sober topological space



topology (point-set topology, point-free topology)

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If π’ͺ(X)\mathcal{O}(X) is the topology on a topological space XX (i.e. its frame of opens), and if a map π’ͺ(X)β†’π’ͺ(1)\mathcal{O}(X) \to \mathcal{O}(1) that preserves finite meets and arbitrary joins (a homomorphism of frames) is considered an instance of β€œseeing a point 1β†’X1 \to X”, then XX is sober precisely if every point we see is really there (i.e., is induced from a continuous function 1β†’X1 \to X), and if we never see double.

The condition that a topological space be sober is an extra condition akin to a separation axiom. In fact with classical logic it is a condition implied by the T 2T_2 separation axiom (Hausdorff implies sober) and implying T 0T_0.

separation axioms
T 2=Hausdorff ⇙ β‡˜ T 1 sober β‡˜ ⇙ T 0=Kolmogorov \array{\\ &&& T_2 = \text{Hausdorff} \\ && \swArrow && \seArrow \\ \, & T_1 && && \text{sober} & \, \\ && \seArrow && \swArrow \\ &&& T_0 = \text{Kolmogorov} \\ }

But the sobriety condition on a topological space has deeper meaning. It means that continuous functions betwen sober topolgical spaces are entirely determined by their inverse image functions on the frames of opens, disregarding the underlying sets of points. Technically this means that the sober topological spaces are precisely the locales among the topological spaces.


A topological space XX is sober if its points are exactly determined by its lattice of open subsets. Different equivalent ways to say this are:

In each case, half of the definition is that XX is T0, the other half states that XX has enough points:


A topological space XX has enough points if the following equivalent conditions hold:



Hausdorff=T 2⇒sober⇒T 0 Hausdorff = T_2 \Rightarrow sober \Rightarrow T_0

As locales with enough points

What makes the concept of sober topological spaces special is that for them the concept of continuous functions may be expressed entirely in terms of the relations between their open subsets, disregarding the underlying set of points of which these open are in fact subsets. In order to express this property (proposition 1 below), we first introduce the following terminology:


Let (X,Ο„ X)(X,\tau_X) and (Y,Ο„ Y)(Y,\tau_Y) be topological spaces. Then a function

Ο„ XβŸ΅Ο„ Y:Ο• \tau_X \longleftarrow \tau_Y \;\colon\; \phi

between their sets of open subsets is called a frame homomorphism if it preserves

  1. arbitrary unions;

  2. finite intersections.

In other words, Ο•\phi is a frame homomorphism if

  1. for every set II and every II-indexed set {U iβˆˆΟ„ Y} i∈I\{U_i \in \tau_Y\}_{i \in I} of elements of Ο„ Y\tau_Y, then

    Ο•(βˆͺi∈IU i)=βˆͺi∈IΟ•(U i)βˆˆΟ„ X, \phi\left(\underset{i \in I}{\cup} U_i\right) \;=\; \underset{i \in I}{\cup} \phi(U_i)\;\;\;\;\in \tau_X \,,
  2. for every finite set JJ and every JJ-indexed set {U jβˆˆΟ„ Y}\{U_j \in \tau_Y\} of elements in Ο„ Y\tau_Y, then

    Ο•(∩j∈JU j)=∩j∈JΟ•(U j)βˆˆΟ„ X. \phi\left(\underset{j \in J}{\cap} U_j\right) \;=\; \underset{j \in J}{\cap} \phi(U_j) \;\;\;\;\in \tau_X \,.

A frame homomorphism Ο•\phi as in def. 2 necessarily also preserves inclusions in that

  • for every inclusion U 1βŠ‚U 2U_1 \subset U_2 with U 1,U 2βˆˆΟ„ YβŠ‚P(Y)U_1, U_2 \in \tau_Y \subset P(Y) then

    Ο•(U 1)βŠ‚Ο•(U 2)βˆˆΟ„ X. \phi(U_1) \subset \phi(U_2) \;\;\;\;\;\;\; \in \tau_X \,.

This is because inclusions are witnessed by unions

(U 1βŠ‚U 2)⇔(U 1βˆͺU 2=U 2) (U_1 \subset U_2) \;\Leftrightarrow\; \left( U_1 \cup U_2 = U_2 \right)

and by finite intersections:

(U 1βŠ‚U 2)⇔(U 1∩U 2=U 1). (U_1 \subset U_2) \;\Leftrightarrow\; \left( U_1 \cap U_2 = U_1 \right) \,.


f:(X,Ο„ X)⟢(Y,Ο„ Y) f \;\colon\; (X,\tau_X) \longrightarrow (Y, \tau_Y)

a continuous function, then its function of pre-images

Ο„ XβŸ΅Ο„ Y:f βˆ’1 \tau_X \longleftarrow \tau_Y \;\colon\; f^{-1}

is a frame homomorphism according to def. 2.

For sober topological spaces the converse holds:


If (X,Ο„ X)(X,\tau_X) and (Y,Ο„ Y)(Y,\tau_Y) are sober topological spaces, then for every frame homomorphism (def. 2)

Ο„ XβŸ΅Ο„ Y:Ο• \tau_X \longleftarrow \tau_Y \;\colon\; \phi

there is a unique continuous function f:X→Yf \colon X \to Y such that ϕ\phi is the function of forming pre-images under ff:

Ο•=f βˆ’1. \phi = f^{-1} \,.

We prove this below, after the following lemma.

Let *=({1},Ο„ *={βˆ…,{1}})\ast = (\{1\}, \tau_\ast = \left\{\emptyset, \{1\}\right\}) be the point topological space.


For (X,Ο„)(X,\tau) a topological space, then there is a bijection between the irreducible closed subspaces of (X,Ο„)(X,\tau) and the frame homomorphisms from Ο„ X\tau_X to Ο„ *\tau_\ast, given bys

Hom Frame(Ο„ X,Ο„ *) βŸΆβ‰ƒ IrrClSub(X) Ο• ↦ X\U βˆ…(Ο•) \array{ Hom_{Frame}(\tau_X, \tau_\ast) &\underoverset{\simeq}{}{\longrightarrow}& IrrClSub(X) \\ \phi &\mapsto& X \backslash U_\emptyset(\phi) }

where U βˆ…(Ο•)U_\emptyset(\phi) is the union of all elements UβˆˆΟ„ xU \in \tau_x such that Ο•(U)=βˆ…\phi(U) = \emptyset:

U βˆ…(Ο•)≔βˆͺUβˆˆΟ„ XΟ•(U)=βˆ…U. U_{\emptyset}(\phi) \coloneqq \underset{{U \in \tau_X} \atop {\phi(U) = \emptyset} }{\cup} U \,.

See also (Johnstone 82, II 1.3).


First we need to show that the function is well defined in that given a frame homomorphism Ο•:Ο„ Xβ†’Ο„ *\phi \colon \tau_X \to \tau_\ast then X\U βˆ…(Ο•)X \backslash U_\emptyset(\phi) is indeed an irreducible closed subspace.

To that end observe that:

(*)(\ast) If there are two elements U 1,U 2βˆˆΟ„ XU_1, U_2 \in \tau_X with U 1∩U 2βŠ‚U βˆ…(Ο•)U_1 \cap U_2 \subset U_{\emptyset}(\phi) then U 1βŠ‚U βˆ…(Ο•)U_1 \subset U_{\emptyset}(\phi) or U 2βŠ‚U βˆ…(Ο•)U_2 \subset U_{\emptyset}(\phi).

This is because

Ο•(U 1)βˆ©Ο•(U 2) =Ο•(U 1∩U 2) βŠ‚Ο•(U βˆ…(Ο•)) =βˆ…, \begin{aligned} \phi(U_1) \cap \phi(U_2) & = \phi(U_1 \cap U_2) \\ & \subset \phi(U_{\emptyset}(\phi)) \\ & = \emptyset \end{aligned} \,,

where the first equality holds because Ο•\phi preserves finite intersections by def. 2, the inclusion holds because Ο•\phi respects inclusions by remark 1, and the second equality holds because Ο•\phi preserves arbitrary unions by def. 2. But in Ο„ *={βˆ…,{1}}\tau_\ast = \{\emptyset, \{1\}\} the intersection of two open subsets is empty precisely if at least one of them is empty, hence Ο•(U 1)=βˆ…\phi(U_1) = \emptyset or Ο•(U 2)=βˆ…\phi(U_2) = \emptyset. But this means that U 1βŠ‚U βˆ…(Ο•)U_1 \subset U_{\emptyset}(\phi) or U 2βŠ‚U βˆ…(Ο•)U_2 \subset U_{\emptyset}(\phi), as claimed.

Now according to prop. \ref{OpenSubsetVersionOfClosedIrreducible} the condition (*)(\ast) identifies the complement X\U βˆ…(Ο•)X \backslash U_{\emptyset}(\phi) as an irreducible closed subspace of (X,Ο„)(X,\tau).

Conversely, given an irreducible closed subset X\U 0X \backslash U_0, define Ο•\phi by

Ο•:U↦{βˆ… |ifUβŠ‚U 0 {1} |otherwise. \phi \;\colon\; U \mapsto \left\{ \array{ \emptyset & \vert \, \text{if} \, U \subset U_0 \\ \{1\} & \vert \, \text{otherwise} } \right. \,.

This does preserve

  1. arbitrary unions

    because Ο•(βˆͺiU i)={βˆ…}\phi(\underset{i}{\cup} U_i) = \{\emptyset\} precisely if βˆͺiU iβŠ‚U 0\underset{i}{\cup}U_i \subset U_0 which is the case precisely if all U iβŠ‚U 0U_i \subset U_0, which means that all Ο•(U i)=βˆ…\phi(U_i) = \emptyset and because βˆͺiβˆ…=βˆ…\underset{i}{\cup}\emptyset = \emptyset;

    while Ο•(βˆͺiU 1)={1}\phi(\underset{i}{\cup}U_1) = \{1\} as soon as one of the U iU_i is not contained in U 0U_0, which means that one of the Ο•(U i)={1}\phi(U_i) = \{1\} which means that βˆͺiΟ•(U i)={1}\underset{i}{\cup} \phi(U_i) = \{1\};

  2. finite intersections

    because if U 1∩U 2βŠ‚U 0U_1 \cap U_2 \subset U_0, then by (*)(\ast) U 1∈U 0U_1 \in U_0 or U 2∈U 0U_2 \in U_0, whence Ο•(U 1)=βˆ…\phi(U_1) = \emptyset or Ο•(U 2)=βˆ…\phi(U_2) = \emptyset, whence with Ο•(U 1∩U 2)=βˆ…\phi(U_1 \cap U_2) = \emptyset also Ο•(U 1)βˆ©Ο•(U 2)=βˆ…\phi(U_1) \cap \phi(U_2) = \emptyset;

    while if U 1∩U 2U_1 \cap U_2 is not contained in U 0U_0 then neither U 1U_1 nor U 2U_2 is contained in U 0U_0 and hence with Ο•(U 1∩U 2)={1}\phi(U_1 \cap U_2) = \{1\} also Ο•(U 1)βˆ©Ο•(U 2)={1}∩{1}={1}\phi(U_1) \cap \phi(U_2) = \{1\} \cap \{1\} = \{1\}.

Hence this is indeed a frame homomorphism τ X→τ *\tau_X \to \tau_\ast.

Finally, it is clear that these two operations are inverse to each other.


of prop. 1

We first consider the special case of frame homomorphisms of the form

Ο„ *βŸ΅Ο„ X:Ο• \tau_\ast \longleftarrow \tau_X \;\colon\; \phi

and show that these are in bijection to the underlying set XX, identified with the continuous functions *β†’(X,Ο„)\ast \to (X,\tau).

By lemma 1, the frame homomorphisms Ο•:Ο„ Xβ†’Ο„ *\phi \colon \tau_X \to \tau_\ast are identified with the irreducible closed subspaces X\U βˆ…(Ο•)X \backslash U_\emptyset(\phi) of (X,Ο„ X)(X,\tau_X). Therefore by assumption of sobriety of (X,Ο„)(X,\tau) there is a unique point x∈Xx \in X with X\U βˆ…=Cl({x})X \backslash U_{\emptyset} = Cl(\{x\}). In particular this means that for U xU_x an open neighbourhood of xx, then U xU_x is not a subset of U βˆ…(Ο•)U_\emptyset(\phi), and so it follows that Ο•(U x)={1}\phi(U_x) = \{1\}. In conclusion we have found a unique x∈Xx \in X such that

Ο•:U↦{{1} |ifx∈U βˆ… |otherwise. \phi \;\colon\; U \mapsto \left\{ \array{ \{1\} & \vert \,\text{if}\, x \in U \\ \emptyset & \vert \, \text{otherwise} } \right. \,.

This is precisely the inverse image function of the continuous function *β†’X\ast \to X which sends 1↦x1 \mapsto x.

Hence this establishes the bijection between frame homomorphisms of the form Ο„ *βŸ΅Ο„ X\tau_\ast \longleftarrow \tau_X and continuous functions of the form *β†’(X,Ο„)\ast \to (X,\tau).

With this it follows that a general frame homomorphism of the form Ο„ XβŸ΅Ο•Ο„ Y\tau_X \overset{\phi}{\longleftarrow} \tau_Y defines a function of sets X⟢fYX \overset{f}{\longrightarrow} Y by composition:

X ⟢f Y (Ο„ *←τ X) ↦ (Ο„ *←τ XβŸ΅Ο•Ο„ Y). \array{ X &\overset{f}{\longrightarrow}& Y \\ (\tau_\ast \leftarrow \tau_X) &\mapsto& (\tau_\ast \leftarrow \tau_X \overset{\phi}{\longleftarrow} \tau_Y) } \,.

By the previous analysis, an element U YβˆˆΟ„ YU_Y \in \tau_Y is sent to {1}\{1\} under this composite precisely if the corresponding point *β†’X⟢fY\ast \to X \overset{f}{\longrightarrow} Y is in U YU_Y, and similarly for an element U XβˆˆΟ„ XU_X \in \tau_X. It follows that Ο•(U Y)βˆˆΟ„ X\phi(U_Y) \in \tau_X is precisely that subset of points in XX which are sent by ff to elements of U YU_Y, hence that Ο•=f βˆ’1\phi = f^{-1} is the pre-image function of ff. Since Ο•\phi by definition sends open subsets of YY to open subsets of XX, it follows that ff is indeed a continuous function. This proves the claim in generality.

Soberification reflection

The category of sober spaces is reflective in the category of all topological spaces; the left adjoint is called the soberification.

This reflection is also induced by the idempotent adjunction between spaces and locales; thus sober spaces are precisely those spaces that are the spaces of points of some locale, and the category of sober spaces is equivalent to the category of locales with enough points.

We now say this in detail.

Recall again the point topological space *≔({1},Ο„ *={βˆ…,{1}})\ast \coloneqq ( \{1\}, \tau_\ast = \left\{ \emptyset, \{1\}\right\} ).


Let (X,Ο„)(X,\tau) be a topological space.

Define SXS X to be the set

SX≔Hom Frame(Ο„ X,Ο„ *) S X \coloneqq Hom_{Frame}( \tau_X, \tau_\ast )

of frame homomorphisms from the frame of opens of XX to that of the point. Define a topology Ο„ SXβŠ‚P(SX)\tau_{S X} \subset P(S X) on this set by declaring it to have one element U˜\tilde U for each element UβˆˆΟ„ XU \in \tau_X and given by

UΛœβ‰”{Ο•βˆˆSX|Ο•(U)={1}}. \tilde U \;\coloneqq\; \left\{ \phi \in S X \,\vert\, \phi(U) = \{1\} \right\} \,.

Consider the function

X ⟢s X SX x ↦ (const x) βˆ’1 \array{ X &\overset{s_X}{\longrightarrow}& S X \\ x &\mapsto& (const_x)^{-1} }

which sends an element x∈Xx \in X to the function which assigns inverse images of the constant function const x:{1}β†’Xconst_x \;\colon\; \{1\} \to X on that element.


The construction (SX,τ SX)(S X, \tau_{S X}) in def. 3 is a topological space, and the function s X:X→SXs_X \colon X \to S X is a continuous function

s X:(X,Ο„ X)⟢(SX,Ο„ SX) s_X \colon (X, \tau_X) \longrightarrow (S X, \tau_{S X})

To see that Ο„ SXβŠ‚P(SX)\tau_{S X} \subset P(S X) is closed under arbitrary unions and finite intersections, observe that the function

Ο„ X ⟢(βˆ’)˜ Ο„ SX U ↦ U˜ \array{ \tau_X &\overset{\widetilde{(-)}}{\longrightarrow}& \tau_{S X} \\ U &\mapsto& \tilde U }

in fact preserves arbitrary unions and finite intersections. Whith this the statement follows by the fact that Ο„ X\tau_X is closed under these operations.

To see that (βˆ’)˜\widetilde{(-)} indeed preserves unions, observe that (e.g. Johnstone 82, II 1.3 Lemma)

p∈βˆͺi∈IU i˜ β‡”βˆƒi∈Ip(U i)={1} ⇔βˆͺi∈Ip(U i)={1} ⇔p(βˆͺi∈IU i)={1} ⇔p∈βˆͺi∈IU i˜, \begin{aligned} p \in \underset{i \in I}{\cup} \widetilde{U_i} \; & \Leftrightarrow \underset{i \in I}{\exists} p(U_i) = \{1\} \\ & \Leftrightarrow \underset{i \in I}{\cup} p(U_i) = \{1\} \\ & \Leftrightarrow p\left(\underset{i \in I}{\cup} U_i\right) = \{1\} \\ & \Leftrightarrow p \in \widetilde{ \underset{i \in I}{\cup} U_i } \end{aligned} \,,

where we used that the frame homomorphism p:τ X→τ *p \colon \tau_X \to \tau_\ast preserves unions. Similarly for intersections, now with II a finite set:

p∈∩i∈IU i˜ β‡”βˆ€i∈Ip(U i)={1} β‡”βˆ©i∈Ip(U i)={1} ⇔p(∩i∈IU i)={1} ⇔p∈∩i∈IU i˜, \begin{aligned} p \in \underset{i \in I}{\cap} \widetilde{U_i} \; & \Leftrightarrow \underset{i \in I}{\forall} p(U_i) = \{1\} \\ & \Leftrightarrow \underset{i \in I}{\cap} p(U_i) = \{1\} \\ & \Leftrightarrow p\left(\underset{i \in I}{\cap} U_i\right) = \{1\} \\ & \Leftrightarrow p \in \widetilde{ \underset{i \in I}{\cap} U_i } \end{aligned} \,,

where now we used that the frame homomorphism pp preserves finite intersections.

To see that s Xs_X is continuous, observe that s X βˆ’1(U˜)=Us_X^{-1}(\tilde U) = U, by construction.


For (X,τ X)(X, \tau_X) a topological space, the function s X:X→SXs_X \colon X \to S X from def. 3 is

  1. an injection precisely if XX is T0;

  2. a bijection precisely if XX is sober.

    In this case s Xs_X is in fact a homeomorphism.


By lemma 1 there is an identification SX≃IrrClSub(X)S X \simeq IrrClSub(X) and via this s Xs_X is identified with the map x↦Cl({x})x \mapsto Cl(\{x\}).

Hence the second statement follows by definition, and the first statement by this prop..

That in the second case s Xs_X is in fact a homeomorphism follows from the definition of the opens U˜\tilde U: they are identified with the opens UU in this case (…expand…).


For (X,Ο„)(X,\tau) a topological space, then the topological space (SX,Ο„ SX)(S X, \tau_{S X}) from def. 3, lemma 2 is sober.

(e.g. Johnstone 82, lemma II 1.7)


Let SX\U˜S X \backslash \tilde U be an irreducible closed subspace of (SX,Ο„ SX)(S X, \tau_{S X}). We need to show that it is the topological closure of a unique element Ο•βˆˆSX\phi \in S X.

Observe first that also X\UX \backslash U is irreducible.

To see this use this prop., saying that irreducibility of X\UX \backslash U is equivalent to U 1∩U 2βŠ‚Uβ‡’(U 1βŠ‚U)or(U 2βŠ‚U)U_1 \cap U_2 \subset U \Rightarrow (U_1 \subset U) or (U_2 \subset U). But if U 1∩U 2βŠ‚UU_1 \cap U_2 \subset U then also U˜ 1∩U˜ 2βŠ‚U˜\tilde U_1 \cap \tilde U_2 \subset \tilde U (as in the proof of lemma 2) and hence by assumption on U˜\tilde U it follows that U˜ 1βŠ‚U˜\tilde U_1 \subset \tilde U or U˜ 2βŠ‚U˜\tilde U_2 \subset \tilde U. By lemma 1 this in turn implies U 1βŠ‚UU_1 \subset U or U 2βŠ‚UU_2 \subset U. In conclusion, this shows that also X\UX \backslash U is irreducible .

By lemma 1 this irreducible closed subspace corresponds to a point p∈SXp \in S X. By that same lemma, this frame homomorphism p:Ο„ Xβ†’Ο„ *p \colon \tau_X \to \tau_\ast takes the value βˆ…\emptyset on all those opens which are inside UU. This means that the topological closure of this point is just SX\U˜S X \backslash \tilde U.

This shows that there exists at least one point of which X\U˜X \backslash \tilde U is the topological closure. It remains to see that there is no other such point.

So let p 1β‰ p 2∈SXp_1 \neq p_2 \in S X be two distinct points. This means that there exists UβˆˆΟ„ XU \in \tau_X with p 1(U)β‰ p 2(U)p_1(U) \neq p_2(U). Equivalently this says that U˜\tilde U contains one of the two points, but not the other. This means that (SX,Ο„ SX)(S X, \tau_{S X}) is T0. By this prop. this is equivalent to there being no two points with the same topological closure.


For (X,Ο„ X)(X, \tau_X) any topological space, for (Y,Ο„ Y sob)(Y,\tau_Y^{sob}) a sober topological space, and for f:(X,Ο„ X)⟢(Y,Ο„ Y)f \colon (X, \tau_X) \longrightarrow (Y, \tau_Y) a continuous function, then it factors uniquely through the soberification s X:(X,Ο„ X)⟢(SX,Ο„ SX)s_X \colon (X, \tau_X) \longrightarrow(S X, \tau_{S X}) from def. 3, lemma 2

(X,Ο„ X) ⟢f (Y,Ο„ Y sob) s X↓ β†— βˆƒ! (SX,Ο„ SX). \array{ (X, \tau_X) &\overset{f}{\longrightarrow}& (Y, \tau_Y^{sob}) \\ {}^{\mathllap{s_X}}\downarrow & \nearrow_{\exists !} \\ (S X , \tau_{S X}) } \,.

By the construction in def. 3, we find that the outer part of the following square commutes:

(X,Ο„ X) ⟢f (Y,Ο„ Y sob) s X↓ β†— ↓ s SX (SX,Ο„ SX) ⟢Sf (SSX,Ο„ SSX). \array{ (X, \tau_X) &\overset{f}{\longrightarrow}& (Y, \tau^{sob}_Y) \\ {}^{\mathllap{s_X}}\downarrow & \nearrow& \downarrow^{\mathrlap{s_{S X}}} \\ (S X, \tau_{S X}) &\underset{S f}{\longrightarrow}& (S S X, \tau_{S S X}) } \,.

By lemma 4 and lemma 3, the right vertical morphism s SXs_{S X} is an isomorphism (a homeomorphism), hence has an inverse morphism. This defines the diagonal morphism, which is the desired factorization.

To see that this factorization is unique, consider two factorizations f˜,fΒ―::(SX,Ο„ SX)β†’(Y,Ο„ Y sob)\tilde f, \overline{f} \colon \colon (S X, \tau_{S X}) \to (Y, \tau_Y^{sob}) and apply the soberification construction once more to the triangles

(X,Ο„ X) ⟢f (Y,Ο„ Y sob) s X↓ β†— f˜,fΒ― (SX,Ο„ SX)AAA↦AAA(SX,Ο„ SX) ⟢Sf (Y,Ο„ Y sob) ≃↓ β†— f˜,fΒ― (SX,Ο„ SX). \array{ (X, \tau_X) &\overset{f}{\longrightarrow}& (Y, \tau_Y^{sob}) \\ {}^{\mathllap{s_X}}\downarrow & \nearrow_{ \tilde f, \overline{f}} \\ (S X , \tau_{S X}) } \phantom{AAA} \mapsto \phantom{AAA} \array{ (S X, \tau_{S X}) &\overset{S f}{\longrightarrow}& (Y, \tau_Y^{sob}) \\ {}^{\simeq}\downarrow & \nearrow_{ \tilde f, \overline{f}} \\ (S X , \tau_{S X}) } \,.

Here on the right we used again lemma 3 to find that the vertical morphism is an isomorphism, and that f˜\tilde f and f¯\overline{f} do not change under soberification, as they already map between sober spaces. But now that the left vertical morphism is an isomorphism, the commutativity of this triangle for both f˜\tilde f and f¯\overline{f} implies that f˜=f¯\tilde f = \overline{f}.

Enough points

A topological space has enough points in the sense of def. 1 if and only if its T 0T_0 quotient is sober. The category of topological spaces with enough points is a reflective subcategory of the category Top of all topological spaces, and a topological space is T0 iff this reflection is sober.

Examples and Non-examples

Further examples of spaces that are not sober includes the following:

More interestingly:

  • For RR a commutative ring, then the space R 2R^2 with the Zariski topology is T 1T_1 but not sober, since every subvariety is an irreducible closed set which is not the closure of a point. Its soberification is, unsurprisingly, the scheme Spec(R[x,y])Spec(R[x,y]), which contains β€œgeneric points” whose closures are the subvarieties.

The following non-example shows that sobriety is not a hereditary separation property, i.e., topological subspaces of sober spaces need not be sober:

  • The Alexandroff topology on a poset is also not, in general, sober. For instance, if PP is the infinite binary tree (the set of finite {0,1}\{0,1\}-words (lists) with the β€œextends” preorder), then the soberification of its Alexandroff topology is Wilson space?, the space of finite or infinite {0,1}\{0,1\}-words (streams).


Revised on June 10, 2017 15:45:36 by Urs Schreiber (