Topological Spaces

Topology

Topological Space

A topology on a set is a collection of subsets of , which we agree to call the “open sets”, such that

1. and are open；
2. the intersection of finitely many open sets is open；
3. arbitrary unions of open sets are open.

The pair is called a topological space.

e.g.

• The topology induced by a metric: in any metric space the collection of all open sets forms a topology.
• A topology is said to be discrete if all subsets are open. It is indiscrete or trivial if the only open sets are and .
• Zariski Topology on : A set is open if it is , , or its complement is the set of zeros of a polynomial with real coefficients.
• A topology is cofinite if all open subsets are , , or the set whose complement is finite.
• A topology is cocountable if all open subsets are , , or the set whose complement is countable.
  

Closedness

In a topological space , a set is called closed if is open. i.e. .

Coarser and Finer

If and are two topologies on then we say that is coarser than if , that is contains fewer open sets than . In this situation, we also say that is finer than .

e.g. Given a set , the trivial topology is the coarsest/weakest topology on and the discrete topology is the finest/strongest topology on .

Metrizable

Topologies need not come from metrics, but if there is, we say that is metrizable.

e.g.

• Suppose that consists of more than one point. Then the trivial topology on is not metrizable.
• discrete yes
• cofinite topology on is metrizable iff is finite

Proof Assume the indiscrete topology on is induced by a metric on . Let with . Then . The set is an open subset of . Since , this set is not empty. And since this set is not all of . But and are the only open sets, yielding a contradiction.

Bases and Sub-bases

Basis of a Topology

A basis for a topology on is a collection such that every set in is the union of some sets from .

e.g. Let be a metric space. Then is a basis for the metric topology on .

Characterization of Basis

Let be a topological space. Then is a basis for iff for any and any with there is such that .

Proof Suppose is a basis, then any open neighbourhood is a union of sets from , so must contain some containing . Conversely, for any nonempty , it is an open neighbourhood of each of its points, so by assumption for each there is such that . Then , which is a union of sets from . Therefore is a basis.

Corollary

Let be a set. Let and be two topologies on with bases and respectively. The followings are equivalent:

• .
• For each and each there is such that .
  

Proof Suppose , since is a basis for , for each and , note that is also in , so there exists some such that . Conversely, for any , because is a basis, we shall write for . For each and each , there exists some such that . Therefore, we have which is a union of sets from , hence .

Proposition

A collection of sets cannot be basis for two distinct topologies.

Proof Suppose that is a basis for both and . Then every set in is a union of sets in . Since , every set in is open in , this implies that . Similarly, we have .

Theorem

Suppose is a set, then is a basis for some topology if and only if both of the following hold:

• is the union of some sets from ;
• If then is the union of some sets from .

In this case, the topology is the unique one generated by :

Proof If is a basis, then it clearly has the above properties, we will prove the converse here. Let . We first prove that it is a topology. Clearly, both and is in , and any union of sets in is in . Now consider the intersection of two sets and in . Then Since each is a union of sets from , we have that is a union of sets from , thus . Therefore, is a topology. Uniqueness is obvious.

Sub-basis

A sub-basis for a topology on is a collection such that every set in is a union of finite intersections of sets from .