description/proof of that for topological group, subgroup as topological subspace is topological group
Topics
About: group
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topological group.
- The reader admits the proposition that any subspace of any Hausdorff topological space is Hausdorff.
- The reader admits the proposition that for any possibly uncountable number of indexed topological spaces or any finite number of topological spaces and their subspaces, the product of the subspaces is the subspace of the product of the base spaces.
Target Context
- The reader will have a description and a proof of the proposition that for any topological group, any subgroup as the topological subspace is a topological group.
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:
\(G'\): \(\in \{\text{ the topological groups }\}\)
\(G\): \(\in \{\text{ the subgroups of } G'\}\), as the topological subspace
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Statements:
\(G\in \{\text{ the topological groups }\}\)
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2: Proof
Whole Strategy: Step 1: see that \(G\) is Hausdorff; Step 2: see that the multiplication and the inversion on \(G\) are continuous; Step 3: conclude the proposition.
Step 1:
\(G\) is Hausdorff, by the proposition that any subspace of any Hausdorff topological space is Hausdorff.
Step 2:
Let \(m: G \times G \to G\) and \(i: G \to G\) be the multiplication and the inversion on \(G\).
Let \(m': G' \times G' \to G'\) and \(i': G' \to G'\) be the multiplication and the inversion on \(G\), which are continuous.
\(m\) and \(i\) are some restrictions of \(m'\) and \(i'\).
\(G \times G\) is a topological subspace of \(G' \times G'\), by the proposition that for any possibly uncountable number of indexed topological spaces or any finite number of topological spaces and their subspaces, the product of the subspaces is the subspace of the product of the base spaces.
\(m\) is continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
\(i\) is continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
Step 3:
So, \(G\) is a topological group.
