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
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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:
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: , with the properties specified below
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Statements:
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2: Proof
Whole Strategy: Step 1: see that satisfies one of some criteria for any collection of open sets to be a basis; Step 2: see that is a neighborhoods basis at ; Step 3: see that the group operations of are continuous; Step 4: see that is Hausdorff; Step 5: see that satisfies one of some criteria for any collection of open sets to be a basis; Step 6: see that is a neighborhoods basis at ; Step 7: see that the group operations of are continuous; Step 8: see that is Hausdorff.
Step 1:
Let us see that satisfies Description 2 of some criteria for any collection of open sets to be a basis.
1) (refer to the definition of union of set)?
, because , so, , there is an such that , , and .
For each , , because , so, , as there is an such that , , , and .
So, 1) holds.
2) for each sets, , and each point, , there is a set, , such that ?
Let , , and .
and where .
As , ; , likewise.
By 5), there is an such that ; there is an such that , likewise.
and .
By 4), there is an such that ; there is an such that , likewise.
So, , so, ; , likewise.
By 2), there is an such that .
So, .
, because .
So, will do.
So, with is a topological space, by the proposition that any basis of any topological space determines the topology.
Step 2:
Let us see that is a neighborhoods basis at .
Let be any neighborhood of .
There is a such that : the point is that is not guaranteed to be taken to be , yet.
. So, there is an such that , by 5).
.
There is an such that , by 4). So, .
So, , but , so, .
Step 3:
Let us see that the group operations of are continuous.
Let us deal with the inverse map.
Let be any neighborhood of .
There is a such that .
There is an such that , by 4).
So, .
There is an such that , by 3).
.
But and , so, .
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 be any neighborhood of .
There is a such that .
.
There is an such that , by 4), so, .
There is an such that , by 3), so, .
But as the inverse map is a homeomorphism, is an open neighborhood of , so, there is an such that , so, .
.
There is an such that , by 4), so, .
So, .
Step 4:
Let us see that is Hausdorff.
Let be any such that .
, so, there is an such that , by 1).
There is a symmetric neighborhood of , , such that , by the proposition that for any group with any topology with any continuous operations (especially, topological group), any neighborhood of , and any positive natural number, there is a symmetric neighborhood of whose power to the natural number is contained in the neighborhood.
Let us see that .
Let us suppose that .
There would be an .
for an .
. But as is symmetric, , so, , a contradiction against .
So, .
There is an such that , and .
So, .
Step 5:
Let us see that satisfies Description 2 of some criteria for any collection of open sets to be a basis.
1) (refer to the definition of union of set)?
, because , so, , there is an such that , , and .
For each , , because , so, , as there is an such that , , , and .
So, 1) holds.
2) for each sets, , and each point, , there is a set, , such that ?
Let , , and .
and where .
As , ; , likewise.
By 5), there is an such that ; there is an such that , likewise.
and .
So, ; , likewise.
By 2), there is an such that .
So, .
, because .
So, will do.
So, with is a topological space, by the proposition that any basis of any topological space determines the topology.
Step 6:
Let us see that is a neighborhoods basis at .
Let be any neighborhood of .
There is a such that : the point is that is not guaranteed to be taken to be , yet.
. So, there is an such that , by 5).
So, , but , so, .
Step 7:
Let us see that the group operations of are continuous.
Let us deal with the inverse map.
Let be any neighborhood of .
There is a such that .
There is an such that , by 4).
So, .
There is an such that , by 3).
.
But and , so, .
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 be any neighborhood of .
There is a such that .
.
There is an such that , by 4), so, .
There is an such that , by 3), so, .
But as the inverse map is a homeomorphism, is an open neighborhood of , so, there is an such that , so, .
.
There is an such that , by 4), so, .
So, .
Step 8:
Let us see that is Hausdorff.
Let be any such that .
, so, there is an such that , by 1).
There is a symmetric neighborhood of , , such that , by the proposition that for any group with any topology with any continuous operations (especially, topological group), any neighborhood of , and any positive natural number, there is a symmetric neighborhood of whose power to the natural number is contained in the neighborhood.
Let us see that .
Let us suppose that .
There would be an .
for an .
. But as is symmetric, , so, , a contradiction against .
So, .
There is an such that , and .
So, .
References
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