description/proof of that for group with topology with continuous operations (especially, topological group), element multiplied by (open) neighborhood of \(1\) from left or right is (open) neighborhood of element
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, topological spaces map.
- The reader knows a definition of neighborhood of point on topological space.
- 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.
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, any topological group), any element multiplied by any (open) neighborhood of \(1\) from left or right is a (open) 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
\(g\): \(\in G\)
\(N_1\): \(\in \{\text{ the neighborhoods of } 1 \text{ on } G\}\)
\(U_1\): \(\in \{\text{ the open neighborhoods of } 1 \text{ on } G\}\)
//
Statements:
\(g N_1 \in \{\text{ the neighborhoods of } g \text{ on } G\}\)
\(\land\)
\(N_1 g \in \{\text{ the neighborhoods of } g \text{ on } G\}\)
\(\land\)
\(g U_1 \in \{\text{ the open neighborhoods of } g \text{ on } G\}\)
\(\land\)
\(U_1 g \in \{\text{ the open neighborhoods of } g \text{ on } G\}\)
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2: Proof
Whole Strategy: Step 1: think of the maps, \(f_l: G \to G, g' \mapsto g g'\) and \(f_r: G \to G, g' \mapsto g' g\), and see that they are some homeomorphisms; Step 2: take an open neighborhood of \(1\), \(U'_1 \subseteq N_1\), and see that \(g \in f_l (U'_1) \subseteq g N_1\) and \(g \in f_r (U'_1) \subseteq N_1 g\); Step 3: see that \(g \in g U_1 = f_l (U_1)\) and \(g \in U_1 g = f_r (U_1)\).
Step 1:
Let us think of the maps, \(f_l: G \to G, g' \mapsto g g'\) and \(f_r: G \to G, g' \mapsto g' g\).
They are some homeomorphisms, 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.
Step 2:
There is an open neighborhood of \(1\), \(U'_1 \subseteq G\), such that \(1 \in U'_1 \subseteq N_1\).
\(g = g 1 \in g U'_1 = f_l (U'_1) \subseteq g N_1\).
But \(f_l (U'_1) \subseteq G\) is an open subset, because \(f_l\) is a homeomorphism.
So, \(g N_1\) is a neighborhood of \(g\).
\(g = 1 g \in U'_1 g = f_r (U'_1) \subseteq N_1 g\).
But \(f_r (U'_1) \subseteq G\) is an open subset, because \(f_r\) is a homeomorphism.
So, \(N_1 g\) is a neighborhood of \(g\).
Step 3:
\(g = g 1 \in g U_1 = f_l (U_1)\).
But \(f_l (U_1) \subseteq G\) is an open subset, because \(f_l\) is a homeomorphism.
So, \(g U_1\) is an open neighborhood of \(g\).
\(g = 1 g \in U_1 g = f_r (U_1)\).
But \(f_r (U_1) \subseteq G\) is an open subset, because \(f_r\) is a homeomorphism.
So, \(U_1 g\) is an open neighborhood of \(g\).
