter on the underlying set $U X$. We say $F$ converges to a point $x$ (in symbols, $F \rightsquigarrow x$) if the neighborhood filter $N_x$ of $x$ is contained in $F$.

Convergence defines a relation $\xi$ from $\beta(U X)$ to $U X$.

###### Proposition

If $X$ is Hausdorff, then the relation $\xi$ is well-defined, or functional (i.e., there is at most one point to which a given ultrafilter $F$ converges).

###### Proof

If $x \neq y$, then there are disjoint neighborhoods $U$, $V$ of $x$ and $y$. We cannot have both $U \in F$ and $V \in F$ (otherwise $\emptyset = U \cap V$ would be an element of $F$), so at most one of the neighborhood filters $N_x, N_y$ can be contained in $F$.

###### Proposition

If $X$ is compact, then the relation $\xi$ is total (i.e., there exists a point to which a given ultrafilter $F$ converges).

###### Proof

If not, then for each $x \in X$ there is an open neighborhood $U_x$ that does not belong to $F$. Then $\neg U_x \in F$. Some finite number of neighborhoods $U_{x_1}, \ldots, U_{x_n}$ covers $X$. Then $\neg U_{x_1} \cap \ldots \cap \neg U_{x_n} = \emptyset \in F$, which is a contradiction.

###### Proposition

If $X$ is compact Hausdorff, then the function $\xi: \beta(U X) \to X$ is continuous.

###### Proof

Let $U$ be an open neighborhood of $x \in X$; we must show that $\xi^{-1}(U)$ contains an open neighborhood of any of its points (i.e., ultrafilters $F$ such that $F \rightsquigarrow x$). Since $X$ is $T_3$ (Hausdorff regular), we may choose a neighborhood $V \in N_x$ whose closure $\bar{V}$ is contained in $U$. Then $\hat{V}$ is an open neighborhood of $F$ in $\beta(U X)$, and we claim $\hat{V} \subseteq \xi^{-1}(U)$.

For this, we must check that if $G \in \hat{V}$ and $G \rightsquigarrow y$, then $y \in U$. But if $y \in \neg U \subseteq \neg \bar{V}$, then $\neg \bar{V} \in N_y$, whence $G \rightsquigarrow y$ implies $\neg \bar{V} \in G$. This contradicts $G \in \hat{V}$, i.e., contradicts $V \in G$, since $V \cap \neg \bar{V} = \emptyset$.

### Spaces of ultrafilters are universal

###### Proposition

If $S$ is a set and $X$ is a compact Hausdorff space, then any function $f: S \to X$ can be extended (along $prin_S: S \to \beta S$) to a continuous function $\hat{f}: \beta S \to X$.

###### Proof

We define $\hat{f}$ to be the composite

$\beta S \stackrel{\beta(f)}{\to} \beta (U X) \stackrel{\xi}{\to} X$

where $\beta(f)$ is continuous by Remark and $\xi$ is continuous by Proposition . It remains to check that the following diagram is commutative:

$\array{ S & \stackrel{f}{\to} & U X & & \\ \mathllap{prin_S} \downarrow & & \mathllap{prin_{U X}} \downarrow & \searrow \mathrlap{1_{U X}} & \\ \beta S & \underset{\beta (f)}{\to} & \beta (U X) & \underset{\xi}{\to} & U X. }$

The square commutes by naturality of $prin$, and commutativity of the triangle simply says that the ultrafilter $prin_{U X}(x)$ converges to $x$, or that $N_x \subseteq prin(x)$, which reduces to the tautology that $x \in V$ for every neighborhood $V \in N_x$.

###### Theorem

For any set $S$, the function $prin_S: S \to \beta S$ is universal among functions from $S$ to compact Hausdorff spaces. Hence the functor $F: Set \to CH$ that takes $S$ to the compact Hausdorff space $\beta S$ is left adjoint to the forgetful functor $CH \to Set$.

###### Proof

Proposition shows that for any function $f: S \to U X$ to a compact Hausdorff space, there exists continuous $g: \beta S \to X$ such that $g \circ prin_S = f$. All that remains is to establish uniqueness of such $g$. But if two maps $g, g': \beta S \to X$ to a Hausdorff space $X$ agree on a dense subspace, in this case the subspace $prin_S : S \hookrightarrow \beta S$ by Proposition , then they must be equal. Indeed, the pullback of the closed diagonal defines a closed subspace $D$ of $\beta S$,

$\array{ D & \to & X \\ \downarrow & & \downarrow \mathrlap{\delta_X} \\ \beta S & \underset{\langle g, g' \rangle}{\to} & X \times X, }$

and $D$ contains a dense subspace $S$, therefore $D = \beta S$; i.e., the equalizer of $g$ and $g'$ is all of $\beta S$, hence these two maps are equal.

### Compact Hausdorff spaces are monadic over sets

We recall hypotheses of Beck’s precise monadicity theorem: a functor $U: C \to D$ is monadic if and only if

1. $U$ has a left adjoint,

2. $U$ reflects isomorphisms: a morphism $f: X \to Y$ of $C$ is an isomorphism if $U f: U X \to U Y$ is an isomorphism in $D$,

3. $D$ has, and $U$ preserves, coequalizers of parallel pairs that are $U$-split. (We say

$X \stackrel{\overset{f}{\to}}{\underset{g}{\to}} Y$

is $U$-split if there is a coequalizer

$U X \stackrel{\overset{U f}{\to}}{\underset{U g}{\to}} U Y \stackrel{h}{\to} Z$

that is split in $D$: there exists $i: Z \to U Y$ and $j: U Y \to U X$ such that $U h \circ i = 1_Z$, $U g \circ j = i \circ h$, and $U f \circ j = 1_{U Y}$.)

In the case where $D = Set$, we have the following useful lemma:

###### Lemma

Suppose given a coequalizer in $Set$

$X \stackrel{\overset{f}{\to}}{\underset{g}{\to}} Y \stackrel{h}{\to} Z$

split by $i: Z \to Y$, $j: Y \to X$ (so that $f j = 1_Y$, $h i = 1_Z$, $g j = i h$). Let $R \hookrightarrow Y \times Y$ be the image of $\langle f, g \rangle : X \to Y \times Y$, and let $\langle p_1, p_2 \rangle : E \to Y \times Y$ be the equivalence relation given by the kernel pair $(p_1, p_2)$ of $h$. Then $E = R \cdot R^{op}$, the relational composite given by taking the image of the span composite $R \times_Y R^{op} \to Y \times Y$.

###### Proof

Clearly $R \subseteq E$ since $h f = h g$, and we have $R^{op} \subseteq E^{op} = E$ and $R \cdot R^{op} \subseteq E \cdot E \subseteq E$ by symmetry and transitivity of $E$. In the other direction, suppose $(y_1, y_3) \in E$, so that $h(y_1) = h(y_3)$. Put $x = j(y_1)$ and $x' = j(y_3)$ (so that $f(x) = y_1$ and $f(x') = y_3$), and put $y_2 = g(x)$. Clearly then $(y_1, y_2) \in R$. Moreover,

$y_2 = g(x) = g j(y_1) = i h(y_1) = i h(y_3) = g j(y_3) = g(x')$

so that $(y_3, y_2) \in R$, or $(y_2, y_3) \in R^{op}$. Hence $(y_1, y_3) \in R \cdot R^{op}$, and we have shown $E \subseteq R \cdot R^{op}$.

###### Theorem

The forgetful functor $U: CH \to Set$ is monadic.

###### Proof

By theorem , $U$ has a left adjoint. Since bijective continuous maps between compact Hausdorff spaces are homeomorphisms, we have that $U$ reflects isomorphisms. Finally, suppose $(f, g)$ is a $U$-split pair of morphisms $X \to Y$ in $CH$; let $h: Y \to Z$ be their coequalizer in $Top$, given by a suitable quotient space. Being a quotient of a compact space, $Z$ is compact. Since $CH$ is a full subcategory of $Top$, the map $h$ is a coequalizer in $CH$ once we prove the following claim:

• Claim: $Z$ is Hausdorff.

Furthermore, since the forgetful functor $Top \to Set$ has a right adjoint (given by taking indiscrete topologies on sets), the underlying function of $h$ (again denoted $h$) is the coequalizer of $(U f, U g)$ in $Set$, so that $U$ would preserve the claimed coequalizer.

In other words, to complete the proof, it suffices to verify the claim. Letting $p_0, p_1: E \stackrel{\to}{\to} Y$ be the kernel pair of $h$ in $Top$, to show $Z = Y/E$ is Hausdorff, it suffices to prove that the equivalence relation $\langle p_0, p_1 \rangle : E \to Y \times Y$ in $Top$ is closed. Let $R \hookrightarrow Y \times Y$ be the image of $\langle f, g \rangle: X \to Y \times Y$. By lemma , the subset $E$ of $U Y \times U Y$ coincides with the subset $R \cdot R^{op} \subseteq U Y \times U Y$. Now $R$ is the image of the compact space $X$ under the continuous map $\langle f, g \rangle$, so $R$ is a closed subset of $Y \times Y$. Similarly $R^{op}$ is a closed subset of $Y \times Y$. Under their subspace topologies, their fiber product $R \times_Y R^{op}$ is compact, and so its image $R \cdot R^{op}$ under the (continuous) span composite $R \times_Y R^{op} \to Y \times Y$ is also closed. This completes the proof.

## Weak versions

Every Hausdorff space, hence every compactum, satisfies the separation axiom $T_0$. As is usual with separation axioms, we can also look for a non-$T_0$ version. A priori, this is a compact preregular space; however, since every such space is regular, we can speak instead of a compact regular space.

In the absence of the axiom of choice, and especially in constructive mathematics, the best definition of compactum seems to be a compact regular locale. That is, it is the category of compact regular locales that has all of the nice properties, forming a nice category of spaces, and that has the desired examples, such as the unit interval. (See the discussion at Tychonoff theorem for an example of how the category of compact Hausdorff topological spaces might fail to be nice; see Frank Waaldijk’s PhD thesis (pdf) for a thorough discussion of what is needed to make the unit interval a compact Hausdorff topological space.)

The monadic definition, in particular, falls quite flat without some form of the axiom of choice; even excluded middle and COSHEP are powerless here. In fact, it is quite consistent to assume that every ultrafilter is principal (a strong denial of the ultrafilter principle), in which case $\beta$ is the identity monad. Then a compactum would be just a set if that were the definition used.

On the other hand, it is the monadic definition that gives an algebraic category with a nice relationship to Set. Without the ultrafilter principle, there is no reason to think that the set-of-points functor from compact regular locales to sets is even continuous.

## Properties

### Stone–Čech compactification

By general nonsense, every $\beta S$, regarded as a free $\beta$-algebra, is a compactum, and the functor

$\beta: Set \to Comp$

is left adjoint to the forgetful functor $Comp \to Set$. Assuming the ultrafilter principle, this functor extends to a functor $\beta: Top \to Comp$ (identifying a set with its discrete space) that is left adjoint to the forgetful functor $Comp \to Top$. This is the Stone–Čech compactification? functor (N.B.: for many authors, Stone–Čech compactification refers to the restriction of this functor to Tychonoff spaces $X$, which are precisely those spaces where the unit $X \to \beta X$ is an embedding so that we have a compactification in the technical sense).

A classical construction of the Stone–Čech compactification starts with the unit interval $I =[0, 1]$ and proceeds to the codensity monad induced from the functor

$\hom(-, I) \colon Top^{op} \to Set.$

The monad is given on objects by $X \mapsto I^{\hom(X, I)}$; this lands in compact Hausdorff spaces under the ultrafilter principle. Let $\bar{X}$ be the closure of the image of the unit $u_X: X \to I^{\hom(X, I)}$; this $\bar{X}$ is compact Hausdorff.

###### Proposition

If $X$ is a Tychonoff space, then the unit $u_X: X \to I^{\hom(X, I)}$ is a subspace embedding, so that $I$ is a cogenerator in the category of Tychonoff spaces. (In particular, $u_C$ is an embedding if $C$ is compact Hausdorff, so $I$ is also a cogenerator in the category of compact Hausdorff spaces.)

The proof is essentially Urysohn's lemma; see also related discussion at Tychonoff space and at uniform space (noting that compact Hausdorff spaces are uniform spaces for a unique uniformity).

###### Theorem

The natural map $i_X: X \to \bar{X}$ is universal among maps from $X$ to compact Hausdorff spaces, thus giving a left adjoint $Top \to Comp$ to the (fully faithful) forgetful functor $U: Comp \to Top$.

###### Proof

Let $f: X \to C$ be a map, where $C$ is a compact Hausdorff space. Since $I$ is a cogenerator in the category of compact Hausdorff spaces, the unit for the codensity monad $M_I$,

$u_C: C \to I^{\hom(C, I)},$

is a continuous injection (and hence a closed subspace embedding, since $C$ is compact Hausdorff). Let $\hat{f} = M_I(f)$, and consider the pullback square

$\array{ \hat{f}^{-1}(C) & \to & I^{\hom(X, I)} \\ \mathllap{\pi} \downarrow & & \downarrow \mathrlap{\hat{f}} \\ C & \underset{u_C}{\to} & I^{\hom(C, I)}. }$

From an evident naturality square for the unit $u$, we have a map $h: X \to \hat{f}^{-1}(C)$, i.e., the map $u_X: X \to I^{\hom(X, I)}$ factors through the closed subspace $\hat{f}^{-1}(C) \hookrightarrow I^{\hom(X, I)}$. Therefore $h$ factors as

$X \stackrel{i_X}{\to} \bar{X} \subseteq \hat{f}^{-1}(C)$

and since $\pi \circ h = f$, we conclude that $f$ factors through $i_X$. And moreover, there is at most one $k: \bar{X} \to C$ such that $k \circ i_X = f$, because $i_X$ maps $X$ onto a dense subspace of $\bar{X}$, and dense subspaces are epic in the category of Hausdorff spaces. This completes the proof.

We have a similar Stone–Čech compactification functor $Loc \to Comp$; we do not need the ultrafilter principle here if $Comp$ is defined in terms of locales.

### Category of compacta

The category $Comp$ of compact Hausdorff spaces and continuous maps is

From the first two properties, it follows that $Comp$ is a pretopos, meaning that $Comp$ enjoys the same finitary exactness properties that hold in a topos; in particular, first-order intuitionistic logic may be enacted within $Comp$.

Last revised on March 26, 2018 at 08:14:47. See the history of this page for a list of all contributions to it.