# General Topology/Separation

Suppose we have a topological space and two distinct points . Suppose further that using the topological structure, we try to separate these points, that is, to find sets which contain one of the points, but not the other. Usually this is not possible. But the topological spaces where it is possible form important classes of topological spaces, which are then said to satisfy *separation axioms*. There is an important hierarchy of separation axioms which are numbered T_{0} until T_{6}, and then there are the R_{0} and R_{1} axioms. The T axioms concern separation in the classical sense, whereas the R axioms concern only separation of topologically distinguishable points.

**Definition (topological distinguishability)**:

Let be a topological space, and let . The two points and are said to be **topologically distinguishable** iff there exists an open set of that contains one point of , but not the other. Otherwise, and are called **topologically indistinguishable**.

**Definition (T _{0} space)**:

Let be a topological space. is said to be a **T _{0} space** if and only if for any two points which are distinct (ie. ) there exists an open set of which either contains or , but not both.

Alternatively, we may define a T_{0} space to be a space where any two points are topologically distinguishable.

The situation is depicted in the following picture:

**Definition (T _{1} space)**:

Let be a topological space. is said to be a **T _{1} space** iff for any two distinct points , there exists an open set of which does not contain but does contain .

The situation is depicted in the following picture:

We introduce our first R axiom:

**Definition (R _{0} space)**:

An R_{0} space is a topological space where for any pair of topologically distinguishable points , there exists both a neighbourhood of not containing and a neighbourhood of not containing .

**Proposition (characterisation of T _{1} spaces)**:

Let be a topological space. The following are equivalent:

- is a T
_{1}space - All cofinite subsets of are open
- All finite subsets of are closed
- The points of are closed
- is simultaneously T
_{0}and R_{0}

**Proof:** If is a T_{1} space and , then is open, because for we find an open neighbourhood of (which may be chosen the union of all such neighbourhoods in order to avoid the axiom of choice) such that , and then

- .

Then every cofinite subset of is open as the intersection of finitely many open sets. Since the closed sets are the complements of the open sets, 3. follows. 3. 4. is trivial. Given distinct points , set and . Then and show that is T_{0} (any two points are topologically distinguishable) and R_{0} (if a statement holds for any two points, then in ptic. for topologically distinguishable points). Finally suppose is simultaneously T_{0} and R_{0}. Let be any two distinct points. Since is T_{0}, are topologically distinguishable, and since is R_{0}, we find neighbourhoods of and of so that and .

**Definition (R _{1} space)**:

A topological space is called R_{1} if and only if for any two topologically distinguishable points of , there exist open neighbourhoods of and of so that .

By far the most common class of spaces that satisfy a separation axiom are the Hausdorff spaces:

**Definition (Hausdorff space)**:

Let be a topological space. is said to be a **Hausdorff space** if and only if for any two distinct points there exists two open sets and of so that , and .

This situation is depicted in the following picture:

**Proposition (separation of finitely many points in Hausdorff spaces)**:

Let be a Hausdorff topological space, and let be distinct points in . Then there exist neighbourhoods of () such that for , we have .

**Proof:** We proceed by induction on . The case is trivial. Suppose the claim holds true for , and let be distinct points in . By induction, choose neighbourhoods of respectively such that for . Since is Hausdorff, we find for all neighbourhoods of resp. such that . Then set and for set . Then is a set of open sets as desired.

Hausdorff spaces are located within the T_{0} until T_{6} hierarchy:

**Definition (T _{2} space)**:

A T_{2} space is a Hausdorff space.

**Proposition (R-axiom characterisation of Hausdorff spaces)**:

Let be a topological space. is Hausdorff if and only if is T_{0} and R_{1}.

**Proof:** If is Hausdorff, it is certainly T_{0} and R_{1}. On the other hand, if it is T_{0}, all points are topologically indistinguishable, so that for *any* two points we find open with , and .

**Proposition (diagonal criterion for being Hausdorff)**:

Let be a topological space, and let be the diagonal. Then is Hausdorff if and only if is closed, where we consider as a topological space with the product topology.

**Proof:** Let be arbitrary. being Hausdorff means that there exist open neighbourhoods and so that . If that is always the case, then whenever is not in and are so chosen, then is a neighbourhood of in , so that the diagonal is closed as the complement of an open set. Conversely, if is closed, let be arbitrary and not equal to each other. Then , and since is open, by definition of the product topology we find open sets and of so that and . From the first of these conditions we deduce that and , and from the second we deduce that , since any would yield .

**Definition (Urysohn space)**:

A topological space is called a **Urysohn space** iff for all distinct points we find closed *neighbourhoods* of resp. so that .

**Proposition (Urysohn spaces are Hausdorff)**:

Let be a Urysohn space. Then is Hausdorff.

**Proof:** Let so that , and pick closed neighbourhoods of resp. so that . Then by def. of a neighbourhood of a point, pick and open s.t. , . Then are neighbourhoods of resp. that satisfy the requirements given by the Hausdorff condition.

**Definition (T _{2½} space)**:

A topological space is a **T _{2½} space** iff it is a Urysohn space.

There are also stronger separation conditions. They are usually formulated in regard to closed sets, and hence they don't really fit in with the T-axiom hierarchy (you'll see what I mean), but combining them with a small T axiom (namely T_{0} or T_{1}), we produce corresponding axioms in the T-axiom hierarchy.

**Definition (regular space)**:

Let be a topological space. is **regular** iff for every closed set and every point , there exist open sets so that , and .

**Proposition (regular spaces are characterised by the existence of closed neighbourhood fundamental systems)**:

Let be a topological space. is regular if and only if for all there exists a fundamental system of consisting of closed sets.

**Proof:** Suppose first that each () has a fundamental system of closed sets. Let be any closed set and let . Now is an open neighbourhood of . Since has a fundamental system consisting of closed neighbourhoods, pick closed so that . Since is a neighbourhood of , pick open so that . Then and do the trick in the definition of a regular space.

Suppose now that is regular, let , and let be any open neighbourhood of . Claim that contains a closed neighbourhood of . Indeed, set . By regularity, pick a neighbourhood of and an open set containing such that . Then is the closed neighbourhood as desired, since it contains the open neighbourhood of .

**Definition (T _{3} space)**:

A **T _{3} space** is a topological space which is T

_{0}and regular.

**Proposition (T _{3} spaces are Urysohn)**:

Let be a T_{3} space. Then is a Urysohn space.

**Proof:** Since is regular, there exists for each point a closed fundamental system of the neighbourhood system. Hence, let be any two distinct points. It suffices to prove that they are separated by disjoint open neighbourhoods. By the T_{0} property wlog. choose an open neighbourhood of so that . Set , which is closed and contains . Since is regular, we may pick open disjoint sets and such that , and .

**Definition (completely regular space)**:

Let be a topological space. is a **completely regular space** iff for all closed sets and points we can find a continuous function such that and .

**Proposition (completely regular spaces are regular)**:

Let by a completely regular space. Then is regular.

**Proof:** Let and let be closed so that . Since is completely regular, pick a continuous function so that and for . Then set and and observe that and are open by the definition of the subspace topology of , that and certainly , , so that is regular.

**Definition (T _{3½} space)**:

A **T _{3½} space** is a topological space that is both T

_{0}and completely regular.

**Proposition (T _{3½} spaces are T_{3})**:

Let be a T_{3½} space. Then is T_{3}.

**Proof:** It is T_{0} by definition and regular since completely regular spaces are regular.

**Definition (normal space)**:

A **normal space** is a topological space so that for all closed disjoint subsets , , there exist open sets such that , and .

**Definition (T _{4} space)**:

A **T _{4}** space is a topological space that is both T

_{1}and normal.

The situation is depicted in the following picture:

**Theorem (Urysohn's lemma)**:

Let be a normal space and let be two closed disjoint subspaces of . Then we can find a continuous function such that for all and for all .

**Proof:** We instead prove that if we are given an open set of and a closed set , then we can find a continuous function such that is zero inside and outside ; the theorem will then follow upon defining and applying our auxiliary result.

Indeed, note that since is normal, we find an open and a closed set (which we suggestively denote by and ) such that . We now repeat this procedure; that is, for every , we construct a sequence

from normality and the sequence that we had gotten in the step (this requires the axiom of dependent choice). Then we define

- ,

where we define to equal or , depending on whether is even or odd, and and . We then have for and for , and further is continuous; indeed, it is continuous at each point, since if we set for a fixed and fix an , then we choose sufficiently large so that and then so that (if , then is a neighbourhood that will map wholly to ) and then either is even, so that maps to (note if and even), or is odd and is a neighbourhood of the desired form.

**Proposition (T _{4} spaces are T_{3½})**:

Let be a T_{4} space. Then is a T_{3½} space.

**Proof:** By Urysohn's lemma, it suffices to note that all points of are closed. But this is the case, since is T_{1}.

**Theorem (Tietze–Urysohn theorem)**:

Let be a normal space, let be closed, and let be a continuous function. Then there exists a continuous function such that .

**Proof:**

## Exercises[edit | edit source]

- Let be a finite set. Prove that there exists exactly one Hausdorff topology on and find out what it is.
- Let be a topological space. Prove that the sets such that is finite, when the empty set is adjoined to it, form a topology on (this topology is called the
**co-finite topology**). Further prove that if is infinite, then together with this topology it is , but not Hausdorff. - Let be a topological space that is equipped with the initial topology stemming from a family of functions . Prove that is a T
_{0}space if and only if for each in , there exists an such that and are topologically distinguishable.