compactum (changes) in nLab

Context

Topology

• Idea
• Definitions
• Compacta as algebras
  • The space of ultrafilters
  • Ultrafilters form a compactum
  • Convergence
  • Spaces of ultrafilters are universal
  • Compact Hausdorff spaces are monadic over sets
• Weak versions
• Properties
  • General
  • Stone–Čech compactification
  • Category of compacta
  • Relation to compactly generated topological spaces
• Related concepts
• References

Proposition

The injection prin X:X→βXprin_X: X \to \beta X exhibits XX as a dense subset of βX\beta X.

Proof

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

Proof

Let F,GF, G be distinct ultrafilters, so there is A⊆SA \subseteq S with A∈FA \in F and ¬A∈G\neg A \in G. Then A^\hat{A} and ¬A^\widehat{\neg A} are disjoint neighborhoods which contain FF and GG respectively.

Proof

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

If 𝒪\mathcal{O} covers UU, it admits a refinement by basic clopens also covering UU, and thus we may assume WLOG that 𝒪\mathcal{O} consists of basic clopens A^\hat{A}. If every finite union of elements of 𝒪\mathcal{O} is a proper subset of βS\beta S, then every finite intersection ¬A 1^∩…∩¬A n^\widehat{\neg A_1} \cap \ldots \cap \widehat{\neg A_n} is nonempty, so that the ¬A\neg A generate a filter, which is contained in some ultrafilter FF. This FF lies outside the union of all the A^\hat{A}‘s.

Definition

Let XX be a topological space, and FF an ultrafilter on the underlying set UXU X. We say FF converges to a point xx (in symbols, F⇝xF \rightsquigarrow x) if the neighborhood filter N xN_x of xx is contained in FF.

Proposition

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

Proof

If x≠yx \neq y, then there are disjoint neighborhoods UU, VV of xx and yy. We cannot have both U∈FU \in F and V∈FV \in F (otherwise ∅=U∩V\emptyset = U \cap V would be an element of FF), so at most one of the neighborhood filters N x,N yN_x, N_y can be contained in FF.

Proposition

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

Proof

If not, then for each x∈Xx \in X there is an open neighborhood U xU_x that does not belong to FF. Then ¬U x∈F\neg U_x \in F. Some finite number of neighborhoods U x 1,…,U x nU_{x_1}, \ldots, U_{x_n} covers XX. Then ¬U x 1∩…∩¬U x n=∅∈F\neg U_{x_1} \cap \ldots \cap \neg U_{x_n} = \emptyset \in F, which is a contradiction.

Proposition

If XX is compact Hausdorff, then the function ξ:β(UX)→X\xi: \beta(U X) \to X is continuous.

Proof

Let UU be an open neighborhood of x∈Xx \in X; we must show that ξ −1(U)\xi^{-1}(U) contains an open neighborhood of any of its points (i.e., ultrafilters FF such that F⇝xF \rightsquigarrow x). Since XX is T 3T_3 (Hausdorff regular), we may choose a neighborhood V∈N xV \in N_x whose closure V¯\bar{V} is contained in UU. Then V^\hat{V} is an open neighborhood of FF in β(UX)\beta(U X), and we claim V^⊆ξ −1(U)\hat{V} \subseteq \xi^{-1}(U).

For this, we must check that if G∈V^G \in \hat{V} and G⇝yG \rightsquigarrow y, then y∈Uy \in U. But if y∈¬U⊆¬V¯y \in \neg U \subseteq \neg \bar{V}, then ¬V¯∈N y\neg \bar{V} \in N_y, whence G⇝yG \rightsquigarrow y implies ¬V¯∈G\neg \bar{V} \in G. This contradicts G∈V^G \in \hat{V}, i.e., contradicts V∈GV \in G, since V∩¬V¯=∅V \cap \neg \bar{V} = \emptyset.

Proposition

If SS is a set and XX is a compact Hausdorff space, then any function f:S→Xf: S \to X can be extended (along prin S:S→βSprin_S: S \to \beta S) to a continuous function f^:βS→X\hat{f}: \beta S \to X.

Proof

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

βS→β(f)β(UX)→ξX\beta S \stackrel{\beta(f)}{\to} \beta (U X) \stackrel{\xi}{\to} X

where β(f)\beta(f) is continuous by Remark 1 and ξ\xi is continuous by Proposition 6. It remains to check that the following diagram is commutative:

S →f UX prin S↓ prin UX↓ ↘1 UX βS →β(f) β(UX) →ξ UX.\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 prinprin, and commutativity of the triangle simply says that the ultrafilter prin UX(x)prin_{U X}(x) converges to xx, or that N x⊆prin(x)N_x \subseteq prin(x), which reduces to the tautology that x∈Vx \in V for every neighborhood V∈N xV \in N_x.

Theorem

For any set SS, the function prin S:S→βSprin_S: S \to \beta S is universal among functions from SS to compact Hausdorff spaces. Hence the functor F:Set→CHF: Set \to CH that takes SS to the compact Hausdorff space βS\beta S is left adjoint to the forgetful functor CH→SetCH \to Set.

Proof

Proposition 7 shows that for any function f:S→UXf: S \to U X to a compact Hausdorff space, there exists continuous g:βS→Xg: \beta S \to X such that g∘prin S=fg \circ prin_S = f. All that remains is to establish uniqueness of such gg. But if two maps g,g′:βS→Xg, g': \beta S \to X to a Hausdorff space XX agree on a dense subspace, in this case the subspace prin S:S↪βSprin_S : S \hookrightarrow \beta S by Proposition 1, then they must be equal. Indeed, the pullback of the closed diagonal defines a closed subspace DD of βS\beta S,

D → X ↓ ↓δ X βS →⟨g,g′⟩ X×X,\array{ D & \to & X \\ \downarrow & & \downarrow \mathrlap{\delta_X} \\ \beta S & \underset{\langle g, g' \rangle}{\to} & X \times X, }

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

Lemma

Suppose given a coequalizer in SetSet

X→g→fY→hZX \stackrel{\overset{f}{\to}}{\underset{g}{\to}} Y \stackrel{h}{\to} Z

split by i:Z→Yi: Z \to Y, j:Y→Xj: Y \to X (so that fj=1 Yf j = 1_Y, hi=1 Zh i = 1_Z, gj=ihg j = i h). Let R↪Y×YR \hookrightarrow Y \times Y be the image of ⟨f,g⟩:X→Y×Y\langle f, g \rangle : X \to Y \times Y, and let ⟨p 1,p 2⟩:E→Y×Y\langle p_1, p_2 \rangle : E \to Y \times Y be the equivalence relation given by the kernel pair (p 1,p 2)(p_1, p_2) of hh. Then E=R⋅R opE = R \cdot R^{op}, the relational composite given by taking the image of the span composite R× YR op→Y×YR \times_Y R^{op} \to Y \times Y.

Proof

Clearly R⊆ER \subseteq E since hf=hgh f = h g, and we have R op⊆E op=ER^{op} \subseteq E^{op} = E and R⋅R op⊆E⋅E⊆ER \cdot R^{op} \subseteq E \cdot E \subseteq E by symmetry and transitivity of EE. In the other direction, suppose (y 1,y 3)∈E(y_1, y_3) \in E, so that h(y 1)=h(y 3)h(y_1) = h(y_3). Put x=j(y 1)x = j(y_1) and x′=j(y 3)x' = j(y_3) (so that f(x)=y 1f(x) = y_1 and f(x′)=y 3f(x') = y_3), and put y 2=g(x)y_2 = g(x). Clearly then (y 1,y 2)∈R(y_1, y_2) \in R. Moreover,

y 2=g(x)=gj(y 1)=ih(y 1)=ih(y 3)=gj(y 3)=g(x′)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)∈R(y_3, y_2) \in R, or (y 2,y 3)∈R op(y_2, y_3) \in R^{op}. Hence (y 1,y 3)∈R⋅R op(y_1, y_3) \in R \cdot R^{op}, and we have shown E⊆R⋅R opE \subseteq R \cdot R^{op}.

Theorem

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

Proof

By theorem 1, UU has a left adjoint. Since bijective continuous maps between compact Hausdorff spaces are homeomorphisms, we have that UU reflects isomorphisms. Finally, suppose (f,g)(f, g) is a UU-split pair of morphisms X→YX \to Y in CHCH; let h:Y→Zh: Y \to Z be their coequalizer in TopTop, given by a suitable quotient space. Being a quotient of a compact space, ZZ is compact. Since CHCH is a full subcategory of TopTop, the map hh is a coequalizer in CHCH once we prove the following claim:

• Claim: ZZ is Hausdorff.

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

In other words, to complete the proof, it suffices to verify the claim. Letting p 0,p 1:E→→Yp_0, p_1: E \stackrel{\to}{\to} Y be the kernel pair of hh in TopTop, to show Z=Y/EZ = Y/E is Hausdorff, it suffices to prove that the equivalence relation ⟨p 0,p 1⟩:E→Y×Y\langle p_0, p_1 \rangle : E \to Y \times Y in TopTop is closed. Let R↪Y×YR \hookrightarrow Y \times Y be the image of ⟨f,g⟩:X→Y×Y\langle f, g \rangle: X \to Y \times Y. By lemma 1, the subset EE of UY×UYU Y \times U Y coincides with the subset R⋅R op⊆UY×UYR \cdot R^{op} \subseteq U Y \times U Y. Now RR is the image of the compact space XX under the continuous map ⟨f,g⟩\langle f, g \rangle, so RR is a closed subset of Y×YY \times Y. Similarly R opR^{op} is a closed subset of Y×YY \times Y. Under their subspace topologies, their fiber product R× YR opR \times_Y R^{op} is compact, and so its image R⋅R opR \cdot R^{op} under the (continuous) span composite R× YR op→Y×YR \times_Y R^{op} \to Y \times Y is also closed. This completes the proof.

Proposition

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

Theorem

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

Proof

Let f:X→Cf: X \to C be a map, where CC is a compact Hausdorff space. Since II is a cogenerator in the category of compact Hausdorff spaces, the unit for the codensity monad M IM_I,

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

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

f^ −1(C) → I hom(X,I) π↓ ↓f^ C →u C I hom(C,I).\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 uu, we have a map h:X→f^ −1(C)h: X \to \hat{f}^{-1}(C), i.e., the map u X:X→I hom(X,I)u_X: X \to I^{\hom(X, I)} factors through the closed subspace f^ −1(C)↪I hom(X,I)\hat{f}^{-1}(C) \hookrightarrow I^{\hom(X, I)}. Therefore hh factors as

X→i XX¯⊆f^ −1(C)X \stackrel{i_X}{\to} \bar{X} \subseteq \hat{f}^{-1}(C)

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