description/proof of that subset of not-necessarily-open topological subspace is open on subspace if it is open on basespace
Topics
About: topological space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
- 4: Note
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 topological subspace that is not necessarily open on the basespace, any subset of the subspace is open on the subspace if it is open on the basespace.
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: Structured Description
Here is the rules of Structured Description.
Entities:
\(T'\): \(\in \{\text{ the topological spaces }\}\)
\(T\): \(\subseteq T'\) with the subspace topology
\(S\): \(\subseteq T\)
//
Statements:
\(S \in \{\text{ the open subsets of } T'\}\)
\(\implies\)
\(S \in \{\text{ the open subsets of } T\}\)
//
2: Natural Language Description
For any topological space, \(T'\), and any topological subspace that is not necessarily open on \(T'\), \(T\), any subset, \(S \subseteq T\), is open on \(T\) if \(S\) is open on \(T'\).
3: Proof
Let us suppose that \(S\) is open on \(T'\).
\(S = S \cap T\), so, by the definition of subspace topology, \(S\) is open on \(T\).
4: Note
The reverse of this proposition does not hold, because supposing \(S\) is open on \(T\), \(S = U \cap T\) where \(U \subseteq T'\) is an open subset on \(T'\) by the definition of subspace topology, but as \(T\) is not necessarily open on \(T'\), \(U \cap T\) is not necessarily open on \(T'\).
While we have shown another proposition, that is really covered by this proposition.
