description/proof of that for group with topology with continuous operations (especially, topological group), set of symmetric neighborhoods of \(1\) is neighborhoods basis at \(1\)
Topics
About: group
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of group.
- The reader knows a definition of topological space.
- The reader knows a definition of continuous map.
- The reader knows a definition of symmetric subset of group.
- The reader knows a definition of neighborhood of point on topological space.
- The reader knows a definition of neighborhoods basis at point.
- The reader admits the proposition that for any group, the inverse of any subset is the image of the subset under the inverse map, and the double inverse of the subset is the subset.
- The reader admits the proposition that for any topological space and its any point, the intersection of any finite number of neighborhoods of the point is a neighborhood of the point.
- The reader admits the proposition that for any group and any subsets, the inverse of the intersection of the subsets is the intersection of the inverses of the subsets.
Target Context
- The reader will have a description and a proof of the proposition that for any group with any topology with any continuous operations (especially, topological group), the set of the symmetric neighborhoods of \(1\) is a neighborhoods basis at \(1\).
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:
\(G\): \(\in \{\text{ the groups }\}\) with any topology such that the group operations are continuous
\(S\): \(= \{\text{ the symmetric neighborhoods of } 1\}\)
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Statements:
\(S \in \{\text{ the neighborhood bases at } 1 \text{ for } G\}\)
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2: Proof
Whole Strategy: Step 1: let \(N_1\) be any neighborhood of \(1\), and take \(N_1 \cap N_1^{-1}\) and see that \(1 \in N_1 \cap N_1^{-1} \subseteq N_1\) and \(N_1 \cap N_1^{-1} \in S\).
Step 1:
Let \(N_1\) be any neighborhood of \(1\).
Let us take \(N_1 \cap N_1^{-1}\).
\(1 \in N_1 \cap N_1^{-1}\), because \(1 \in N_1\) and \(1 \in N_1^{-1}\).
\(N_1 \cap N_1^{-1} \subseteq N_1\).
So, \(1 \in N_1 \cap N_1^{-1} \subseteq N_1\).
Let \(f_2: G \to G\) be the inverse map, which is a homeomorphism.
\(N_1^{-1} = f_2 (N_1)\), by the proposition that for any group, the inverse of any subset is the image of the subset under the inverse map, and the double inverse of the subset is the subset.
So, \(N_1^{-1}\) is a neighborhood of \(1\), because as \(f_2\) is a homeomorphism, an open neighborhood of \(1\) contained in \(N_1\) is mapped to an open neighborhood of \(1\) contained in \(N_1^{-1} = f_2 (N_1)\).
\(N_1 \cap N_1^{-1}\) is a neighborhood of \(1\), by the proposition that for any topological space and its any point, the intersection of any finite number of neighborhoods of the point is a neighborhood of the point.
\((N_1 \cap N_1^{-1})^{-1} = N_1^{-1} \cap {N_1^{-1}}^{-1}\), by the proposition that for any group and any subsets, the inverse of the intersection of the subsets is the intersection of the inverses of the subsets, \(= N_1^{-1} \cap N_1\), by the proposition that for any group, the inverse of any subset is the image of the subset under the inverse map, and the double inverse of the subset is the subset, \(= N_1 \cap N_1^{-1}\), so, \(N_1 \cap N_1^{-1}\) is symmetric.
So, \(N_1 \cap N_1^{-1} \in S\).
