regular space

Regular spaces



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

see also differential topology, algebraic topology, functional analysis and topological homotopy theory


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topological homotopy theory

Regular spaces


A regular space is a topological space (or variation, such as a locale) that has, in a certain sense, enough regular open subspaces. The condition of regularity is one of the separation axioms satsified by every metric space (and in this case, by every pseudometric space).


Fix a topological space XX.

The classical definition is this: that if a point and a closed set are disjoint, then they are separated by neighbourhoods. In detail, this means:


Given any point aa and closed set FF, if aFa \notin F, then there exist a neighbourhood VV of aa and a neighbourhood GG of FF such that VGV \cap G is empty.

In many contexts, it is more helpful to change perspective, from a closed set that aa does not belong to, to an open set that aa does belong to. Then the definition reads:


Given any point aa and neighbourhood UU of aa, there exist a neighbourhood VV of aa and an open set GG such that VG=V \cap G = \empty but UG=XU \cup G = X.

You can think of VV as being half the size of UU, with GG the exterior of VV. (In a metric space, or even in a uniform space, this can be made into a proof.)

If we apply the regularity condition twice, then we get what at first might appear to be a stronger result:


Given any point aa and neighbourhood UU of aa, there exist a neighbourhood WW of aa and an open set GG such that Cl(W)Cl(G)=Cl(W) \cap Cl(G) = \empty but UG=XU \cup G = X (where ClCl indicates topological closure).

Proof of equivalence

Find VV and GG as above. Now apply the regularity axiom to aa and the interior Int(V)Int(V) of VV to get WW (and HH).

In terms of the classical language of separation axioms, this says that aa and FF are separated by closed neighbourhoods.

Sometimes one includes in the definition that a regular space must be T 0T_0:

Definition (of T₀)

A space is T 0T_0 if, given any two points, if each neighbourhood of either is a neighbourhood of the other, then they are equal.

Other authors use the weaker definition above but call a regular T 0T_0 space a T 3T_3 space, but then that term is also used for a (merely) regular space. An unambiguous term for the weaker condition is an R 2R_2 space, but hardly anybody uses that.

We have


Every T 3T_3 space is Hausdorff.


Suppose every neighbourhood of aa meets every neighbourhood of bb; by T 0T_0 (and symmetry), it's enough to show that each neighbourhood UU of aa is a neighbourhood of bb. Use regularity to get VV and GG. Then GG cannot be a neighbourhood of bb, so UU is.

Since every Hausdorff space is T 0T_0, a less ambiguous term for a T 3T_3 space is a regular Hausdorff space.

It is possible to describe the regularity condition fairly simply entirely in terms of the algebra of open sets. First notice the relevance above of the condition that Cl(V)UCl(V) \subset U; we write VUV \subset\!\!\!\!\subset U in that case and say that VV is well inside UU. We now rewrite this condition in terms of open sets and regularity in terms of this condition.


Given sets U,VU, V, then VUV \subset\!\!\!\!\subset U iff there exists an open set GG such that VG=V \cap G = \empty but UG=XU \cup G = X. Then XX is regular iff, given any open set UU, UU is the union of all of the open sets that are well inside UU.

This definition is suitable for locales. As the definition of a Hausdroff locale is rather more complicated, one often speaks of compact regular locales where classically one would have spoken of compact Hausdorff spaces. (The theorem that compact regular T 0T_0 spaces and compact Hausdorff spaces are the same works also for locales, and every locale is T 0T_0, so compact regular locales and compact Hausdorff locales are the same.)

The condition that a space XX be regular is related to the regular open sets in XX, that is those open sets GG such that GG is the interior of its own closure. (In the Heyting algebra of open subsets of XX, this means precisely that GG is its own double negation; this immediately generalises the concept to locales.) Basically, we start with a neighbourhood UU of xx and reduce that to a closed neighbourhood Cl(V)Cl(V) of xx. Then Int(Cl(V))Int(Cl(V)) is a regular open neighbourhood of xx.

This gives us another way to characterise regular spaces, as follows:


Given a neighbourhood UU of xx, there is a closed neighbourhood of xx that is contained in UU. (Equivalently, xx has a regular open neighbourhood, or indeed any neighbourhood, well inside UU.) In other words, the closed neighbourhoods of xx form a local base (a base of the neighbourhood filter) at xx.


It is not sufficient that the regular open neighbourhoods themselves form a local base of each point; see Counterexample . It's the closures of the regular open neighbourhoods (which are arbitrary closed neighbourhoods) that form the basis. But compare semiregular spaces below.

In constructive mathematics, Definition is good; then everything else follows without change, except for the equivalence with . Even then, the classical separation axioms hold for a regular space; they just are not sufficient.


Following up on Definition , we have:


For any regular space XX, the regular open sets form a basis for the topology of XX.


For any closed neighbourhood VV of xXx \in X, the interior Int(V)Int(V) is a regular open neighbourhood of xx. Using Definition finishes the proof.

This suggests a slightly weaker condition, that of a semiregular space:

Definition (of semiregular)

The regular open sets form a basis for the topology of XX.

As we've seen above, a regular T 0T_0 space (T 3T_3) is Hausdorff (T 2T_2); we can also remove the T 0T_0 condition from the latter to get R 1R_1:

Definition (of R₁)

Given points aa and bb, if every neighbourhood of aa meets every neighbourhood of bb, then every neighbourhood of aa is a neighbourhood of bb.

It is immediate that T 2R 1T 0T_2 \equiv R_1 \wedge T_0, and the proof above that T 3T 2T_3 \Rightarrow T_2 becomes a proof that R 2R 1R_2 \Rightarrow R_1; that is, every regular space is R 1R_1. An R 1R_1 space is also called preregular (in HAF) or reciprocal (in convergence space theory).

A bit stronger than regularity is complete regularity; a bit stronger than T 3T_3 is T 312T_{3\frac{1}{2}}. The difference here is that for a completely regular space we require that aa and FF be separated by a function, that is by a continuous real-valued function. See Tychonoff space for more. This strengthening implies (Tychonoff Embedding Theorem) that the space embeds into a product of metric spaces.

For locales, there is also a weaker notion called weak regularity?, which uses the notion of fiberwise closed sublocale? instead of ordinary closed sublocales.



Let (X,d)(X,d) be a metric space regarded as a topological space via its metric topology. Then this is a normal Hausdorff space, in particular hence a regular Hausdorff space.


We need to show is that given two disjoint closed subsets C 1,C 2XC_1, C_2 \subset X then their exists disjoint open neighbourhoods U C 1C 1U_{C_1} \subset C_1 and U C 2C 2U_{C_2} \supset C_2.

Consider the function

d(S,):X d(S,-) \colon X \to \mathbb{R}

which computes distances from a subset SXS \subset X, by forming the infimum of the distances to all its points:

d(S,x)inf{d(s,x)|sS}. d(S,x) \coloneqq inf\left\{ d(s,x) \vert s \in S \right\} \,.

Then the unions of open balls

U C 1x 1C 1B x 1 (d(C 2,x 1)) U_{C_1} \coloneqq \underset{x_1 \in C_1}{\cup} B^\circ_{x_1}( d(C_2,x_1) )


U C 2x 2C 2B x 2 (d(C 1,x 2)). U_{C_2} \coloneqq \underset{x_2 \in C_2}{\cup} B^\circ_{x_2}( d(C_1,x_2) ) \,.

have the required properties.


The real numbers equipped with their K-topology K\mathbb{R}_K are a Hausdorff topological space which is not a regular Hausdorff space (hence in particular not a normal Hausdorff space).


By construction the K-topology is finer than the usual euclidean metric topology. Since the latter is Hausdorff, so is K\mathbb{R}_K. It remains to see that K\mathbb{R}_K contains a point and a disjoint closed subset such that they do not have disjoint open neighbourhoods.

But this is the case essentially by construction: Observe that

\K=(,1/2)((1,1)\K)(1/2,) \mathbb{R} \backslash K \;=\; (-\infty,-1/2) \cup \left( (-1,1) \backslash K \right) \cup (1/2, \infty)

is an open subset in K\mathbb{R}_K, whence

K=\(\K) K = \mathbb{R} \backslash ( \mathbb{R} \backslash K )

is a closed subset of K\mathbb{R}_K.

But every open neighbourhood of {0}\{0\} contains at least (ϵ,ϵ)\K(-\epsilon, \epsilon) \backslash K for some positive real number ϵ\epsilon. There exists then n 0n \in \mathbb{N}_{\geq 0} with 1/n<ϵ1/n \lt \epsilon and 1/nK1/n \in K. An open neighbourhood of KK needs to contain an open interval around 1/n1/n, and hence will have non-trivial intersection with (ϵ,ϵ)(-\epsilon, \epsilon). Therefore {0}\{0\} and KK may not be separated by disjoint open neighbourhoods, and so K\mathbb{R}_K is not normal.


Let ((0,1)×(0,1)){0}\bigl((0, 1)\times(0, 1)\bigr)\cup\{0\} be equipped with the Euclidean topology on (0,1)×(0,1)(0, 1)\times(0,1) and have the sets of the form (0,1/2)×(0,ε){0}(0, 1/2)\times(0, \varepsilon)\cup\{0\} (for ε(0,1)\varepsilon \in (0, 1)) as a basis of open neighbourhoods for the point 00.

  • This space is not regular since we cannot separate 00 from [1/2,1)×(0,1)[1/2, 1)\times(0,1)
  • Every point p=(p 1,p 2)0p = (p_1, p_2) \neq 0 has the euclidean balls of centre pp and radius ε(0,p 1)\varepsilon \in (0, p_1) as regular neighbourhood basis.
  • The provided basis for the neighbourhoods of 00 already is a system of regular open sets.

Therefore, this space has the property that every point has a neighbourhood basis of regular open sets (and consequently, the space is semiregular, an even weaker property), but 00 does not have a neighbourhood basis of closed sets (and consequently, the space is not regular). The problem is that, while every basic neighbourhood of 00 (and therefore every neighbourhood of 00 whatsoever) contains a regular open neighbourhood of 00, none of these basic neighbourhoods contains the closure of any of these regular open neighbourhoods (or any other closed neighbourhood of 00).


the main separation axioms

T 0T_0Kolmogorovgiven two distinct points, at least one of them has an open neighbourhood not containing the other pointevery irreducible closed subset is the closure of at most one point
T 1T_1given two distinct points, both have an open neighbourhood not containing the other pointall points are closed
T 2T_2Hausdorffgiven two distinct points, they have disjoint open neighbourhoodsthe diagonal is a closed map
T >2T_{\gt 2}T 1T_1 and…all points are closed and…
T 3T_3regular 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 4T_4normal 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

A uniform space is automatically regular and even completely regular, at least in classical mathematics. In constructive mathematics this may not be true, and there is an intermediate notion of interest called uniform regularity.

Every regular space comes with a naturally defined (point-point) apartness relation: we say x#yx # y if there is an open set containing xx but not yy. This can be defined for any topological space and is obviously irreflexive, but in a regular space it is symmetric and a comparison, hence an apartness. For symmetry, if xUx\in U and yUy\notin U, let VV be an open set containing xx and GG an open set such that VG=V\cap G = \emptyset and GU=XG\cup U = X; then yGy\in G (since yUy\notin U) while xGx\notin G (since xVx\in V). With the same notation, to prove comparison, for any zz we have either zGz\in G, in which case z#xz # x, or zUz\in U, in which case z#yz # y. Note that this argument is valid constructively; indeed, classically, the much weaker R 0R_0 separation axiom is enough to make this relation symmetric, and it is a comparison on any topological space whatsoever.

Note that if a space is localically strongly Hausdorff (a weaker condition than regularity), then it has an apartness relation defined by x#yx \# y if there are disjoint open sets containing xx and yy. If XX is regular, then this coincides with the above-defined apartness.

Last revised on December 11, 2019 at 12:08:38. See the history of this page for a list of all contributions to it.