nLab Hausdorff space

Context

Topology

topology

algebraic topology

Contents

Idea

A topological space (or more generally, a convergence space) is Hausdorff if convergence is unique. The concept can also be defined for locales (see Definition 3 below) and categorified (see Beyond topological spaces below). A Hausdorff space is often called $T_2$, since this condition came second in the original list of four separation axioms (there are more now) satisfied by metric spaces.

Hausdorff spaces are a kind of nice topological space; they do not form a particularly nice category of spaces themselves, but many such nice categories consist of only Hausdorff spaces. In fact, Felix Hausdorff's original definition of ‘topological space’ actually required the space to be Hausdorff, hence the name. Certainly homotopy theory (up to weak homotopy equivalence) needs only Hausdorff spaces. It is also common in analysis to assume that all spaces encountered are Hausdorff; if necessary, this can be arranged since every space has a Hausdorff quotient (in fact, the Hausdorff spaces form a reflective subcategory of Top), although usually an easier method is available than this sledgehammer.

Definitions

There are many equivalent ways of characterizing a space $S$ as Hausdorff. The traditional definition is this:

Definition

Given points $x$ and $y$ of $S$, if $x \neq y$, then there exist open neighbourhoods $U$ of $x$ and $V$ of $y$ in $S$ that are disjoint: such that their intersection $U \cap V$ is the empty set (or explicitly, such that $x' \ne y'$ whenever $x' \in U$ and $y' \in V$).

That is, any two distinct points can be separated by open neighbourhoods, and it is simply a mundane way of saying that $\ne$ is open in the product topology on $S \times S$.

Here is a classically equivalent definition that is more suitable for constructive mathematics:

Definition

Given points $x$ and $y$ of $S$, if every neighbourhood $U$ of $x$ in $S$ meets every neighbourhood $V$ of $y$ in $S$ (which means that $U \cap V$ is inhabited), then $x = y$.

This is the mundane way of saying that $=$ is closed in $S \times S$.

Another way of saying this, which makes sense also for locales, is the following:

Definition

The diagonal embedding $S \to S \times S$ is a proper map (or equivalently a closed map, since any closed subspace inclusion is proper).

This way of stating the definition generalizes to topos theory and thus to many other contexts; but it is not always a faithful generalization of the classical notion for topological spaces. See Beyond topological spaces below for more.

Here is an equivalent definition (constructively equivalent to Definition 2) that makes sense for arbitrary convergence spaces:

Definition

Given a net (or equivalently, a proper filter) in $S$, if it converges to both $x$ and $y$, then $x = y$.

That is, convergence in a Hausdorff space is unique.

Properties

The topology on a compact Hausdorff space is given precisely by the (existent because compact, unique because Hausdorff) limit of each ultrafilter on the space. Accordingly, compact Hausdorff topological spaces are (perhaps surprisingly) described by a (large) algebraic theory. In fact, the category of compact Hausdorff spaces is monadic (over Set); the monad in question maps each set to the set ultrafilters on it. (The results of this paragraph require the ultrafilter theorem, a weak form of the axiom of choice; see ultrafilter monad.)

A compact Hausdorff locale (or space) is necessarily regular; a regular locale (or $T_0$ space) is necessarily Hausdorff. Accordingly, locale theory usually speaks of ‘compact regular’ locales instead of ‘compact Hausdorff’ locales, since the definition of regularity is easier and more natural. Then a version of the previous paragraph works for compact regular locales without the ultrafilter theorem, and indeed constructively over any topos.

Proposition

Arguably, the desire to make spaces Hausdorff ($T_2$) in analysis is really a desire to make them $T_0$; nearly every space that arises in analysis is at least regular, and a regular $T_0$ space must be Hausdorff. Forcing a space to be $T_0$ is like forcing a category to be skeletal; indeed, forcing a preorder to be a partial order is a special case of both (see specialisation topology for how). It may be nice to assume, when working with a particular space, that it is $T_0$ but not to assume, when working with a particular underlying set, that every topology on it is $T_0$.

Whatever one thinks of that, there is a non-$T_0$ version of Hausdorff space, an $R_1$ space. (The symbol here comes from being a weak version of a regular space; in general a $T_i$ space is precisely both $R_{i-1}$ and $T_0$). This is also called a preregular space (in HAF) and a reciprocal space (in convergence theory).

Definition (of $R_1$)

Given points $a$ and $b$, if every neighbourhood of $a$ meets every neighbourhood of $b$, then every neighbourhood of $a$ is a neighbourhood of $b$. Equivalently, if any net (or proper filter) converges to both $a$ and $b$, then every net (or filter) that converges to $a$ also converges to $b$.

There is also a notion of sequentially Hausdorff space:

Definition (of sequentially Hausdorff)

Whenever a sequence converges to both $x$ and $y$, then $x = y$.

Some forms of predicative mathematics find this concept more useful. Hausdorffness implies sequential Hausdorffness, but the converse is false even for sequential spaces (although it is true for first-countable spaces).

The reader can now easily define a sequentially $R_1$ space.

Beyond topological spaces

Hausdorff locales

The most obvious definition for a locale $X$ to be Hausdorff is that its diagonal $X\to X\times X$ is a closed (and hence proper) inclusion. However, if $X$ is a sober space regarded as a locale, this might not coincide with the condition for $X$ to be Hausdorff as a space, since the product $X\times X$ in the category of locales might not coincide with the product in the category of spaces. But it does coincide if $X$ is a locally compact locale, so in that case the two notions of Hausdorff are the same.

Separated toposes and schemes

This notion of a Hausdorff locale is a special case of that of Hausdorff topos in topos theory. This also includes notions such as a separated scheme etc. The corresponding relative notion (over an arbitrary base topos) is that of separated geometric morphism. For schemes see separated morphism of schemes.

In constructive mathematics

In constructive mathematics, the Hausdorff notion multifurcates further, due to the variety of possible meanings of closed subspace. If we ask the diagonal to be weakly closed, then in the spatial case, this means that it contains all its limit points, giving Definition 2 above. But if we ask the diagonal to be strongly closed, i.e. the complement of an open set, then in the spatial case this means that there is a tight inequality $\ne$ (the exterior? of $=$) relative to which Definition 1 holds. (We use $\ne$ twice in that definition: in the hypothesis that $x \ne y$ and in the conclusion that $x' \ne y'$.)

It is natural to call these conditions weakly Hausdorff and strongly Hausdorff, but one should be aware of terminological clashes: in classical mathematics there is a different notion of a weak Hausdorff space, whereas (strong) Hausdorffness for locales has by some authors been called “strongly Hausdorff” only to contrast it with Hausdorffness for spaces.

As a simple example, consider a discrete space $X$ regarded as a locale. Since it is locally compact, the locale product $X\times X$ coincides with the space product (a theorem that is valid constructively); but nevertheless we have:

1. The diagonal $X\to X\times X$ always has an open complement.
2. Definition 2 above always holds, since $\{x\}$ and $\{y\}$ are neighborhoods of $x$ and $y$, and if they intersect then $x=y$. That is, $X$ is spatially weakly Hausdorff.
3. The diagonal $X\to X\times X$ is the complement of an open subset (i.e. $X$ is spatially strongly Hausdorff) if and only if equality in $X$ is stable under double negation, in other words if $X$ admits a tight inequality relation.
4. The locale diagonal $\Delta:X\to X\times X$ is a closed sublocale (i.e. $X$ is localically strongly Hausdorff) if and only if $X$ has decidable equality. For closedness of $\Delta$ means that $\Delta_\ast(U\cup \Delta^\ast(V)) \subseteq \Delta_\ast(U) \cup V$ for any $U\in O(X)$ and $V\in O(X\times X)$, where $x\in \Delta^\ast(V)$ means $(x,x)\in V$, while $(x,y)\in \Delta_\ast(U)$ means $(x=y)\to (x\in U)$. Now let $U = \emptyset$ and $V = \{ (x,x) \mid x\in X \}$. Then $(x,y) \in \Delta_\ast(U\cup \Delta^\ast(V))$ means $(x=y) \to (\bot \vee (x=x))$, which is a tautology; while $(x,y) \in \Delta_\ast(U) \cup V$ means $((x=y)\to \bot) \vee (x=y)$, i.e. $\neg(x=y) \vee (x=y)$.
5. I don’t know what it means for $X$ to be localically weakly Hausdorff. (Weak closure in locales is very inexplicit.)

In particular, the statement “all discrete locales are localically strongly Hausdorff” is equivalent to excluded middle.

However, non-discrete spaces can constructively be localically strongly Hausdorff without having decidable equality. For instance, any regular space is also regular as a locale, and hence localically strongly Hausdorff. We can also say:

Theorem

In any topological space $X$, let $x\#y$ mean that there exist opens $U,V$ with $x\in U$ and $y\in V$ and $U\cap V = \emptyset$; then $\#$ is always an inequality relation. If the spatial product $X\times X$ coincides with the locale product (such as if $X$ is locally compact), then $X$ is localically strongly Hausdorff if and only if $\#$ is an apartness relation and every open set in $X$ is $\#$-open (i.e. for any $x\in U$ and $y\in X$ we have $y\in U \vee x\#y$).

Proof

Note that $\#$ is, as a subset $W_\# \subseteq X\times X$, the exterior of the diagonal in the product topology, mentioned above. If $X$ is localically strongly Hausdorff, then $W_\#$ must be the open set of which the diagonal is the complementary closed sublocale, since it is the largest open set disjoint from the diagonal.

To say that the diagonal is its complementary closed sublocale implies in particular that for any open set $U\subseteq X$, the open set $(U\times U) \cup W_\#$ is the largest open subset of $X\times X$ whose intersection with the diagonal is contained in $U\cap U = U$. Specifically, therefore, $(U\times U) \cup W_\#$ contains $U\times X$ (since $U\times X$ is an open subset of $X\times X$ whose intersection with the diagonal is $U$). That is, if $x\in U$ and $y\in X$, then either $(x,y)\in U\times U$ (i.e. $y\in U$) or $(x,y)\in W_\#$ (i.e. $x\#y$). This shows that $U$ is $\#$-open.

To show that $\#$ is an apartness, note that for any $x$ the set $\{ z \mid x\# z \}$ is open, since it is the preimage of $W_\#$ under a section of the second projection $X\times X \to X$. Thus, it is $\#$-open, which is to say that if $x\# z$ then for any $y$ either $x\#y$ or $y\#z$, which is the missing comparison axiom for $\#$ to be an apartness.

Conversely, suppose $\#$ is an apartness and every open set is $\#$-open (i.e. the apartness topology refines the given topology on $X$). Let $A\subseteq X\times X$ be an open set; we must show that $A\cup W_\#$ is the largest open subset of $X\times X$ whose intersection with the diagonal is contained in $A\cap \Delta_X$. In other words, suppose $U\times V$ is a basic open in $X\times X$ and $(U\times V)\cap \Delta_X$ (which is $U\cap V$) is contained in $A\cap \Delta_X$; we must show $U\times V\subseteq A\cup W_\#$. In terms of elements, we assume that if $x\in U$ and $x\in V$ then $(x,x)\in A$, and we must show that if $x\in U$ and $y\in V$ then $(x,y)\in A \vee x\#y$.

Assuming $x\in U$ and $y\in V$, since $U$ and $V$ are $\#$-open we have either $y\in U$ or $x\# y$, and either $x\in V$ or $x\#y$. Since we are done if $x\#y$, it suffices to assume $y\in U$ and $x\in V$. Therefore, by assumption, $(x,x)\in A$ and $(y,y)\in A$. Since $A$ is open in the product topology, we have opens $U',V'$ with $x\in U'$ and $y\in V'$ and $U'\times U'\subseteq A$ and $V'\times V' \subseteq A$. But now $\#$-openness of $U'$ and $V'$ tells us again that either $x\#y$ (in which case we are done) or $y\in U'$ and $x\in V'$. In the latter case, $(x,y)\in U'\times U'$ (and also $V'\times V'$), and hence is also in $A$.

Note that the apartness $\#$ need not be tight, and in particular $X$ need not be spatially Hausdorff. In particular, if $X$ might not even be $T_0$: since localic Hausdorffness is (of course) only a property of the open-set lattice, it only “sees” the sobrification? and in particular the $T_0$ quotient. However, this is all that can go wrong: if $\neg(x\# y)$, then by $\#$-openness every open set containing $x$ must also contain $y$ and vice versa, so if $X$ is $T_0$ then $x=y$ and $\#$ is tight.

If the locale product $X\times X$ does not coincide with the spatial product, then the “only if” direction of the above proof still works, if we define $W_\#$ to be the open part of the locale product $X\times X$ given by $W_\# = \bigvee \{ U\otimes V \mid U\cap V = \emptyset \}$. A different proof is to recall that by this theorem, an apartness relation is the same as a (strongly) closed equivalence relation on a discrete locale, and the quotient of such an equivalence relation is the $\#$-topology. Thus, if $X$ is localically strongly Hausdorff, its diagonal is a closed equivalence relation, which yields by pullback a closed equivalence relation on the discrete locale $X_d$ on the same set of points. This is the kernel pair of the canonical surjection $X_d \to X$, and hence its quotient (the $\#$-topology) maps to $X$, i.e. refines the topology of $X$.

References

Comments on the relation to topos theory are for instance in

• S. Niefield, A note on the locally Hausdorff property, Cahiers TGDC (1983) (numdam)

Revised on February 3, 2017 12:19:40 by Mike Shulman (76.167.222.204)