A description/proof of that subset of non-open sub topological space is open on subspace if it is open on base topological space
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 subspace topology.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space and any sub topological space that is not open on the base space, any subset of the subspace is open on the subspace if it is open on the base space.
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, \(T\), and any sub topological space that is not open on \(T\), \(T_s\), any subset of \(T_s\), \(S \subseteq T_s\), is open on \(T_s\) if \(S\) is open on \(T\).
2: Proof
Suppose \(S\) is open on \(T\). \(S = S \cap T_s\), so, by the definition of subspace topology, \(S\) is open on \(T_s\).
3: Note
The reverse of this proposition does not hold, because supposing \(S\) is open on \(T_s\), \(S = U \cap T_s\) where \(U\) is an open set on \(T\) by the definition of subspace topology, but as \(T_s\) is not open on \(T\), \(U \cap T_s\) is not necessarily open on \(T\); for the case in which \(T_s\) is open on \(T\), see another proposition.
