description/proof of that for topological group and \(2\) disjoint compact subsets, there is symmetric neighborhood of \(1\) s.t. compact subsets multiplied by neighborhood from left and by inverse of neighborhood from right are disjoint
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 compact subset of topological space.
- The reader knows a definition of symmetric subset of group.
- The reader admits the proposition that for any Hausdorff topological space and its any 2 disjoint compact subsets, there are some disjoint open subsets each of which contains one of the compact subsets.
- The reader admits 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.
- The reader admits the proposition that for any group, the intersection of any symmetric subsets is symmetric.
- 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.
Target Context
- The reader will have a description and a proof of the proposition that for any topological group and any \(2\) disjoint compact subsets, there is a symmetric neighborhood of \(1\) such that the compact subsets multiplied by the neighborhood from left and multiplied by the inverse of the neighborhood from right are disjoint.
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:
\(G\): \(\in \{\text{ the topological groups }\}\)
\(K'\): \(\in \{\text{ the compact subsets of } G\}\)
\(K''\): \(\in \{\text{ the compact subsets of } G\}\)
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Statements:
\(K' \cap K'' = \emptyset\)
\(\implies\)
\(\exists N_1 \subseteq G \in \{\text{ the symmetric neighborhoods of } 1 \} (N_1 K' {N_1}^{-1} \cap N_1 K'' {N_1}^{-1} = \emptyset)\)
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2: Proof
Whole Strategy: Step 1: take some open neighborhoods of \(K'\) and \(K''\), \(U'\) and \(U''\), such that \(U' \cap U'' = \emptyset\); Step 2: take some symmetric neighborhoods of \(1\), \(N'_1\) and \(N''_1\), such that \(N'_1 K' {N'_1}^{-1} \subseteq U'\) and \(N''_1 K'' {N''_1}^{-1} \subseteq U''\); Step 3: take \(N_1 := N'_1 \cap N''_1\), and see that \(N_1\) is a symmetric neighborhood of \(1\) and \(N_1 K' {N_1}^{-1} \cap N_1 K'' {N_1}^{-1} = \emptyset\).
Step 1:
There are some open neighborhoods of \(K'\) and \(K''\), \(U' \subseteq G\) and \(U'' \subseteq G\), such that \(K' \subseteq U'\) and \(K'' \subseteq U''\) and \(U' \cap U'' = \emptyset\), by the proposition that for any Hausdorff topological space and its any 2 disjoint compact subsets, there are some disjoint open subsets each of which contains one of the compact subsets.
Step 2:
There are some symmetric neighborhoods of \(1\), \(N'_1\) and \(N''_1\), such that \(N'_1 K' {N'_1}^{-1} \subseteq U'\) and \(N''_1 K'' {N''_1}^{-1} \subseteq U''\), by 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.
Step 3:
Let us take \(N_1 := N'_1 \cap N''_1\).
\(N_1\) is a symmetric neighborhood of \(1\), by the proposition that for any group, the intersection of any symmetric subsets is symmetric and 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.
As \(N_1 \subseteq N'_1\), \(N_1 K' {N_1}^{-1} \subseteq N'_1 K' {N'_1}^{-1} \subseteq U'\).
As \(N_1 \subseteq N''_1\), \(N_1 K'' {N_1}^{-1} \subseteq N''_1 K'' {N''_1}^{-1} \subseteq U''\).
So, \(N_1 K' {N_1}^{-1} \cap N_1 K'' {N_1}^{-1} \subseteq U' \cap U'' = \emptyset\), which implies that \(N_1 K' {N_1}^{-1} \cap N_1 K'' {N_1}^{-1} = \emptyset\).
