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
- 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 knows a definition of union of set.
- The reader admits some criteria for any collection of open sets to be a basis.
- The reader admits the proposition that any basis of any topological space determines the topology.
- The reader admits 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.
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:
\(G\): \(\in \{\text{ the groups }\}\)
\(B_1\): \(\subseteq Pow (G)\), with the properties specified below
\(B\): \(= \cup_{g \in G} g B_1\)
\(B'\): \(= \cup_{g \in G} B_1 g\)
//
Statements:
(
0) \(B_1 \neq \emptyset \land \forall S_1 \in B_1 (1 \in S_1)\)
\(\land\)
1) \(\forall g \in G \text{ such that } g \neq 1 (\exists S_1 \in B_1 (g \notin S_1))\)
\(\land\)
2) \(\forall S_1, S'_1 \in B_1 (\exists S''_1 \in B_1 (S''_1 \subseteq S_1 \cap S'_1))\)
\(\land\)
3) \(\forall S_1 \in B_1 (\exists S'_1 \in B_1 ({S'_1}^{-1} S'_1 \subseteq S_1))\)
\(\land\)
4) \(\forall S_1 \in B_1, \forall g \in G (\exists S'_1 \in B_1 (S'_1 \subseteq g S_1 g^{-1}))\)
\(\land\)
5) \(\forall S_1 \in B_1, \forall s \in S_1 (\exists S'_1 \in B_1 (S'_1 s \subseteq S_1))\)
)
\(\implies\)
(
(
\(B \in \{\text{ the topological bases for } G\}\)
\(\land\)
\(G \in \{\text{ the topological groups with } B\}\)
\(\land\)
\(\forall g \in G (g B \in \{\text{ the neighborhoods bases at } g \text{ on } G\})\)
)
\(\land\)
(
\(B' \in \{\text{ the topological bases for } G\}\)
\(\land\)
\(G \in \{\text{ the topological groups with } B'\}\)
\(\land\)
\(\forall g \in G (B g \in \{\text{ the neighborhoods bases at } g \text{ on } G\})\)
)
)
//
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 \(g B_1\) 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 \(B_1 g\) 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) \(G = \cup B\) (refer to the definition of union of set)?
\(1 \in \cup B\), because \(B_1 \subseteq B\), so, \(\cup B_1 \subseteq \cup B\), there is an \(S_1 \in B_1\) such that \(1 \in S_1\), \(S_1 \subseteq \cup B_1\), and \(1 \in S_1 \subseteq \cup B_1 \subseteq \cup B\).
For each \(g \in G\), \(g \in \cup B\), because \(g B_1 \subseteq B\), so, \(\cup g B_1 \subseteq \cup B\), as there is an \(S_1 \in B_1\) such that \(1 \in S_1\), \(g \in g S_1 \in g B_1\), \(g S_1 \subseteq \cup g B_1\), and \(g \in g S_1 \subseteq \cup g B_1 \subseteq \cup B\).
So, 1) holds.
2) for each sets, \(S_j, S_l \in B\), and each point, \(g \in S_j \cap S_l\), there is a set, \(S_m \in B\), such that \(g \in S_m \subseteq S_j \cap S_l\)?
Let \(S_j \in g_j B_1\), \(S_l \in g_l B_1\), and \(S_m \in g_m B_1\).
\(S_j = g_j S_{1, j}\) and \(S_l = g_j S_{1, l}\) where \(S_{1, j}, S_{1, l} \in B_1\).
As \(g \in g_j S_{1, j}\), \({g_j}^{-1} g \in S_{1, j}\); \({g_l}^{-1} g \in S_{1, l}\), likewise.
By 5), there is an \(S'_{1, j} \in B_1\) such that \(S'_{1, j} {g_j}^{-1} g \subseteq S_{1, j}\); there is an \(S'_{1, l} \in B_1\) such that \(S'_{1, l} {g_l}^{-1} g \subseteq S_{1, l}\), likewise.
\(S'_{1, j} \subseteq S_{1, j} g^{-1} g_j\) and \(S'_{1, l} \subseteq S_{1, l} g^{-1} g_l\).
By 4), there is an \(S''_{1, j} \in B_1\) such that \(S''_{1, j} \subseteq g^{-1} g_j S'_{1, j} {g_j}^{-1} g\); there is an \(S''_{1, l} \in B_1\) such that \(S''_{1, l} \subseteq g^{-1} g_l S'_{1, l} {g_l}^{-1} g\), likewise.
So, \(S''_{1, j} \subseteq g^{-1} g_j S_{1, j} g^{-1} g_j {g_j}^{-1} g = g^{-1} g_j S_{1, j}\), so, \(g S''_{1, j} \subseteq g g^{-1} g_j S_{1, j} = g_j S_{1, j}\); \(g S''_{1, l} \subseteq g_l S_{1, l}\), likewise.
By 2), there is an \(S_1 \in B_1\) such that \(S_1 \subseteq S''_{1, j} \cap S''_{1, l}\).
So, \(g S_1 \subseteq g (S''_{1, j} \cap S''_{1, l}) = (g S''_{1, j}) \cap (g S''_{1, l}) \subseteq g_j S_{1, j} \cap g_l S_{1, l} = S_j \cap S_l\).
\(g \in g S_1\), because \(1 \in S_1\).
So, \(g S_1 = S_m\) 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 \(g B_1\) is a neighborhoods basis at \(g\).
Let \(N_g\) be any neighborhood of \(g\).
There is a \(g' S_1 \in B\) such that \(g \in g' S_1 \subseteq N_g\): the point is that \(g'\) is not guaranteed to be taken to be \(g\), yet.
\(g'^{-1} g \in S_1\). So, there is an \(S'_1 \in B_1\) such that \(S'_1 g'^{-1} g \subseteq S_1\), by 5).
\(S'_1 g'^{-1} g = g'^{-1} g g^{-1} g' S'_1 g'^{-1} g\).
There is an \(S''_1 \in B_1\) such that \(S''_1 \subseteq g^{-1} g' S'_1 g'^{-1} g\), by 4). So, \(g'^{-1} g S''_1 \subseteq g'^{-1} g g^{-1} g' S'_1 g'^{-1} g = S'_1 g'^{-1} g \subseteq S_1\).
So, \(g' g'^{-1} g S''_1 \subseteq g' S_1\), but \(g' g'^{-1} g S''_1 = g S''_1\), so, \(g \in g S''_1 \subseteq g' S_1 \subseteq N_g\).
Step 3:
Let us see that the group operations of \(G\) are continuous.
Let us deal with the inverse map.
Let \(N_{g^{-1}} \subseteq G\) be any neighborhood of \(g^{-1}\).
There is a \(g^{-1} S_1 \in B\) such that \(g^{-1} \in g^{-1} S_1 \subseteq N_{g^{-1}}\).
There is an \(S'_1 \in B_1\) such that \(S'_1 \subseteq g^{-1} S_1 g\), by 4).
So, \(S'_1 g^{-1} \subseteq g^{-1} S_1\).
There is an \(S''_1 \in B_1\) such that \({S''_1}^{-1} S''_1 \subseteq S'_1\), by 3).
\({S''_1}^{-1} S''_1 g^{-1} \subseteq S'_1 g^{-1}\).
But \({S''_1}^{-1} S''_1 g^{-1} = (g {S''_1}^{-1} S''_1)^{-1}\) and \(S''_1 \subseteq {S''_1}^{-1} S''_1\), so, \((g S''_1)^{-1} \subseteq S'_1 g^{-1} \subseteq g^{-1} S_1 \subseteq N_{g^{-1}}\).
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 \(N_{g_1 g_2} \subseteq G\) be any neighborhood of \(g_1 g_2\).
There is a \(g_1 g_2 S_1 \in B\) such that \(g_1 g_2 \in g_1 g_2 S_1 \subseteq N_{g_1 g_2}\).
\(g_1 g_2 S_1 = g_1 g_2 S_1 g_2^{-1} g_2\).
There is an \(S'_1 \in B_1\) such that \(S'_1 \subseteq g_2 S_1 g_2^{-1}\), by 4), so, \(g_1 S'_1 g_2 \subseteq g_1 g_2 S_1 g_2^{-1} g_2 = g_1 g_2 S_1\).
There is an \(S''_1 \in B_1\) such that \({S''_1}^{-1} S''_1 \subseteq S'_1\), by 3), so, \(g_1 {S''_1}^{-1} S''_1 g_2 \subseteq g_1 S'_1 g_2 \subseteq g_1 g_2 S_1\).
But as the inverse map is a homeomorphism, \({S''_1}^{-1}\) is an open neighborhood of \(1\), so, there is an \(S'''_1 \in B_1\) such that \(S'''_1 \subseteq {S''_1}^{-1}\), so, \(g_1 S'''_1 S''_1 g_2 \subseteq g_1 {S''_1}^{-1} S''_1 g_2\).
\(g_1 S'''_1 S''_1 g_2 = g_1 S'''_1 g_2 g_2^{-1} S''_1 g_2\).
There is an \(S''''_1 \in B_1\) such that \(S''''_1 \subseteq g_2^{-1} S''_1 g_2\), by 4), so, \(g_1 S'''_1 g_2 S''''_1 \subseteq g_1 S'''_1 S''_1 g_2\).
So, \(g_1 S'''_1 g_2 S''''_1 \subseteq g_1 S'''_1 S''_1 g_2 \subseteq g_1 {S''_1}^{-1} S''_1 g_2 \subseteq g_1 g_2 S_1 \subseteq N_{g_1 g_2}\).
Step 4:
Let us see that \(G\) is Hausdorff.
Let \(g_1, g_2 \in G\) be any such that \(g_1 \neq g_2\).
\({g_1}^{-1} g_2 \neq 1\), so, there is an \(S_1 \in B_1\) such that \({g_1}^{-1} g_2 \notin S_1\), by 1).
There is a symmetric neighborhood of \(1\), \(N_1 \subseteq G\), such that \(N_1^2 \subseteq S_1\), 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 \(N_1 \cap {g_1}^{-1} g_2 N_1 = \emptyset\).
Let us suppose that \(N_1 \cap {g_1}^{-1} g_2 N_1 \neq \emptyset\).
There would be an \(n \in N_1 \cap {g_1}^{-1} g_2 N_1\).
\(n = {g_1}^{-1} g_2 n'\) for an \(n' \in N_1\).
\({g_1}^{-1} g_2 = n n'^{-1}\). But as \(N_1\) is symmetric, \(n'^{-1} \in N_1^{-1} = N_1\), so, \({g_1}^{-1} g_2 \in N_1^2 \subseteq S_1\), a contradiction against \({g_1}^{-1} g_2 \notin S_1\).
So, \(N_1 \cap {g_1}^{-1} g_2 N_1 = \emptyset\).
There is an \(S'_1 \in B_1\) such that \(S'_1 \subseteq N_1\), and \(S'_1 \cap {g_1}^{-1} g_2 S'_1 \subseteq N_1 \cap {g_1}^{-1} g_2 N_1 = \emptyset\).
So, \(g_1 (S'_1 \cap {g_1}^{-1} g_2 S'_1) = g_1 S'_1 \cap g_2 S'_1 = \emptyset\).
Step 5:
Let us see that \(B'\) satisfies Description 2 of some criteria for any collection of open sets to be a basis.
1) \(G = \cup B\) (refer to the definition of union of set)?
\(1 \in \cup B'\), because \(B_1 \subseteq B'\), so, \(\cup B_1 \subseteq \cup B'\), there is an \(S_1 \in B_1\) such that \(1 \in S_1\), \(S_1 \subseteq \cup B_1\), and \(1 \in S_1 \subseteq \cup B_1 \subseteq \cup B'\).
For each \(g \in G\), \(g \in \cup B'\), because \(B_1 g \subseteq B'\), so, \(\cup B_1 g \subseteq \cup B'\), as there is an \(S_1 \in B_1\) such that \(1 \in S_1\), \(g \in S_1 g \in B_1 g\), \(S_1 g \subseteq \cup B_1 g\), and \(g \in S_1 g \subseteq \cup B_1 g \subseteq \cup B'\).
So, 1) holds.
2) for each sets, \(S_j, S_l \in B'\), and each point, \(g \in S_j \cap S_l\), there is a set, \(S_m \in B'\), such that \(g \in S_m \subseteq S_j \cap S_l\)?
Let \(S_j \in B_1 g_j\), \(S_l \in B_1 g_l\), and \(S_m \in B_1 g_m\).
\(S_j = S_{1, j} g_j\) and \(S_l = S_{1, l} g_j\) where \(S_{1, j}, S_{1, l} \in B_1\).
As \(g \in S_{1, j} g_j\), \(g {g_j}^{-1} \in S_{1, j}\); \(g {g_l}^{-1} \in S_{1, l}\), likewise.
By 5), there is an \(S'_{1, j} \in B_1\) such that \(S'_{1, j} g {g_j}^{-1} \subseteq S_{1, j}\); there is an \(S'_{1, l} \in B_1\) such that \(S'_{1, l} g {g_l}^{-1} \subseteq S_{1, l}\), likewise.
\(S'_{1, j} \subseteq S_{1, j} g_j g^{-1}\) and \(S'_{1, l} \subseteq S_{1, l} g_l g^{-1}\).
So, \(S'_{1, j} g \subseteq S_{1, j} g_j g^{-1} g = S_{1, j} g_j\); \(S'_{1, l} g \subseteq S_{1, l} g_l\), likewise.
By 2), there is an \(S_1 \in B_1\) such that \(S_1 \subseteq S'_{1, j} \cap S'_{1, l}\).
So, \(S_1 g \subseteq (S'_{1, j} \cap S'_{1, l}) g = (S'_{1, j} g) \cap (S'_{1, l} g) \subseteq (S_{1, j} g_j) \cap (S_{1, l} g_l) = S_j \cap S_l\).
\(g \in S_1 g\), because \(1 \in S_1\).
So, \(S_1 g = S_m\) 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 \(B_1 g\) is a neighborhoods basis at \(g\).
Let \(N_g\) be any neighborhood of \(g\).
There is a \(S_1 g' \in B'\) such that \(g \in S_1 g' \subseteq N_g\): the point is that \(g'\) is not guaranteed to be taken to be \(g\), yet.
\(g g'^{-1} \in S_1\). So, there is an \(S'_1 \in B_1\) such that \(S'_1 g g'^{-1} \subseteq S_1\), by 5).
So, \(S'_1 g g'^{-1} g' \subseteq S_1 g'\), but \(S'_1 g g'^{-1} g' = S'_1 g\), so, \(g \in S'_1 g \subseteq S_1 g' \subseteq N_g\).
Step 7:
Let us see that the group operations of \(G\) are continuous.
Let us deal with the inverse map.
Let \(N_{g^{-1}} \subseteq G\) be any neighborhood of \(g^{-1}\).
There is a \(S_1 g^{-1} \in B'\) such that \(g^{-1} \in S_1 g^{-1} \subseteq N_{g^{-1}}\).
There is an \(S'_1 \in B_1\) such that \(S'_1 \subseteq g S_1 g^{-1}\), by 4).
So, \(g^{-1} S'_1 \subseteq S_1 g^{-1}\).
There is an \(S''_1 \in B_1\) such that \({S''_1}^{-1} S''_1 \subseteq S'_1\), by 3).
\(g^{-1} {S''_1}^{-1} S''_1 \subseteq g^{-1} S'_1\).
But \(g^{-1} {S''_1}^{-1} S''_1 = ({S''_1}^{-1} S''_1 g)^{-1}\) and \(S''_1 \subseteq {S''_1}^{-1} S''_1\), so, \((S''_1 g)^{-1} \subseteq g^{-1} S'_1 \subseteq S_1 g^{-1} \subseteq N_{g^{-1}}\).
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 \(N_{g_1 g_2} \subseteq G\) be any neighborhood of \(g_1 g_2\).
There is a \(S_1 g_1 g_2 \in B'\) such that \(g_1 g_2 \in S_1 g_1 g_2 \subseteq N_{g_1 g_2}\).
\(S_1 g_1 g_2 = g_1 {g_1}^{-1} S_1 g_1 g_2\).
There is an \(S'_1 \in B_1\) such that \(S'_1 \subseteq {g_1}^{-1} S_1 g_1\), by 4), so, \(g_1 S'_1 g_2 \subseteq g_1 {g_1}^{-1} S_1 g_1 g_2 = S_1 g_1 g_2\).
There is an \(S''_1 \in B_1\) such that \({S''_1}^{-1} S''_1 \subseteq S'_1\), by 3), so, \(g_1 {S''_1}^{-1} S''_1 g_2 \subseteq g_1 S'_1 g_2 \subseteq S_1 g_1 g_2\).
But as the inverse map is a homeomorphism, \({S''_1}^{-1}\) is an open neighborhood of \(1\), so, there is an \(S'''_1 \in B_1\) such that \(S'''_1 \subseteq {S''_1}^{-1}\), so, \(g_1 S'''_1 S''_1 g_2 \subseteq g_1 {S''_1}^{-1} S''_1 g_2\).
\(g_1 S'''_1 S''_1 g_2 = g_1 S'''_1 g_1^{-1} g_1 S''_1 g_2\).
There is an \(S''''_1 \in B_1\) such that \(S''''_1 \subseteq g_1 S'''_1 g_1^{-1}\), by 4), so, \(S''''_1 g_1 S''_1 g_2 \subseteq g_1 S'''_1 S''_1 g_2\).
So, \(S''''_1 g_1 S''_1 g_2 \subseteq g_1 S'''_1 S''_1 g_2 \subseteq g_1 {S''_1}^{-1} S''_1 g_2 \subseteq S_1 g_1 g_2 \subseteq N_{g_1 g_2}\).
Step 8:
Let us see that \(G\) is Hausdorff.
Let \(g_1, g_2 \in G\) be any such that \(g_1 \neq g_2\).
\(g_1 {g_2}^{-1} \neq 1\), so, there is an \(S_1 \in B_1\) such that \(g_1 {g_2}^{-1} \notin S_1\), by 1).
There is a symmetric neighborhood of \(1\), \(N_1 \subseteq G\), such that \(N_1^2 \subseteq S_1\), 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 \(N_1 \cap N_1 g_1 {g_2}^{-1} = \emptyset\).
Let us suppose that \(N_1 \cap N_1 g_1 {g_2}^{-1} \neq \emptyset\).
There would be an \(n \in N_1 \cap N_1 g_1 {g_2}^{-1}\).
\(n = n' g_1 {g_2}^{-1}\) for an \(n' \in N_1\).
\(g_1 {g_2}^{-1} = n'^{-1} n\). But as \(N_1\) is symmetric, \(n'^{-1} \in N_1^{-1} = N_1\), so, \(g_1 {g_2}^{-1} \in N_1^2 \subseteq S_1\), a contradiction against \(g_1 {g_2}^{-1} \notin S_1\).
So, \(N_1 \cap N_1 g_1 {g_2}^{-1} = \emptyset\).
There is an \(S'_1 \in B_1\) such that \(S'_1 \subseteq N_1\), and \(S'_1 \cap S'_1 g_1 {g_2}^{-1} \subseteq N_1 \cap N_1 g_1 {g_2}^{-1} = \emptyset\).
So, \((S'_1 \cap S'_1 g_1 {g_2}^{-1}) g_2 = S'_1 g_2 \cap S'_1 g_1 {g_2}^{-1} g_2 = S'_1 g_2 \cap S'_1 g_1 = \emptyset\).
