# Topology/Separation Axioms

## Separation Axioms for Topological Spaces edit

A topology on a space is a collection of subsets called *open*. We can then ask questions such as "can we separate any two distinct points in the space by enclosing them in two disjoint open sets?" For the real line with its usual topology, the answer is obviously yes, but there are spaces for which this is not so. It turns out that many properties of continous maps one could take for granted depend, in fact, on one of the conditions stated below holding. Topological spaces are classified according to which such conditions, called *separation axioms*, happen to hold.

Let be a topological space, and let *x,y* be any two distinct points in that space. The following conditions, ordered from least to most restrictive, are ones we may wish to place on :

*T*_{0}- For every
*x, y*, there exists an open set*O*that contains one point of the pair, but not the other.

*T*_{1}- For every
*x, y*, there exist open sets and , such that contains*x*but not*y*and contains*y*but not*x*.

*T*_{2}- For every
*x, y*, there exist*disjoint open sets*and , such that contains*x*and contains*y*. spaces are also called**Hausdorff**spaces.

*T*_{2½}- For every
*x, y*, there exist*disjoint closed neighborhoods*and of*x*and*y*respectively.

**regular**- If
*C*is a closed set, and*z*is a point not in*C*, there exist disjoint open sets and , such that contains*C*and contains*z*. A topological space that is both regular and is called*T*_{3}

**completely regular**- If
*C*is a closed set, and*z*is a point not in*C*, there exists a*continuous function*such that*f*(*z*)=0 and for any , we have*f*(*w*)=1 (i.e.*f*(*C*)={1}). A topological space that is both completely regular and is called*T*_{3½}

**normal**- If and are
*disjoint closed sets*, there exist disjoint open sets and , such that contains and contains . A topological space that is both normal and is called*T*_{4}

**completely normal**- Let and be
*separated sets*, meaning that . Then there exist disjoint open sets and , such that contains and contains . A topological space that is both completely normal and is called*T*_{5}

**perfectly normal**- If and are disjoint closed sets, there exists a continuous function such that and . A topological space that is both perfectly normal and is called
*T*_{6}

NOTE: Many authors treat regular, completely regular, normal, completely normal, and perfectly normal spaces as synonyms for the corresponding *T _{i}* property.

## Relations among the Separation Properties edit

The *T _{i}* separation properties (axioms) form a hierarchy, such that if

*i>j*, then property

*T*implies property

_{i}*T*. When property

_{j}*T*implies

_{i+1}*T*, which in turn implies

_{x}*T*, and

_{i}*T*was proposed after

_{x}*T*and

_{i}*T*,

_{i+1}*T*is designated

_{x}*T*. Other implications of these properties include:

_{i½}- Complete regularity always implies regularity, even without assuming
*T*;_{1} *T*alone suffices to make a regular space_{0}*T*. The full_{3}*T*property is unnecessary;_{1}- Perfect normality implies complete normality, which in turn implies normality;
- A topological space is completely normal if and only if every subspace is normal.

## Exercises edit

- Suppose that a topological space is . Given , show that is open to conclude that is closed.
- Given a topological space , and given that for all , is closed, show that is .

## Some Important Theorems edit

### Theorem 3.1.2 edit

Let be a space and let be a sequence in . Then either does not converge in or it converges to a unique limit.

**Proof**

Assume that converges to two distinct values and .

Since is , there are disjoint open sets and such that and .

Now by definition of convergence, there is an integer such that implies . Similarly there is an integer such that implies .

Take an integer that is greater than both and , so that is in both and , contradicting the fact that the two sets are disjoint. Therefore cannot converge to both and .

### Theorem 3.1.3 edit

If is a metric space, then is normal.

**Proof**

Given any , and any point , define the distance, from to by , where is the distance function supplied by definition of a metric space. Observe that is continuous.

Fix closed, disjoint subsets of , and define by (Note well--defined, since for any , we have iff in the closure of .) Observe is 1 on , -1 on , and in the open interval elsewhere. Also, continuous by the continuity of . Therefore, and are the preimages of open sets (open in , that is) and therefore open, and they're disjoint as the preimages of disjoint sets.

### Theorem 3.1.4 edit

If is a metric space, then is Hausdorff.**Proof**

Let and be two distinct points, and let . Then and are open sets which are disjoint since if there is a point within both open balls, then , a contradiction.

### Urysohn's Lemma edit

A topological space is normal if and only if for any disjoint closed sets and , there exists a continuous function such that and .

*Proof*

In order to prove Urysohn's Lemma, we first prove the following result:

Let X be a topological space. X is normal if and only if for every closed set U, and open set V containing U, there is an open set S containing U whose closure is within V.

Suppose that X is normal. If V is X, then X is a set containing U whose closure is within V. Otherwise, the complement of V is a nonempty closed set, which is disjoint from U. Thus, by the normality of X, there are two disjoint open sets A and B, where A contains U and B contains the complement of V. The closure of A does not meet B because all points in B have a neighborhood that is entirely within B and thus does not meet A (since they are disjoint), so all points in B are not within the closure of A. Thus, the set A is a set containing U, and whose closure does not meet B, and therefore does not meet the complement of V, and therefore is entirely contained in V. Conversely, take any two disjoint closed sets U and V. The complement of V is an open set containing the closed set U. Therefore, there is a set containing U whose closure is within the complement of V, which is the same thing as being disjoint from V. Then the complement of V is an open set containing the closed set Cl( ). Therefore, there is a set containing such that Cl( ) is within the complement of V i. e. is disjoint from V. Then and the complement of are open sets which respectively contain U and V, which are disjoint.

Now we prove Urysohn's Lemma.

Let X be a normal space, and let U and V be two closed sets. Set to be U, and set to be X.

Let be a set containing U_0 whose closure is contained in . In general, inductively define for all natural numbers n and for all natural numbers

to be a set containing whose closure is contained within the complement of . This defines where p is a rational number in the interval [0,1] expressible in the form where a and n are whole numbers.

Now define the function [0,1] to be f(p)=inf{x| }.

Consider any element x within the normal space X, and and consider any open interval (a,b) around f(x). There exists rational numbers c and d in that open interval expressible in the form where p and n are whole numbers, such that c<f(x)<d. If c<0, then replace it with 0, and if d>1, then replace it with 1. Then the intersection of the complement of the set and the set is an open neighborhood of f(x) with an image within (a,b), proving that the function is continuous.

Conversely, suppose that for any two disjoint closed sets, there is a continuous function f from X to [0,1] such that f(x)=0 when x is an element of U, and that f(x)=1 when x is an element of V. Then since the disjoint sets [0,.5) and (.5,1] are open and under the subspace topology, the inverses , which contains X, and , which contains Y, are also open and disjoint.

## Exercises edit

It is instructive to build up a series of spaces, such that each member belongs to one class, but not the next.

- The indiscrete topology is not .
- If is the unit interval , and , then this space is but not .
- Consider an arbitrary infinite set . Let and every finite subset be closed sets, and call the open sets . Determine whether , is a topological space which is but not . Hint: Consider the intersection of any two open, non-empty sets.

Verify that

- implies
- implies
- implies (hint: use theorem 3.1.1)