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 neighborhood of in , the set is a neighborhood of in .

Theorem

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

  • is continuous at every point in .
  • is continuous on .
  • is closed in for every closed set in .
  • whenever for a basis of .
  • whenever for a subbasis of .

Theorem

Thrm 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.

Definition

Def 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

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

Thrm If is

Proposition

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

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