nLab Hausdorff space

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 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.

the main separation axioms

number	name	statement	reformulation
$T_0$	Kolmogorov	given two distinct points, at least one of them has an open neighbourhood not containing the other point	every irreducible closed subset is the closure of at most one point
$T_1$		given two distinct points, both have an open neighbourhood not containing the other point	all points are closed
$T_2$	Hausdorff	given two distinct points, they have disjoint open neighbourhoods	the diagonal is a closed map
$T_{\gt 2}$		$T_1$ and…	all points are closed and…
$T_3$	regular Hausdorff	…given a point and a closed subset not containing it, they have disjoint open neighbourhoods	…every neighbourhood of a point contains the closure of an open neighbourhood
$T_4$	normal Hausdorff	…given two disjoint closed subsets, they have disjoint open neighbourhoods	…every neighbourhood of a closed set also contains the closure of an open neighbourhood … every pair of disjoint closed subsets is separated by an Urysohn function

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 ) that makes sense for arbitrary convergence spaces or preconvergence 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.

Examples

Proposition

a CW-complex is a Hausdorff space.

Properties

In terms of lifting properties

The separation conditions $T_0$ to $T_4$ may equivalently be understood as lifting properties against certain maps of finite topological spaces, among others.

This is discussed at separation axioms in terms of lifting properties, to which we refer for further details. Here we just briefly indicate the corresponding lifting diagrams.

In the following diagrams, the relevant finite topological spaces are indicated explicitly by illustration of their underlying point set and their open subsets:

• points (elements) are denoted by $\bullet$ with subscripts indicating where the points map to;

• boxes are put around open subsets,

• an arrow $\bullet_u \to \bullet_c$ means that $\bullet_c$ is in the topological closure of $\bullet_u$.

In the lifting diagrams for $T_2-T_4$ below, an arrow out of the given topological space $X$ is a map that determines (classifies) a decomposition of $X$ into a union of subsets with properties indicated by the picture of the finite space.

Notice that the diagrams for $T_2$-$T_4$ below do not in themselves imply $T_1$.

Proposition

(Lifting property encoding $T_0$)
The following lifting property in Top equivalently encodes the separation axiom $T_0$:

Proposition

(Lifting property encoding $T_1$)
The following lifting property in Top equivalently encodes the separation axiom $T_1$:

Proposition

(Lifting property encoding $T_2$)
The following lifting property in Top equivalently encodes the separation axiom $T_2$:

Proposition

(Lifting property encoding $T_3$)
The following lifting property in Top equivalently encodes the separation axiom $T_3$:

Proposition

(Lifting property encoding $T_4$)
The following lifting property in Top equivalently encodes the separation axiom $T_4$:

Hausdorff reflection

Proposition

(Hausdorff reflection)

For every topological space $X$ there exists a Hausdorff topological space $H X$ and a continuous function

$$h_X \;\colon\; X \longrightarrow H X$$

which is the “closest approximation from the left” to $X$ by a Hausdorff topological space, in that for $Y$ any Hausdorff topological space, then continuous functions of the form

$$f \;\colon\; X \longrightarrow Y$$

are in bijection with continuous function of the form

$$\tilde f \;\colon\; H X \longrightarrow Y$$

and such that the bijection is constituted by

$$f = \tilde f \circ h_X \;\colon\; X \overset{h_X}{\longrightarrow} H X \overset{\tilde f}{\longrightarrow} Y \,.$$

Here $H X$ (or more precisely $h_X$) is also called the Hausdorff reflection (or Hausdorffication or similar) of $X$.

Moreover, the operation $H(-)$ extends to continuous functions $f \colon X \to Y$

$$(X \overset{f}{\to} Y) \;\mapsto\; (H X \overset{H f}{\to} H Y)$$

by setting

$$H f \;\colon\; [x] \mapsto [f(x)] \,,$$

where $[x]$ denotes the equivalence class under $\sim_X$ of any $x \in X$.

Finally, the comparison map is compatible with this in that the follows squares commute:

$$\array{ X &\overset{f}{\longrightarrow}& Y \\ {}^{\mathllap{h_X}}\downarrow && \downarrow^{\mathrlap{h_Y}} \\ H X &\underset{H f}{\longrightarrow}& H Y } \,.$$

Proof

There are various ways to construct $h_X$, see below prop. and prop.

Remark

In the language of category theory the Hausdorff reflection of prop. says that

1. $H$ is a functor $H \;\colon\; Top \longrightarrow Top_{Haus}$ from the category Top of topological spaces to the full subcategory $Top_{Haus} \overset{\iota}{\hookrightarrow} Top$ of Hausdorff topological spaces;

2. $h_X \colon X \to H X$ is a natural transformation from the identity functor on Top to the functor $\iota \circ H$

3. Hausdorff topological spaces form a reflective subcategory of all topological spaces in that $H$ is left adjoint to the inclusion functor $\iota$

   $$Top_{Haus} \underoverset{\underset{\iota}{\hookrightarrow}}{\overset{H}{\longleftarrow}}{\bot} Top \,.$$

There are various ways to see the existence and to construct the Hausdorff reflection. The following is maybe the quickest way to see the existence, even though it leaves the actual construction rather implicit.

Proposition

Let $(X,\tau)$ be a topological space and consider the equivalence relation $\sim$ on the underlying set $X$ for which $x \sim y$ precisely if for every surjective continuous function $f \colon X \to Y$ into any Hausdorff topological space $Y$ we have $f(x) = f(y)$.

Then the set of equivalence classes

$$H X \coloneqq X /{\sim}$$

equipped with the quotient topology is a Hausdorff topological space and the quotient map $h_X \;\colon\; X \to X/{\sim}$ exhibits the Hausdorff reflection of $X$, according to prop. .

Proof

First observe that every continuous function $f \colon X \longrightarrow Y$ into a Hausdorff space $Y$ factors uniquely via $h_X$ through a continuous function $\tilde f$

$$f = \tilde f \circ h_X$$

where

$$\tilde f \colon [x] \mapsto f(x) \,.$$

That this is well defined and continuous follows directly from the definitions.

What remains to be seen is that $H X$ is indeed a Hausdorff space. Hence assume that $[x] \neq [y] \in H X$. By construction of $H X$ this means that there exists a Hausdorff space $Y$ and a surjective continuous function $f \colon X \longrightarrow Y$ such that $f(x) \neq f(y) \in Y$. Accordingly, since $Y$ is Hausdorff, there exist disjoint open neighbourhoods $U_x, U_y \in \tau_Y$. Moreover, by the previous statement there exists a continuous function $\tilde f \colon H X \to Y$ with $\tilde f([x]) = f(x)$ and $\tilde f([y]) = f(y)$. Since, by the nature of continuous functions, the pre-images $\tilde f^{-1}( U_x ), \tilde f^{-1}(U_y) \subset H X$ are still disjoint and open, we have found disjoint neighbourhoods of $[x]$ and $[y]$. Hence $H X$ is Hausdorff.

Some readers may find the following a more direct way of constructing the Hausdorff reflection:

Proposition

For $(Y,\tau_Y)$ a topological space, write $r_Y \subset Y \times Y$ for the transitive closure of the relation given by the topological closure $Cl(\Delta_Y)$ of the image of the diagonal $\Delta_Y \colon Y \hookrightarrow Y \times Y$.

$$r_Y \coloneqq Trans(Cl(Delta_Y)) \,.$$

Now for $(X,\tau_X)$ a topological space, define by induction for each ordinal number $\alpha$ an equivalence relation $r^\alpha$ on $X$ as follows, where we write $q^\alpha \colon X \to H^\alpha(X)$ for the corresponding quotient topological space projection:

We start the induction with the trivial equivalence relation:

• $r^0_X \coloneqq \Delta_X$;

For a successor ordinal we set

• $r_X^{\alpha+1} \coloneqq \left\{ (a,b) \in X \times X \,\vert\, (q^\alpha(a), q^\alpha(b)) \in r_{H^\alpha(X)} \right\}$

and for a limit ordinal $\alpha$ we set

• $r_X^\alpha \coloneqq \underset{\beta \lt \alpha}{\cup} r_X^\beta$.

Then:

1. there exists an ordinal $\alpha$ such that $r_X^\alpha = r_X^{\alpha+1}$

2. for this $\alpha$ then $H^\alpha(X) = H(X)$ is the Hausdorff reflection from prop. .

(vanMunster 14, section 4)

Monadicity

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.

Sobriety

Using classical logic (but not in constructive logic) every Hausdorff space is a sober topological space: Hausdorff implies sober.

Relation to compact spaces

Proposition

compact subspaces of Hausdorff spaces are closed.

Proposition

maps from compact spaces to Hausdorff spaces are closed and proper

Dense subspaces and Epimorphisms

Proposition

In the category of Hausdorff topological spaces (with continuous functions between them), the inclusion of a dense subspace

$$A \overset{\;\;i\;\;}{\hookrightarrow} X$$

is an epimorphism.

Here is a proof in the language of category theory:

Proof

We have to show that for $(f,g)$ any pair of parallel morphisms out of $X$

$$A \overset{\;\;i\;\;}{\hookrightarrow} X \underoverset {\;\;g\;\;} {\;\;f\;\;} {\rightrightarrows} Y$$

into a Hausdorff space $Y$, the equality $f \circ i = g \circ i$ implies $f = g$. Equivalently, that $f \circ i = g \circ i$ implies that $1_X \colon X \to X$ is the equalizer of the pair $(f, g)$. But the equalizer $E \to X$ is formed by taking a pullback of the diagonal map $\Delta \colon Y \to Y \times Y$ along $(f, g) \colon X \to Y \times Y$:

$$\array{ E & \to & Y \\ \downarrow & & \downarrow \mathrlap{\Delta} \\ X & \underset{(f, g)}{\to} & Y \times Y. }$$

Since $Y$ is Hausdorff, the subset $\Delta: Y \to Y \times Y$ is closed, and the pullback $E \hookrightarrow X$ of this closed subset along the continuous map $h = (f, g)$, which is $E = h^{-1}(\Delta)$, is also closed. Since $E$ is a closed subset of $X$ and contains a dense subspace $i: A \hookrightarrow X$, it must be all of $X$ (as a subset of itself).

Note, incidentally, that $X$ itself doesn’t have to be Hausdorff for the argument to go through.

Alternatively, here is a proof in the language of basic topology:

Proof

We have to show that for $(f,g)$ any pair of parallel morphisms out of $X$

$$A \overset{\;\;i\;\;}{\hookrightarrow} X \underoverset {\;\;g\;\;} {\;\;f\;\;} {\rightrightarrows} Y$$

into a Hausdorff space $Y$, the equality $f \circ i = g \circ i$ implies $f = g$. With classical logic we may equivalently show the contrapositive: That $f \neq g$ implies $f \circ i \neq g \circ i$.

So assume that $f \neq g$. This means that there exists $x \in X$ with $f(x) \neq g(x)$. Since $Y$ is Hausdorff, there exist then disjoint open neighbourhoods $O_{f(x)},\;O_{g(x)} \subset Y$, i.e. $f(x) \in O_{f(x)}$ and $g(x) \in O_{g(x)}$ with $O_{f(x)} \cap O_{g(x)} = \varnothing$.

But their preimages must intersect at least in $x \in f^{-1}\big( O_{f(x)} \big) \cap g^{-1}\big( O_{g(x)} \big)$. Since this intersection is an open subset (as preimages of open subsets are open by definition of continuous functions, and since finite intersections of open subsets are open by the definition of topological spaces) there exists a point $a \in A$ with $i(a) \in f^{-1}\big( O_{f(x)} \big) \cap g^{-1}\big( O_{g(x)} \big)$ (by definition of dense subset). But since then $f(i(a)) \in O_{f(x)}$ and $g(i(a)) \in O_{g(x)}$ while $O_{f(x)}$ is disjoint from $O_{g(x)}$, it follows that $f(i(a)) \neq g(i(a))$. This means that $f \circ i \neq g \circ i$.

In fact the converse holds: any epimorphism in the category of Hausdorff spaces has dense image (e.g. LL 18).

Related notions

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 Cartesian product $X\times X$ in the category Loc of locales might not coincide with the product in the category Top of topological spaces (the Tychonoff product). 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 is formally similar to 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 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 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 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 (Kolmogorov 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 $X$.

Related concepts

• non-Hausdorff topological space

• locally Hausdorff topological space

• Kolmogorov topological space

• nice topological space

References

Due to

• Felix Hausdorff, 1914

See also

• Wikipedia, Hausdorff space

A detailed discussion of Hausdorff reflection is in

• Bart van Munster, The Hausdorff quotient, 2014 (pdf)

On epimorphisms of Hausdorff spaces:

• Jérôme Lapuyade-Lahorgue, The epimorphisms of the category Haus are exactly the image-dense morphisms (arXiv:1810.00778)

Comments on the relation to topos theory:

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