description/proof of that for group with topology with continuous operations (especially, topological group) and compact subset contained in open subset, there is symmetric neighborhood of \(1\) s.t. compact subset multiplied by neighborhood from left and by inverse of neighborhood from right is contained in open subset
Topics
About: group
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of group.
- The reader knows a definition of topological space.
- The reader knows a definition of continuous, topological spaces map.
- The reader knows a definition of symmetric subset of group.
- The reader knows a definition of neighborhood of point on topological space.
- 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 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.
- 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 open subset multiplied by any subset from left or right is an open subset.
- The reader admits the proposition that for any group, the intersection of any symmetric subsets is symmetric.
Target Context
- The reader will have a description and a proof of the proposition that for any group with topology with continuous operations (especially, topological group) and any compact subset contained in any open subset, there is a symmetric neighborhood of \(1\) such that the compact subset multiplied by the neighborhood from left and multiplied by the inverse of the neighborhood from right is contained in the open subset.
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:
\(G\): \(\in \{\text{ the groups with topologies with continuous operations }\}\)
\(K\): \(\in \{\text{ the compact subsets of } G\}\)
\(U\): \(\in \{\text{ the open subsets of } G\}\) such that \(K \subseteq U\)
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Statements:
\(\exists N_1 \subseteq G \in \{\text{ the symmetric neighborhoods of } 1\} (N_1 K {N_1}^{-1} \subseteq U)\)
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2: Proof
Whole Strategy: Step 1: for each \(k \in K\), take a symmetric neighborhood of \(1\), \(N'_{1, k}\), such that \(N'_{1, k} k {N'_{1, k}}^{-1} \subseteq U\), and take a symmetric neighborhood of \(1\), \(N_{1, k}\), such that \(N_{1, k}^2 \subseteq N'_{1, k}\); Step 2: see that \(N_{1, k} k {N_{1, k}}^{-1}\) is a neighborhood of \(k\); Step 3: take a finite cover of \(K\), \(\{N_{1, k_j} k_j {N_{1, k_j}}^{-1} \vert j \in \{1, ..., n\}\}\); Step 4: take \(N_1 := N_{1, k_1} \cap ... \cap N_{1, k_n}\), and see that \(N_1\) is a symmetric neighborhood of \(1\) and \(N_1 K {N_1}^{-1} \subseteq U\).
Step 1:
Let \(k \in K\) be any.
\(U\) is an open neighborhood of \(k\).
There is a symmetric neighborhood of \(1\), \(N'_{1, k} \subseteq G\), such that \(N'_{1, k} k {N'_{1, k}}^{-1} \subseteq U\), 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.
There is a symmetric neighborhood of \(1\), \(N_{1, k} \subseteq G\), such that \(N_{1, k}^2 \subseteq N'_{1, k}\), 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.
Note that inevitably \(N_{1, k} \subseteq N'_{1, k}\), because if there was a \(g \in N_{1, k} \setminus N'_{1, k}\), \(g = 1 g \in {N_{1, k}}^2\), a contradiction against \(N_{1, k}^2 \subseteq N'_{1, k}\).
Step 2:
Let us see that \(N_{1, k} k {N_{1, k}}^{-1}\) is a neighborhood of \(k\).
There is an open neighborhood of \(1\), \(U_{1, k}\), such that \(1 \in U_{1, k} \subseteq N_{1, k}\).
\(k = 1 k 1 \in U_{1, k} k {N_{1, k}}^{-1} \subseteq N_{1, k} k {N_{1, k}}^{-1}\), but \(U_{1, k} k {N_{1, k}}^{-1} = U_{1, k} (k {N_{1, k}}^{-1})\) is an open subset, by the proposition that for any group with any topology with any continuous operations (especially, topological group), any open subset multiplied by any subset from left or right is an open subset, which means that \(N_{1, k} k {N_{1, k}}^{-1}\) is a neighborhood of \(k\).
Step 3:
\(\{U_{1, k} k {N_{1, k}}^{-1} \vert k \in K\}\) is an open cover of \(K\).
As \(K\) is a compact subset, there is a finite subcover, \(\{U_{1, k_j} k_j {N_{1, k_j}}^{-1} \vert j \in \{1, ..., n\}\}\).
So, \(\{N_{1, k_j} k_j {N_{1, k_j}}^{-1} \vert j \in \{1, ..., n\}\}\) is a finite cover of \(K\).
Step 4:
Let us take \(N_1 := N_{1, k_1} \cap ... \cap N_{1, k_n}\), which is a symmetric 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 and the proposition that for any group, the intersection of any symmetric subsets is symmetric.
Let us see that \(N_1 K {N_1}^{-1} \subseteq U\).
Let \(g \in N_1 K {N_1}^{-1}\) be any.
\(g \in N_1 k {N_1}^{-1}\) for a \(k \in K\).
As \(N_1 \subseteq N_{1, k_j}\) for each \(j \in \{1, ..., n\}\), \({N_1}^{-1} \subseteq {N_{1, k_j}}^{-1}\) and \(N_1 k {N_1}^{-1} \subseteq N_{1, k_j} k {N_{1, k_j}}^{-1}\).
But \(k \in N_{1, k_l} k_l {N_{1, k_l}}^{-1}\) for an \(l \in \{1, ..., n\}\), because \(\{N_{1, k_j} k_j {N_{1, k_j}}^{-1} \vert j \in \{1, ..., n\}\}\) covers \(K\).
So, \(N_{1, k_l} k {N_{1, k_l}}^{-1} \subseteq N_{1, k_l} N_{1, k_l} k_l {N_{1, k_l}}^{-1} {N_{1, k_l}}^{-1} = {N_{1, k_l}}^2 k_l {{N_{1, k_l}}^{-1}}^2 = {N_{1, k_l}}^2 k_l {{N_{1, k_l}}}^2 \subseteq N'_{1, k_l} k_l N'_{1, k_l} = N'_{1, k_l} k_l {N'_{1, k_l}}^{-1} \subseteq U\).
So, \(g \in N_1 k {N_1}^{-1} \subseteq U\).
So, \(N_1 K {N_1}^{-1} \subseteq U\).
