description/proof of that for topological group, neighborhoods basis at \(1\) satisfies these properties and point multiplied by neighborhoods basis at \(1\) is neighborhoods basis at point
Topics
About: topological group
The table of contents of this article
Starting Context
- The reader knows a definition of topological group.
- The reader knows a definition of neighborhoods basis at point on topological space.
- The reader knows a definition of inverse of subset of group.
- 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 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.
- 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 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 topological group, any neighborhoods basis at \(1\) satisfies these properties and each point multiplied by the neighborhoods basis at \(1\) is a neighborhoods basis at the point.
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:
\(G\): \(\in \{\text{ the topological groups }\}\)
\(B_1\): \(\in \{\text{ the neighborhoods bases at } 1 \text{ on } G\}\)
//
Statements:
1) \(\forall g \in G \text{ such that } g \neq 1 (\exists N_1 \in B_1 (g \notin N_1))\)
\(\land\)
2) \(\forall N_1, N'_1 \in B_1 (\exists N''_1 \in B_1 (N''_1 \subseteq N_1 \cap N'_1))\)
\(\land\)
3) \(\forall N_1 \in B_1 (\exists N'_1 \in B_1 ({N'_1}^{-1} N'_1 \subseteq N_1))\)
\(\land\)
4) \(\forall N_1 \in B_1, \forall g \in G (\exists N'_1 \in B_1 (N'_1 \subseteq g N_1 g^{-1}))\)
\(\land\)
5) \(\forall N_1 \in B_1, \forall n \in N_1 \text{ such that } \exists U_n \in \{\text{ the open neighborhoods of } n\} (n \in U_n \subseteq N_1) (\exists N'_1 \in B_1 (N'_1 n \subseteq N_1))\)
\(\land\)
\(\forall g \in G (g B_1 \in \{\text{ the neighborhoods bases at } g \text{ on } G\})\)
\(\land\)
\(\forall g \in G (B_1 g \in \{\text{ the neighborhoods bases at } g \text{ on } G\})\)
//
\(g B_1\) means \(\{g S_1 \vert S_1 \in B_1\}\); \(B_1 g\) means \(\{S_1 g \vert S_1 \in B_1\}\).
2: Proof
Whole Strategy: use the Hausdorff-ness and the continuousness of the operations; Step 1: see that 1) holds; Step 2: see that 2) holds; Step 3: see that 3) holds; Step 4: see that 4) holds; Step 5: see that 5) holds; Step 6: see that \(g B_1\) is a neighborhoods basis at \(g\); Step 7: see that \(B_1 g\) is a neighborhoods basis at \(g\).
Step 1:
Let us see that 1) holds.
As \(G\) is Hausdorff, there is an open neighborhood of \(1\), \(U_1 \subseteq G\), and an open neighborhood of \(g\), \(U_g \subseteq G\), such that \(U_1 \cap U_g = \emptyset\).
There is a \(N_1 \in B_1\) such that \(N_1 \subseteq U_1\), by the definition of neighborhoods basis at point on topological space.
\(N_1 \cap U_g = \emptyset\), so, \(g \notin N_1\).
Step 2:
Let us see that 2) holds.
\(N_1 \cap N'_1\) 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.
So, there is a \(N''_1 \in B_1\) such that \(N''_1 \subseteq N_1 \cap N'_1\), by the definition of neighborhoods basis at point on topological space.
Step 3:
Let us see that 3) holds.
There is a (symmetric) neighborhood of \(1\), \(N''_1 \subseteq G\), such that \({N''_1}^{-1} 1 N''_1 \subseteq N_1\), by 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: \(g\) in the proposition is taken to be \(1\) and \({N''_1}^{-1} 1 N''_1 = N''_1 1 {N''_1}^{-1}\) because \(N''_1\) is symmetric.
But \({N''_1}^{-1} 1 N''_1 = {N''_1}^{-1} N''_1\).
There is an \(N'_1 \in B_1\) such that \(N'_1 \subseteq N''_1\), by the definition of neighborhoods basis at point on topological space.
\({N'_1}^{-1} \subseteq {N''_1}^{-1}\), 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, \({N'_1}^{-1} N'_1 \subseteq {N''_1}^{-1} N''_1\).
So, \({N'_1}^{-1} N'_1 \subseteq N_1\).
Step 4:
Let us see that 4) holds.
The conjugation map by \(g\), \(f: G \to G, g' \mapsto g g' g^{-1}\), is a homeomorphism, 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.
\(1 \in g N_1 g^{-1}\), because \(1 \in N_1\) and \(g 1 g^{-1} = 1\).
So, \(g N_1 g^{-1}\) is a neighborhood of \(1\): while \(N_1\) contains an open neighborhood of \(1\), \(U_1\), \(g N_1 g^{-1} = f (N_1)\) contains \(f (U_1)\), which is an open neighborhood of \(1\): \(f (1) = 1\).
So, there is an \(N'_1 \in B_1\) such that \(N'_1 \subseteq g N_1 g^{-1}\), by the definition of neighborhoods basis at point on topological space.
Step 5:
Let us see that 5) holds.
The multiplication-by-element-from-right map by \(n^{-1}\), \(f: G \to G, g' \mapsto g' n^{-1}\), is a homeomorphism, 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.
\(1 \in U_n n^{-1} \subseteq N_1 n^{-1}\), because \(n \in U_n\) and \(n n^{-1} = 1\).
So, \(N_1 n^{-1}\) is a neighborhood of \(1\): \(N_1 n^{-1} = f (N_1)\) contains \(U_n n^{-1} = f (U_n)\), which is an open neighborhood of \(1\).
So, there is an \(N'_1 \in B_1\) such that \(N'_1 \subseteq N_1 n^{-1}\), by the definition of neighborhoods basis at point on topological space.
So, \(N'_1 n \subseteq N_1 n^{-1} n = N_1\).
Step 6:
Let \(g \in G\) be any.
Let \(N_g \subseteq G\) be any neighborhood of \(g\).
\(g^{-1} N_g\) is a neighborhood of \(1\), because while the multiplication-by-\(g^{-1}\)-from-left-map is a homeomorphism 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, an open neighborhood of \(g\) contained in \(N_g\) is mapped into \(g^{-1} N_g\) as an open neighborhood of \(1\).
So, there is an \(S_1 \in B_1\) such that \(1 \in S_1 \subseteq g^{-1} N_g\).
So, \(g \in g S_1 \subseteq g g^{-1} N_g = N_g\), while \(g S_1\) is a neighborhood of \(g\), because while the multiplication-by-\(g\)-from-left-map is a homeomorphism 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, an open neighborhood of \(1\) contained in \(S_1\) is mapped into \(g S_1\) as an open neighborhood of \(g\).
Step 7:
Let \(g \in G\) be any.
Let \(N_g \subseteq G\) be any neighborhood of \(g\).
\(N_g g^{-1}\) is a neighborhood of \(1\), because while the multiplication-by-\(g^{-1}\)-from-right-map is a homeomorphism 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, an open neighborhood of \(g\) contained in \(N_g\) is mapped into \(N_g g^{-1}\) as an open neighborhood of \(1\).
So, there is an \(S_1 \in B_1\) such that \(1 \in S_1 \subseteq N_g g^{-1}\).
So, \(g \in S_1 g \subseteq N_g g^{-1} g = N_g\), while \(S_1 g\) is a neighborhood of \(g\), because while the multiplication-by-\(g\)-from-right-map is a homeomorphism 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, an open neighborhood of \(1\) contained in \(S_1\) is mapped into \(S_1 g\) as an open neighborhood of \(g\).
