description/proof of that for topological group and finite number of subsets whose closures are compact, closure of product of subsets is contained in product of closures of subsets and is compact
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 finite product of subsets of group.
- The reader knows a definition of closure of subset of topological space.
- The reader knows a definition of compact subset of topological space.
- The reader admits the proposition that for any group with topology with continuous operations (especially, topological group), the inverse of any compact subset is compact and the finite product of any compact subsets is compact.
- The reader admits the proposition that any compact subset of any Hausdorff topological space is closed.
- The reader admits the proposition that the compactness of any topological subset as a subset equals the compactness as a subspace.
- The reader admits the proposition that any subset on any topological subspace is closed if and only if there is a closed set on the base space whose intersection with the subspace is the subset.
- The reader admits the proposition that any closed subset of any compact topological space is compact.
- The reader admits the proposition that for any topological space, any compact subset of any subspace is compact on the base space.
Target Context
- The reader will have a description and a proof of the proposition that for any topological group and any finite number of subsets whose closures are compact, the closure of the product of the subsets is contained in the product of the closures of the subsets and is compact.
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:
\(G\): \(\in \{\text{ the topological groups }\}\)
\(J\): \(\in \{\text{ the finite index sets }\}\), \(= \{j_1, ..., j_n\}\)
\(\{S_j \subseteq G \vert j \in J\}\):
//
Statements:
\(\forall j \in J (\overline{S_j} \in \{\text{ the compact subsets of } G\})\)
\(\implies\)
(
\(\overline {S_{j_1} ... S_{j_n}} \subseteq \overline{S_{j_1}} ... \overline{S_{j_n}}\)
\(\implies\)
\(\overline {S_{j_1} ... S_{j_n}} \in \{\text{ the compact subsets of } G\}\)
)
//
2: Note
Proof requires \(G\) to be Hausdorff, so, \(G\) is required to be a topological group, not just a group with topology with continuous operations.
3: Proof
Whole Strategy: Step 1: see that \(S_{j_1} ... S_{j_n} \subseteq \overline{S_{j_1}} ... \overline{S_{j_n}}\); Step 2: see that \(\overline{S_{j_1}} ... \overline{S_{j_n}}\) is compact and closed; Step 3: see that \(\overline {S_{j_1} ... S_{j_n}} \subseteq \overline{S_{j_1}} ... \overline{S_{j_n}}\); Step 4: conclude the proposition.
Step 1:
\(S_{j_1} ... S_{j_n} \subseteq \overline{S_{j_1}} ... \overline{S_{j_n}}\), because \(S_{j_l} \subseteq \overline{S_{j_l}}\).
Step 2:
\(\overline{S_{j_1}} ... \overline{S_{j_n}}\) is compact, by the proposition that for any group with topology with continuous operations (especially, topological group), the inverse of any compact subset is compact and the finite product of any compact subsets is compact.
\(\overline{S_{j_1}} ... \overline{S_{j_n}}\) is closed, by the proposition that any compact subset of any Hausdorff topological space is closed.
Step 3:
\(\overline {S_{j_1} ... S_{j_n}} \subseteq \overline{S_{j_1}} ... \overline{S_{j_n}}\), because \(\overline {S_{j_1} ... S_{j_n}}\) is the intersection of the closed subsets that contain \(S_{j_1} ... S_{j_n}\) while \(\overline{S_{j_1}} ... \overline{S_{j_n}}\) is one of such closed subsets by Step 1 and Step 2.
Step 4:
\(\overline{S_{j_1}} ... \overline{S_{j_n}}\) as the topological subspace is a compact topological space, by the proposition that the compactness of any topological subset as a subset equals the compactness as a subspace.
\(\overline {S_{j_1} ... S_{j_n}}\) is closed on \(G\) and on \(\overline{S_{j_1}} ... \overline{S_{j_n}}\) as the topological subspace, by the proposition that any subset on any topological subspace is closed if and only if there is a closed set on the base space whose intersection with the subspace is the subset: \(\overline {S_{j_1} ... S_{j_n}} = \overline {S_{j_1} ... S_{j_n}} \cap \overline{S_{j_1}} ... \overline{S_{j_n}}\).
\(\overline {S_{j_1} ... S_{j_n}}\) is a compact subset of \(\overline{S_{j_1}} ... \overline{S_{j_n}}\), by the proposition that any closed subset of any compact topological space is compact.
\(\overline {S_{j_1} ... S_{j_n}}\) is a compact subset of \(G\), by the proposition that for any topological space, any compact subset of any subspace is compact on the base space.
