A description/proof of that for topological space, intersection of basis and subspace is basis for subspace
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topological space.
- The reader knows a definition of basis of topological space.
- The reader knows a definition of subspace topology.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space, the intersection of any basis and any subspace is a basis for the subspace.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For any topological space, \(T\), any basis, \(B = \{B_\alpha\}\), and any subset, \(S \subseteq T\), with the subspace topology, the intersection of the basis and the subset, \(\{B_\alpha \cap S\}\), is a basis for the subspace.
2: Proof
For any open set, \(U \subseteq S\), on \(S\), \(U = U' \cap S\) where \(U' \subseteq T\) is open on \(T\). For any point, \(p \in U\), as \(p \in U'\), there is a \(B_\alpha\) such that \(p \in B_\alpha \subseteq U'\) by the definition of basis. As \(p \in S\), \(p \in B_\alpha \cap S \subseteq U' \cap S = U\). So, for any open set, \(U \subseteq S\), and any point, \(p \in U\), there is a \(B_\alpha \cap S\) such that \(p \in B_\alpha \cap S \subseteq U\), which means that \(\{B_\alpha \cap S\}\) is a basis for \(S\) by the definition of basis.
