Topology/Separation Axioms

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Separation Axioms for Topological SpacesEdit

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  :

T0
For every x, y, there exists an open set O that contains one point of the pair, but not the other.
T1
For every x, y, there exist open sets   and  , such that   contains x but not y and   contains y but not x.
T2
For every x, y, there exist disjoint open sets   and  , such that   contains x and   contains y.   spaces are also called Hausdorff spaces.
T
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 T3.
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.
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 T4.
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 T5.
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 T6.

NOTE: Many authors treat regular, completely regular, normal, completely normal, and perfectly normal spaces as synonyms for the corresponding Ti property.

Relations among the Separation PropertiesEdit

The Ti separation properties (axioms) form a hierarchy, such that if i>j, then property Ti implies property Tj. When property Ti+1 implies Tx, which in turn implies Ti, and Tx was proposed after Ti and Ti+1, Tx is designated T. Other implications of these properties include:

  • Complete regularity always implies regularity, even without assuming T1;
  • T0 alone suffices to make a regular space T3. The full T1 property is unnecessary;
  • Perfect normality implies complete normality, which in turn implies normality;
  • A topological space is completely normal if and only if every subspace is normal.

ExercisesEdit

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

Some Important TheoremsEdit

Theorem 3.1.2Edit

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 3.1.3Edit

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 3.1.4Edit

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 LemmaEdit

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.

ExercisesEdit

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)


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