Continuous Maps on Topological Spaces

Continuous Function

Let and be topological spaces and a function. is called continuous if for every open set in , is open in .
is called continuous at if, for any open neighborhood of in , the set is a neighborhood of in .

e.g.

• The identity map is continuous;
• Any constant map is continuous;
• The composition of two continuous maps is continuous;
  

Theorem

Let and be topological spaces. Consider a function . The following are equivalent:

1. is continuous on ;
2. is continuous at every point in ;
3. is closed in for every closed set in ;
4. whenever for a basis of ;
5. whenever for a subbasis of ;
6. for all ;
   

Proof 2 is obvious; 3 comes from definition of closed sets; 4 and 5 are because preimages preserve unions and intersections. We prove 6 here. Suppose is continuous. Since is closed, is closed in . Note that , so , hence , thus . For the other direction, suppose for all . Let be a closed subset of . Then , thus , so is closed in . Therefore, is continuous.

Theorem

Let be a function between two topological spaces and . Assume that ,where is open in for each index . Then is continuous if and only if is continuous for every .

Open and Closed Map

A map is open if for every open set , is open in . Similarly, it is closed if for every closed set , is closed in .

Projection Map

Define the following function as projection maps:

Thrm Let , and be topological spaces. A function is continuous if and only if the components and are continuous.
Proof Recall that is continuous iff is open in for any . Since we complete the proof.

Corollary The product topology on is the coarsest topology for which both and are continuous.

Compactness and Continuous Functions

Continuous Image of a Compact Set is Compact

Let and be topological spaces. If is compact and is continuous, then is compact.

Proposition

The inverse of a bijective continuous function , with compact, is continuous, so it is a homeomorphism.

Proof Suppose is a continuous bijection. Consider the inverse function . Let be open, then is open in because is continuous. Since is compact, is compact. Since is a bijection, is also a bijection. Therefore, is closed. Hence, is continuous.

Corollary Let be a compact topological space and continuous, where is endowed with the standard topology. Then achieves its maximum and minimum value on .

Lemma Let and be topological spaces and let be endowed with the product topology. Assume that is compact. For , if is open with , then there is an open neighborhood of such that .

Thrm If and are compact topological spaces, then is also compact in the product topology.

Heine–Borel Theorem

Any closed interval is a compact subset of . More generally, a subset of is compact if and only if, it is bounded and is a close subset of .

Proof