2025-07-13

1200: For Group, This Set of Subsets Constitutes Topological Basis Constituting Topological Group with This as Neighborhoods Basis at Point

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description/proof of that for group, this set of subsets constitutes topological basis constituting topological group with this as 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 group, this set of subsets constitutes a topological basis constituting a topological group with this as a neighborhoods basis at each 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 groups }
B1: Pow(G), with the properties specified below
B: =gGgB1
B: =gGB1g
//

Statements:
(
0) B1S1B1(1S1)

1) gG such that g1(S1B1(gS1))

2) S1,S1B1(S1B1(S1S1S1))

3) S1B1(S1B1(S11S1S1))

4) S1B1,gG(S1B1(S1gS1g1))

5) S1B1,sS1(S1B1(S1sS1))
)

(
(
B{ the topological bases for G}

G{ the topological groups with B}

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

(
B{ the topological bases for G}

G{ the topological groups with B}

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


2: Proof


Whole Strategy: Step 1: see that B satisfies one of some criteria for any collection of open sets to be a basis; Step 2: see that gB1 is a neighborhoods basis at g; Step 3: see that the group operations of G are continuous; Step 4: see that G is Hausdorff; Step 5: see that B satisfies one of some criteria for any collection of open sets to be a basis; Step 6: see that B1g is a neighborhoods basis at g; Step 7: see that the group operations of G are continuous; Step 8: see that G is Hausdorff.

Step 1:

Let us see that B satisfies Description 2 of some criteria for any collection of open sets to be a basis.

1) G=B (refer to the definition of union of set)?

1B, because B1B, so, B1B, there is an S1B1 such that 1S1, S1B1, and 1S1B1B.

For each gG, gB, because gB1B, so, gB1B, as there is an S1B1 such that 1S1, ggS1gB1, gS1gB1, and ggS1gB1B.

So, 1) holds.

2) for each sets, Sj,SlB, and each point, gSjSl, there is a set, SmB, such that gSmSjSl?

Let SjgjB1, SlglB1, and SmgmB1.

Sj=gjS1,j and Sl=gjS1,l where S1,j,S1,lB1.

As ggjS1,j, gj1gS1,j; gl1gS1,l, likewise.

By 5), there is an S1,jB1 such that S1,jgj1gS1,j; there is an S1,lB1 such that S1,lgl1gS1,l, likewise.

S1,jS1,jg1gj and S1,lS1,lg1gl.

By 4), there is an S1,jB1 such that S1,jg1gjS1,jgj1g; there is an S1,lB1 such that S1,lg1glS1,lgl1g, likewise.

So, S1,jg1gjS1,jg1gjgj1g=g1gjS1,j, so, gS1,jgg1gjS1,j=gjS1,j; gS1,lglS1,l, likewise.

By 2), there is an S1B1 such that S1S1,jS1,l.

So, gS1g(S1,jS1,l)=(gS1,j)(gS1,l)gjS1,jglS1,l=SjSl.

ggS1, because 1S1.

So, gS1=Sm will do.

So, G with B is a topological space, by the proposition that any basis of any topological space determines the topology.

Step 2:

Let us see that gB1 is a neighborhoods basis at g.

Let Ng be any neighborhood of g.

There is a gS1B such that ggS1Ng: the point is that g is not guaranteed to be taken to be g, yet.

g1gS1. So, there is an S1B1 such that S1g1gS1, by 5).

S1g1g=g1gg1gS1g1g.

There is an S1B1 such that S1g1gS1g1g, by 4). So, g1gS1g1gg1gS1g1g=S1g1gS1.

So, gg1gS1gS1, but gg1gS1=gS1, so, ggS1gS1Ng.

Step 3:

Let us see that the group operations of G are continuous.

Let us deal with the inverse map.

Let Ng1G be any neighborhood of g1.

There is a g1S1B such that g1g1S1Ng1.

There is an S1B1 such that S1g1S1g, by 4).

So, S1g1g1S1.

There is an S1B1 such that S11S1S1, by 3).

S11S1g1S1g1.

But S11S1g1=(gS11S1)1 and S1S11S1, so, (gS1)1S1g1g1S1Ng1.

The inverse map is a bijection and its inverse is itself, so, the inverse map is a homeomorphism.

Let us deal with the multiplication map.

Let Ng1g2G be any neighborhood of g1g2.

There is a g1g2S1B such that g1g2g1g2S1Ng1g2.

g1g2S1=g1g2S1g21g2.

There is an S1B1 such that S1g2S1g21, by 4), so, g1S1g2g1g2S1g21g2=g1g2S1.

There is an S1B1 such that S11S1S1, by 3), so, g1S11S1g2g1S1g2g1g2S1.

But as the inverse map is a homeomorphism, S11 is an open neighborhood of 1, so, there is an S1B1 such that S1S11, so, g1S1S1g2g1S11S1g2.

g1S1S1g2=g1S1g2g21S1g2.

There is an S1B1 such that S1g21S1g2, by 4), so, g1S1g2S1g1S1S1g2.

So, g1S1g2S1g1S1S1g2g1S11S1g2g1g2S1Ng1g2.

Step 4:

Let us see that G is Hausdorff.

Let g1,g2G be any such that g1g2.

g11g21, so, there is an S1B1 such that g11g2S1, by 1).

There is a symmetric neighborhood of 1, N1G, such that N12S1, by the proposition that for any group with any topology with any continuous operations (especially, topological group), any neighborhood of 1, and any positive natural number, there is a symmetric neighborhood of 1 whose power to the natural number is contained in the neighborhood.

Let us see that N1g11g2N1=.

Let us suppose that N1g11g2N1.

There would be an nN1g11g2N1.

n=g11g2n for an nN1.

g11g2=nn1. But as N1 is symmetric, n1N11=N1, so, g11g2N12S1, a contradiction against g11g2S1.

So, N1g11g2N1=.

There is an S1B1 such that S1N1, and S1g11g2S1N1g11g2N1=.

So, g1(S1g11g2S1)=g1S1g2S1=.

Step 5:

Let us see that B satisfies Description 2 of some criteria for any collection of open sets to be a basis.

1) G=B (refer to the definition of union of set)?

1B, because B1B, so, B1B, there is an S1B1 such that 1S1, S1B1, and 1S1B1B.

For each gG, gB, because B1gB, so, B1gB, as there is an S1B1 such that 1S1, gS1gB1g, S1gB1g, and gS1gB1gB.

So, 1) holds.

2) for each sets, Sj,SlB, and each point, gSjSl, there is a set, SmB, such that gSmSjSl?

Let SjB1gj, SlB1gl, and SmB1gm.

Sj=S1,jgj and Sl=S1,lgj where S1,j,S1,lB1.

As gS1,jgj, ggj1S1,j; ggl1S1,l, likewise.

By 5), there is an S1,jB1 such that S1,jggj1S1,j; there is an S1,lB1 such that S1,lggl1S1,l, likewise.

S1,jS1,jgjg1 and S1,lS1,lglg1.

So, S1,jgS1,jgjg1g=S1,jgj; S1,lgS1,lgl, likewise.

By 2), there is an S1B1 such that S1S1,jS1,l.

So, S1g(S1,jS1,l)g=(S1,jg)(S1,lg)(S1,jgj)(S1,lgl)=SjSl.

gS1g, because 1S1.

So, S1g=Sm will do.

So, G with B is a topological space, by the proposition that any basis of any topological space determines the topology.

Step 6:

Let us see that B1g is a neighborhoods basis at g.

Let Ng be any neighborhood of g.

There is a S1gB such that gS1gNg: the point is that g is not guaranteed to be taken to be g, yet.

gg1S1. So, there is an S1B1 such that S1gg1S1, by 5).

So, S1gg1gS1g, but S1gg1g=S1g, so, gS1gS1gNg.

Step 7:

Let us see that the group operations of G are continuous.

Let us deal with the inverse map.

Let Ng1G be any neighborhood of g1.

There is a S1g1B such that g1S1g1Ng1.

There is an S1B1 such that S1gS1g1, by 4).

So, g1S1S1g1.

There is an S1B1 such that S11S1S1, by 3).

g1S11S1g1S1.

But g1S11S1=(S11S1g)1 and S1S11S1, so, (S1g)1g1S1S1g1Ng1.

The inverse map is a bijection and its inverse is itself, so, the inverse map is a homeomorphism.

Let us deal with the multiplication map.

Let Ng1g2G be any neighborhood of g1g2.

There is a S1g1g2B such that g1g2S1g1g2Ng1g2.

S1g1g2=g1g11S1g1g2.

There is an S1B1 such that S1g11S1g1, by 4), so, g1S1g2g1g11S1g1g2=S1g1g2.

There is an S1B1 such that S11S1S1, by 3), so, g1S11S1g2g1S1g2S1g1g2.

But as the inverse map is a homeomorphism, S11 is an open neighborhood of 1, so, there is an S1B1 such that S1S11, so, g1S1S1g2g1S11S1g2.

g1S1S1g2=g1S1g11g1S1g2.

There is an S1B1 such that S1g1S1g11, by 4), so, S1g1S1g2g1S1S1g2.

So, S1g1S1g2g1S1S1g2g1S11S1g2S1g1g2Ng1g2.

Step 8:

Let us see that G is Hausdorff.

Let g1,g2G be any such that g1g2.

g1g211, so, there is an S1B1 such that g1g21S1, by 1).

There is a symmetric neighborhood of 1, N1G, such that N12S1, by the proposition that for any group with any topology with any continuous operations (especially, topological group), any neighborhood of 1, and any positive natural number, there is a symmetric neighborhood of 1 whose power to the natural number is contained in the neighborhood.

Let us see that N1N1g1g21=.

Let us suppose that N1N1g1g21.

There would be an nN1N1g1g21.

n=ng1g21 for an nN1.

g1g21=n1n. But as N1 is symmetric, n1N11=N1, so, g1g21N12S1, a contradiction against g1g21S1.

So, N1N1g1g21=.

There is an S1B1 such that S1N1, and S1S1g1g21N1N1g1g21=.

So, (S1S1g1g21)g2=S1g2S1g1g21g2=S1g2S1g1=.


References


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