If is a topological space and , then the subspace topology on is We call a topological subspace of .
e.g. If we consider as a subspace of then the open sets in consist of all sets where is an open subset of . In particular, is open in the subspace for every .
Lemma
Suppose that is a metric space with corresponding topology . If then subspace topology on corresponds to the topology on that arises from the metric space .
Proof
Proposition
Closed subset in closed subspace is closed in the original topological space.
Proof Let be a closed subspace of , and is closed in . Then for some open set in . It follows that . Thus, Since is closed in , is open. Therefore is open, and is closed in .
Universal Property of the Subspace Topology
Let be a topological space and a subspace (with the subspace topology). Then
The inclusion map is continuous;
Finite Product Spaces
Product Topology
Suppose that and are two topological spaces. Then the product topology on is the topology with basis
We call the product topology on .
Remark
Note that may not be a topology on . For example.
Moreover, the product topology is independent of choice of the bases. This is because if and are bases for and , respectively, then the topology generated by is the same as the topology generated by . (This is a direct result of proposition)
the topology generated by A0 ⇥ B0 is the same as the topology generated by
A ⇥ B
Thrm Let and be topological spaces with respective bases and . Then