A description/proof of that in nest of topological subspaces, openness of subset on subspace does not depend on superspace
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 in any nest of topological subspaces, the openness of any subset on any subspace does not depend on the superspace of which the subspace is regarded to be a 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, \(T\), and any nest of topological subspaces, \(T_1, T_2\), such that \(T_2 \subseteq T_1 \subseteq T\), if any subset, \(S \subseteq T_2\), is open or not open with \(T_2\) regarded as a subspace of \(T_1\), \(S\) is open or not open respectively with \(T_2\) regarded as a subspace of \(T\); if \(S\) is open or not open with \(T_2\) regarded as a subspace of \(T\), \(S\) is open or not open respectively with \(T_2\) regarded as a subspace of \(T_1\).
2: Proof
Suppose that \(S\) is open with \(T_2\) regarded as a subspace of \(T_1\). There is an open set, \(U_1 \subseteq T_1\), such that \(S = U_1 \cap T_2\). As \(T_1\) is a subspace of \(T\), there is an open set, \(U \subseteq T\), such that \(U_1 = U \cap T_1\). So, \(S = U \cap T_1 \cap T_2 = U \cap T_2\), so, \(S\) is open with \(T_2\) regarded as a subspace of \(T\).
Suppose that \(S\) is not open with \(T_2\) regarded as a subspace of \(T_1\). There is no open set, \(U_1 \subseteq T_1\), such that \(S = U_1 \cap T_2\). As \(T_1\) is a subspace of \(T\), there is no open set, \(U \subseteq T\), such that \(U_1 = U \cap T_1\) and \(S = U_1 \cap T_2 = U \cap T_1 \cap T_2 = U \cap T_2\). So, \(S\) is not open with \(T_2\) regarded as a subspace of \(T\).
Suppose that \(S\) is open with \(T_2\) regarded as a subspace of \(T\). There is an open set, \(U \subseteq T\), such that \(S = U \cap T_2\). But \(S = U \cap T_1 \cap T_2\). As \(T_1\) is a subspace of \(T\), \(U \cap T_1\) is open on \(T_1\), so, \(S\) is open with \(T_2\) regarded as a subspace of \(T_1\).
Suppose that \(S\) is not open with \(T_2\) regarded as a subspace of \(T\). There is no open set, \(U \subseteq T\), such that \(S = U \cap T_2\). But \(U \cap T_2 = U \cap T_1 \cap T_2\). As \(T_1\) is a subspace of \(T\), any open set on \(T_1\) has to be \(U \cap T_1\), but as there is no such \(U\), \(S\) is not open with \(T_2\) regarded as a subspace of \(T_1\).
