Contents

# Contents

## Idea

If $\mathcal{O}(X)$ is the topology on a topological space $X$ (i.e. its frame of opens), and if a map $\mathcal{O}(X) \to \mathcal{O}(1)$ that preserves finite meets and arbitrary joins (a homomorphism of frames) is considered an instance of “seeing a point $1 \to X$”, then $X$ is sober precisely if every point we see is really there (i.e., is induced from a continuous function $1 \to X$), 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 $T_2$ separation axiom (Hausdorff implies sober) and implying $T_0$.

separation axioms
$\array{\\ &&& T_2 = \text{Hausdorff} \\ && \swArrow && \seArrow \\ \, & T_1 && && \text{sober} & \, \\ && \seArrow && \swArrow \\ &&& T_0 = \text{Kolmogorov} \\ }$

(Note that this diagram is not a pullback – there are $T_1$ sober spaces which are not Hausdorff.)

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.

## Definition

A topological space $X$ 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 $X$ is T0, the other half states that $X$ has enough points:

###### Definition

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

## Properties

### Separation

$Hausdorff = T_2 \Rightarrow sober \Rightarrow T_0$

### 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 below), we first introduce the following terminology:

###### Definition

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

$\tau_X \longleftarrow \tau_Y \;\colon\; \phi$

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

1. arbitrary unions;

In other words, $\phi$ is a frame homomorphism if

1. for every set $I$ and every $I$-indexed set $\{U_i \in \tau_Y\}_{i \in I}$ of elements of $\tau_Y$, then

$\phi\left(\underset{i \in I}{\cup} U_i\right) \;=\; \underset{i \in I}{\cup} \phi(U_i)\;\;\;\;\in \tau_X \,,$
2. for every finite set $J$ and every $J$-indexed set $\{U_j \in \tau_Y\}$ of elements in $\tau_Y$, then

$\phi\left(\underset{j \in J}{\cap} U_j\right) \;=\; \underset{j \in J}{\cap} \phi(U_j) \;\;\;\;\in \tau_X \,.$
###### Remark

A frame homomorphism $\phi$ as in def. necessarily also preserves inclusions in that

• for every inclusion $U_1 \subset U_2$ with $U_1, U_2 \in \tau_Y \subset P(Y)$ then

$\phi(U_1) \subset \phi(U_2) \;\;\;\;\;\;\; \in \tau_X \,.$

This is because inclusions are witnessed by unions

$(U_1 \subset U_2) \;\Leftrightarrow\; \left( U_1 \cup U_2 = U_2 \right)$

and by finite intersections:

$(U_1 \subset U_2) \;\Leftrightarrow\; \left( U_1 \cap U_2 = U_1 \right) \,.$
###### Example

For

$f \;\colon\; (X,\tau_X) \longrightarrow (Y, \tau_Y)$

a continuous function, then its function of pre-images

$\tau_X \longleftarrow \tau_Y \;\colon\; f^{-1}$

is a frame homomorphism according to def. .

For sober topological spaces the converse holds:

###### Proposition

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

$\tau_X \longleftarrow \tau_Y \;\colon\; \phi$

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

$\phi = f^{-1} \,.$

We prove this below, after the following lemma.

Let $\ast = (\{1\}, \tau_\ast = \left\{\emptyset, \{1\}\right\})$ be the point topological space.

###### Lemma

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

$\array{ Hom_{Frame}(\tau_X, \tau_\ast) &\underoverset{\simeq}{}{\longrightarrow}& IrrClSub(X) \\ \phi &\mapsto& X \backslash U_\emptyset(\phi) }$

where $U_\emptyset(\phi)$ is the union of all elements $U \in \tau_x$ such that $\phi(U) = \emptyset$:

$U_{\emptyset}(\phi) \coloneqq \underset{{U \in \tau_X} \atop {\phi(U) = \emptyset} }{\cup} U \,.$

###### Proof

First we need to show that the function is well defined in that given a frame homomorphism $\phi \colon \tau_X \to \tau_\ast$ then $X \backslash U_\emptyset(\phi)$ is indeed an irreducible closed subspace.

To that end observe that:

$(\ast)$ If there are two elements $U_1, U_2 \in \tau_X$ with $U_1 \cap U_2 \subset U_{\emptyset}(\phi)$ then $U_1 \subset U_{\emptyset}(\phi)$ or $U_2 \subset U_{\emptyset}(\phi)$.

This is because

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

Now according to this prop., the condition $(\ast)$ identifies the complement $X \backslash U_{\emptyset}(\phi)$ as an irreducible closed subspace of $(X,\tau)$.

Conversely, given an irreducible closed subset $X \backslash U_0$, define $\phi$ by

$\phi \;\colon\; U \mapsto \left\{ \array{ \emptyset & \vert \, \text{if} \, U \subset U_0 \\ \{1\} & \vert \, \text{otherwise} } \right. \,.$

This does preserve

1. arbitrary unions

because $\phi(\underset{i}{\cup} U_i) = \{\emptyset\}$ precisely if $\underset{i}{\cup}U_i \subset U_0$ which is the case precisely if all $U_i \subset U_0$, which means that all $\phi(U_i) = \emptyset$ and because $\underset{i}{\cup}\emptyset = \emptyset$;

while $\phi(\underset{i}{\cup}U_1) = \{1\}$ as soon as one of the $U_i$ is not contained in $U_0$, which means that one of the $\phi(U_i) = \{1\}$ which means that $\underset{i}{\cup} \phi(U_i) = \{1\}$;

2. finite intersections

because if $U_1 \cap U_2 \subset U_0$, then by $(\ast)$ $U_1 \in U_0$ or $U_2 \in U_0$, whence $\phi(U_1) = \emptyset$ or $\phi(U_2) = \emptyset$, whence with $\phi(U_1 \cap U_2) = \emptyset$ also $\phi(U_1) \cap \phi(U_2) = \emptyset$;

while if $U_1 \cap U_2$ is not contained in $U_0$ then neither $U_1$ nor $U_2$ is contained in $U_0$ and hence with $\phi(U_1 \cap U_2) = \{1\}$ also $\phi(U_1) \cap \phi(U_2) = \{1\} \cap \{1\} = \{1\}$.

Hence this is indeed a frame homomorphism $\tau_X \to \tau_\ast$.

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

###### Proof

of prop.

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

$\tau_\ast \longleftarrow \tau_X \;\colon\; \phi$

and show that these are in bijection to the underlying set $X$, identified with the continuous functions $\ast \to (X,\tau)$.

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

$\phi \;\colon\; U \mapsto \left\{ \array{ \{1\} & \vert \,\text{if}\, x \in U \\ \emptyset & \vert \, \text{otherwise} } \right. \,.$

This is precisely the inverse image function of the continuous function $\ast \to X$ which sends $1 \mapsto x$.

Hence this establishes the bijection between frame homomorphisms of the form $\tau_\ast \longleftarrow \tau_X$ and continuous functions of the form $\ast \to (X,\tau)$.

With this it follows that a general frame homomorphism of the form $\tau_X \overset{\phi}{\longleftarrow} \tau_Y$ defines a function of sets $X \overset{f}{\longrightarrow} Y$ by composition:

$\array{ X &\overset{f}{\longrightarrow}& Y \\ (\tau_\ast \leftarrow \tau_X) &\mapsto& (\tau_\ast \leftarrow \tau_X \overset{\phi}{\longleftarrow} \tau_Y) } \,.$

By the previous analysis, an element $U_Y \in \tau_Y$ is sent to $\{1\}$ under this composite precisely if the corresponding point $\ast \to X \overset{f}{\longrightarrow} Y$ is in $U_Y$, and similarly for an element $U_X \in \tau_X$. It follows that $\phi(U_Y) \in \tau_X$ is precisely that subset of points in $X$ which are sent by $f$ to elements of $U_Y$, hence that $\phi = f^{-1}$ is the pre-image function of $f$. Since $\phi$ by definition sends open subsets of $Y$ to open subsets of $X$, it follows that $f$ 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 $\ast \coloneqq ( \{1\}, \tau_\ast = \left\{ \emptyset, \{1\}\right\} )$.

###### Definition

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

Define $S X$ to be the set

$S X \coloneqq Hom_{Frame}( \tau_X, \tau_\ast )$

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

$\tilde U \;\coloneqq\; \left\{ \phi \in S X \,\vert\, \phi(U) = \{1\} \right\} \,.$

Consider the function

$\array{ X &\overset{s_X}{\longrightarrow}& S X \\ x &\mapsto& (const_x)^{-1} }$

which sends an element $x \in X$ to the function which assigns inverse images of the constant function $const_x \;\colon\; \{1\} \to X$ on that element.

###### Lemma

The construction $(S X, \tau_{S X})$ in def. is a topological space, and the function $s_X \colon X \to S X$ is a continuous function

$s_X \colon (X, \tau_X) \longrightarrow (S X, \tau_{S X})$
###### Proof

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

$\array{ \tau_X &\overset{\widetilde{(-)}}{\longrightarrow}& \tau_{S X} \\ U &\mapsto& \tilde U }$

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

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

\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 $p \colon \tau_X \to \tau_\ast$ preserves unions. Similarly for intersections, now with $I$ a finite set:

\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 $p$ preserves finite intersections.

To see that $s_X$ is continuous, observe that $s_X^{-1}(\tilde U) = U$, by construction.

###### Lemma

For $(X, \tau_X)$ a topological space, the function $s_X \colon X \to S X$ from def. is

1. an injection precisely if $X$ is T0;

2. a bijection precisely if $X$ is sober.

In this case $s_X$ is in fact a homeomorphism.

###### Proof

By lemma there is an identification $S X \simeq IrrClSub(X)$ and via this $s_X$ is identified with the map $x \mapsto Cl(\{x\})$.

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

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

###### Lemma

For $(X,\tau)$ a topological space, then the topological space $(S X, \tau_{S X})$ from def. , lemma is sober.

###### Proof

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

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

To see this use this prop., saying that irreducibility of $X \backslash U$ is equivalent to $U_1 \cap U_2 \subset U \Rightarrow (U_1 \subset U) or (U_2 \subset U)$. But if $U_1 \cap U_2 \subset U$ then also $\tilde U_1 \cap \tilde U_2 \subset \tilde U$ (as in the proof of lemma ) and hence by assumption on $\tilde U$ it follows that $\tilde U_1 \subset \tilde U$ or $\tilde U_2 \subset \tilde U$. By lemma this in turn implies $U_1 \subset U$ or $U_2 \subset U$. In conclusion, this shows that also $X \backslash U$ is irreducible .

By lemma this irreducible closed subspace corresponds to a point $p \in S X$. By that same lemma, this frame homomorphism $p \colon \tau_X \to \tau_\ast$ takes the value $\emptyset$ on all those opens which are inside $U$. This means that the topological closure of this point is just $S X \backslash \tilde U$.

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

So let $p_1 \neq p_2 \in S X$ be two distinct points. This means that there exists $U \in \tau_X$ with $p_1(U) \neq p_2(U)$. Equivalently this says that $\tilde U$ contains one of the two points, but not the other. This means that $(S X, \tau_{S X})$ is T0. By this prop. this is equivalent to there being no two points with the same topological closure.

###### Proposition

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

$\array{ (X, \tau_X) &\overset{f}{\longrightarrow}& (Y, \tau_Y^{sob}) \\ {}^{\mathllap{s_X}}\downarrow & \nearrow_{\exists !} \\ (S X , \tau_{S X}) } \,.$
###### Proof

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

$\array{ (X, \tau_X) &\overset{f}{\longrightarrow}& (Y, \tau^{sob}_Y) \\ {}^{\mathllap{s_X}}\downarrow & \nearrow& \downarrow^{\mathrlap{s_{S X}}} \\ (S X, \tau_{S X}) &\underset{S f}{\longrightarrow}& (S S X, \tau_{S S X}) } \,.$

By lemma and lemma , the right vertical morphism $s_{S X}$ 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 $\tilde f, \overline{f} \colon \colon (S X, \tau_{S X}) \to (Y, \tau_Y^{sob})$ and apply the soberification construction once more to the triangles

$\array{ (X, \tau_X) &\overset{f}{\longrightarrow}& (Y, \tau_Y^{sob}) \\ {}^{\mathllap{s_X}}\downarrow & \nearrow_{ \tilde f, \overline{f}} \\ (S X , \tau_{S X}) } \phantom{AAA} \mapsto \phantom{AAA} \array{ (S X, \tau_{S X}) &\overset{S f}{\longrightarrow}& (Y, \tau_Y^{sob}) \\ {}^{\simeq}\downarrow & \nearrow_{ \tilde f, \overline{f}} \\ (S X , \tau_{S X}) } \,.$

Here on the right we used again lemma to find that the vertical morphism is an isomorphism, and that $\tilde f$ and $\overline{f}$ 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 $\tilde f$ and $\overline{f}$ implies that $\tilde f = \overline{f}$.

### Enough points

A topological space has enough points in the sense of def. if and only if its $T_0$ 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:

• Any nontrivial indiscrete space is not sober, since it is not $T_0$.

More interestingly:

• For $R$ a commutative ring, then the space $R^2$ with the Zariski topology is $T_1$ 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 $Spec(R[x,y])$, which contains “generic 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 $P$ is the infinite binary tree (the set of finite $\{0,1\}$-words (lists) with the “extends” preorder), then the soberification of its Alexandroff topology is Wilson space?, the space of finite or infinite $\{0,1\}$-words (streams).

## References

Last revised on September 20, 2018 at 15:40:17. See the history of this page for a list of all contributions to it.