description/proof of that for group with topology with continuous operations (especially, topological group), element, and neighborhood of element, there is symmetric neighborhood of \(1\) s.t. element multiplied from left by neighborhood of \(1\) and multiplied from right by inverse of neighborhood of \(1\) is contained in neighborhood of element
Topics
About: group
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of neighborhood of point on topological space.
- The reader knows a definition of symmetric subset of group.
- The reader knows a definition of product topology.
- The reader admits the proposition that for any group with any topology with any continuous operations (especially, topological group) and each element, the inversion map, the multiplication-by-element-from-left-or-right map, and the conjugation-by-element map are homeomorphisms.
- The reader admits the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.
- The reader admits the proposition that for any topological group, the set of the symmetric neighborhoods of \(1\) is a neighborhood basis at \(1\).
- The reader admits the proposition that the product map of any finite number of continuous maps is continuous by the product topologies.
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), any element, and any neighborhood of the element, there is a symmetric neighborhood of \(1\) such that the element multiplied from left by the neighborhood of \(1\) and multiplied from right by the inverse of the neighborhood of \(1\) is contained in the neighborhood of the element.
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
\(g\): \(\in G\)
\(N_g\): \(\in \{\text{ the neighborhoods of } g \text{ on } G\}\)
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Statements:
\(\exists N_1 \in \{\text{ the symmetric neighborhoods of } 1 \text{ on } G\} (N_1 g {N_1}^{-1} \subseteq N_g)\)
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2: Proof
Whole Strategy: Step 1: think of the continuous \(f: G \times G \to G, (g_1, g_2) \mapsto g_1 g g_2\) and see that \(f ((1, 1)) = g\); Step 2: take an open neighborhood of \((1, 1)\), \(U_{(1, 1)}\), such that \(f (U_{(1, 1)}) \subseteq N_g\) and take an open neighborhood of \(1\), \(U_1\), such that \(U_1 \times U_1 \subseteq U_{(1, 1)}\); Step 3: take a symmetric neighborhood of \(1\), \(N_1 \subseteq U_1\), and see that \(f (N_1 \times N_1) \subseteq N_g\) and \(f (N_1 \times N_1) = N_1 g {N_1}^{-1}\).
Step 1:
Let us think of \(f: G \times G \to G, (g_1, g_2) \mapsto g_1 g g_2\).
\(f\) is the composition of the maps, \(: G \times G \to G \times G, (g_1, g_2) \mapsto (g_1, g g_2)\) and \(: G \times G \to G, (g_1, g_2) \mapsto g_1 g_2\).
The former is continuous, because \(: g_2 \mapsto g g_2\) is continuous, by the proposition that for any group with any topology with any continuous operations (especially, topological group) and each element, the inversion map, the multiplication-by-element-from-left-or-right map, and the conjugation-by-element map are homeomorphisms, and so \(: G \times G \to G \times G, (g_1, g_2) \mapsto (g_1, g g_2)\) is continuous, by the proposition that the product map of any finite number of continuous maps is continuous by the product topologies.
The latter is continuous, because it is the multiplication operation.
So, \(f\) is continuous as the composition of the continuous maps, by the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.
\(f ((1, 1)) = 1 g 1 = g\).
Step 2:
As \(f\) is continuous, there is an open neighborhood of \((1, 1)\), \(U_{(1, 1)} \subseteq G \times G\), such that \(f (U_{(1, 1)}) \subseteq N_g\).
By the definition of product topology, there is an open neighborhood of \(1\), \(U_1 \subseteq G\), such that \(U_1 \times U_1 \subseteq U_{(1, 1)}\): while there are some open neighborhoods of \(1\), \({U_1}', {U_1}'' \subseteq G\), such that \({U_1}' \times {U_1}'' \subseteq U_{(1, 1)}\), we can take \(U_1 := {U_1}' \cap {U_1}''\).
Step 3:
There is a symmetric neighborhood of \(1\), \(N_1 \subseteq G\), such that \(N_1 \subseteq U_1\), by the proposition that for any topological group, the set of the symmetric neighborhoods of \(1\) is a neighborhood basis at \(1\).
As \(N_1 \times N_1 \subseteq U_{(1, 1)}\), \(f (N_1 \times N_1) \subseteq f (U_{(1, 1)}) \subseteq N_g\).
Let us see that \(f (N_1 \times N_1) = N_1 g N_1\).
For each \(g' \in f (N_1 \times N_1)\), \(g' = g_1 g g_2\) where \(g_1, g_2 \in N_1\), so, \(g' \in N_1 g N_1\); for each \(g' \in N_1 g N_1\), \(g' = g_1 g g_2\) where \(g_1, g_2 \in N_1\), but \(g_1 g g_2 = f (g_1, g_2)\), so, \(g' \in f (N_1 \times N_1)\).
But as \(N_1\) is symmetric, \(N_1 g N_1 = N_1 g {N_1}^{-1}\).
So, \(N_1 g {N_1}^{-1} = f (N_1 \times N_1) \subseteq N_g\).
