2025-07-06

1198: For Topological Group, Neighborhoods Basis at 1 Satisfies These Properties and Point Multiplied by Neighborhoods Basis at 1 Is Neighborhoods Basis at Point

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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



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: { the topological groups }
B1: { the neighborhoods bases at 1 on G}
//

Statements:
1) gG such that g1(N1B1(gN1))

2) N1,N1B1(N1B1(N1N1N1))

3) N1B1(N1B1(N11N1N1))

4) N1B1,gG(N1B1(N1gN1g1))

5) N1B1,nN1 such that Un{ the open neighborhoods of n}(nUnN1)(N1B1(N1nN1))

gG(gB1{ the neighborhoods bases at g on G})

gG(B1g{ the neighborhoods bases at g on G})
//

gB1 means {gS1|S1B1}; B1g means {S1g|S1B1}.


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 gB1 is a neighborhoods basis at g; Step 7: see that B1g 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, U1G, and an open neighborhood of g, UgG, such that U1Ug=.

There is a N1B1 such that N1U1, by the definition of neighborhoods basis at point on topological space.

N1Ug=, so, gN1.

Step 2:

Let us see that 2) holds.

N1N1 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 N1B1 such that N1N1N1, 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, N1G, such that N111N1N1, 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 N111N1=N11N11 because N1 is symmetric.

But N111N1=N11N1.

There is an N1B1 such that N1N1, by the definition of neighborhoods basis at point on topological space.

N11N11, 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, N11N1N11N1.

So, N11N1N1.

Step 4:

Let us see that 4) holds.

The conjugation map by g, f:GG,gggg1, 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.

1gN1g1, because 1N1 and g1g1=1.

So, gN1g1 is a neighborhood of 1: while N1 contains an open neighborhood of 1, U1, gN1g1=f(N1) contains f(U1), which is an open neighborhood of 1: f(1)=1.

So, there is an N1B1 such that N1gN1g1, 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 n1, f:GG,ggn1, 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.

1Unn1N1n1, because nUn and nn1=1.

So, N1n1 is a neighborhood of 1: N1n1=f(N1) contains Unn1=f(Un), which is an open neighborhood of 1.

So, there is an N1B1 such that N1N1n1, by the definition of neighborhoods basis at point on topological space.

So, N1nN1n1n=N1.

Step 6:

Let gG be any.

Let NgG be any neighborhood of g.

g1Ng is a neighborhood of 1, because while the multiplication-by-g1-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 Ng is mapped into g1Ng as an open neighborhood of 1.

So, there is an S1B1 such that 1S1g1Ng.

So, ggS1gg1Ng=Ng, while gS1 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 S1 is mapped into gS1 as an open neighborhood of g.

Step 7:

Let gG be any.

Let NgG be any neighborhood of g.

Ngg1 is a neighborhood of 1, because while the multiplication-by-g1-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 Ng is mapped into Ngg1 as an open neighborhood of 1.

So, there is an S1B1 such that 1S1Ngg1.

So, gS1gNgg1g=Ng, while S1g 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 S1 is mapped into S1g as an open neighborhood of g.


References


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