A description/proof of that subset of open sub topological space is open on subspace iff 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 open on the base space, any subset of the subspace is open on the subspace if and only 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 open on \(T\), \(T_s\), any subset of \(T_s\), \(S \subseteq T_s\), is open on \(T_s\) if and only 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 \(T_s\).
Suppose \(S\) is open on \(T_s\). \(S = U \cap T_s\) where \(U\) is an open set on \(T\) by the definition of subspace topology. As \(T_s\) is open on \(T\), \(S\) is open on \(T\) as the intersection of 2 open sets on \(T\).
3: Note
\(T_s\) must be open on \(T\) for this proposition, as the latter half of the proof requires it; for the case in which \(T_s\) is not open on \(T\), see another proposition.
