description/proof of that for group with topology with continuous operations (especially, topological group), open subset multiplied by subset from left or right is 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 admits the proposition that for any group with any topology with any continuous operations (especially, topological group) and each element, the inversion map, the multiplication-by-element-from-left-or-right map, and the conjugation-by-element map are homeomorphisms.
Target Context
- The reader will have a description and a proof of 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.
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:
\(G\): \(\in \{\text{ the groups with topologies with continuous operations }\}\)
\(U\): \(\in \{\text{ the open subsets of } G\}\)
\(S\): \(\in \{\text{ the subsets of } G\}\)
//
Statements:
\(S U \in \{\text{ the open subsets of } G\}\)
\(\land\)
\(U S \in \{\text{ the open subsets of } G\}\)
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2: Proof
Whole Strategy: Step 1: see that \(S U = \cup_{s \in S} s U\); Step 2: see that \(f: G \to G, g \mapsto s g\) is a homeomorphism, and see that \(s U = f (U)\); Step 3: see that \(S U \subseteq G\) is open; Step 4: see that \(U S = \cup_{s \in S} U s\); Step 5: see that \(f: G \to G, g \mapsto g s\) is a homeomorphism, and see that \(U s = f (U)\); Step 6: see that \(U S \subseteq G\) is open.
Step 1:
Let us see that \(S U = \cup_{s \in S} s U\).
For each \(s u \in S U\), \(s u \in s U \subseteq \cup_{s \in S} s U\).
For each \(s u \in \cup_{s \in S} s U\), \(s u \in S U\).
Step 2:
For each \(s \in S\), \(f: G \to G, g \mapsto s g\) is a homeomorphism, by the proposition that for any group with any topology with any continuous operations (especially, topological group) and each element, the inversion map, the multiplication-by-element-from-left-or-right map, and the conjugation-by-element map are homeomorphisms.
\(s U = f (U)\), because for each \(s u \in s U\), \(s u = f (u) \in f (U)\); for each \(s u \in f (U)\), \(s u \in s U\).
So, \(s U\) is an open subset of \(G\).
Step 3:
So, \(S U = \cup_{s \in S} s U\) is an open subset of \(G\).
Step 4:
Let us see that \(U S = \cup_{s \in S} U s\).
For each \(u s \in U S\), \(u s \in U s \subseteq \cup_{s \in S} U s\).
For each \(u s \in \cup_{s \in S} U s\), \(u s \in U S\).
Step 5:
For each \(s \in S\), \(f: G \to G, g \mapsto g s\) is a homeomorphism, by the proposition that for any group with any topology with any continuous operations (especially, topological group) and each element, the inversion map, the multiplication-by-element-from-left-or-right map, and the conjugation-by-element map are homeomorphisms.
\(U s = f (U)\), because for each \(u s \in U s\), \(u s = f (u) \in f (U)\); for each \(u s \in f (U)\), \(u s \in U s\).
So, \(U s\) is an open subset of \(G\).
Step 6:
So, \(U S = \cup_{s \in S} U s\) is an open subset of \(G\).
