description/proof of that for group with topology with continuous operations (especially, topological group), closure of subset is contained in subset multiplied by any neighborhood of \(1\) from left or right
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 closure of subset of topological space.
- 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), the set of the symmetric neighborhoods of \(1\) is a neighborhoods basis at \(1\).
- The reader admits the proposition that for any group with any topology with any continuous operations (especially, any topological group), any element multiplied by any (open) neighborhood of \(1\) from left or right is a (open) neighborhood of the element.
- The reader admits the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
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, any topological group), the closure of any subset is contained in the subset multiplied by any neighborhood of \(1\) from left or right.
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 any topology such that the group operations are continuous
\(S\): \(\subseteq G\)
\(N'_1\): \(\in \{\text{ the neighborhoods of } 1 \text{ on } G\}\)
//
Statements:
\(\overline{S} \subseteq S N'_1\)
\(\land\)
\(\overline{S} \subseteq N'_1 S\)
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2: Proof
Whole Strategy: Step 1: take a symmetric neighborhood of \(1\) contained in \(N'_1\), \(N_1\); Step 2: for each \(g \in \overline{S}\), see that \(g \in S N_1\); Step 3: for each \(g \in \overline{S}\), see that \(g \in N_1 S\).
Step 1:
There is a symmetric neighborhood of \(1\) contained in \(N'_1\), \(N_1 \subseteq G\), such that \(1 \in N_1 \subseteq N'_1\), by the proposition that for any group with any topology with any continuous operations (especially, topological group), the set of the symmetric neighborhoods of \(1\) is a neighborhoods basis at \(1\).
Step 2:
Let \(g \in \overline{S}\) be any.
\(g N_1 \subseteq G\) is a neighborhood of \(g\), by the proposition that for any group with any topology with any continuous operations (especially, any topological group), any element multiplied by any (open) neighborhood of \(1\) from left or right is a (open) neighborhood of the element.
\(g N_1 \cap S \neq \emptyset\), by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
So, there is an \(s \in g N_1 \cap S\), so, there is an \(n \in N_1\) such that \(g n = s\).
\(g = s n^{-1}\), but \(n^{-1} \in {N_1}^{-1} = N_1\), so, \(g \in s N_1 \subseteq S N_1\).
So, \(\overline{S} \subseteq S N_1 \subseteq S N'_1\).
Step 3:
Let \(g \in \overline{S}\) be any.
\(N_1 g \subseteq G\) is a neighborhood of \(g\), by the proposition that for any group with any topology with any continuous operations (especially, any topological group), any element multiplied by any (open) neighborhood of \(1\) from left or right is a (open) neighborhood of the element.
\(N_1 g \cap S \neq \emptyset\), by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
So, there is an \(s \in N_1 g \cap S\), so, there is an \(n \in N_1\) such that \(n g = s\).
\(g = n^{-1} s\), but \(n^{-1} \in {N_1}^{-1} = N_1\), so, \(g \in N_1 s \subseteq N_1 S\).
So, \(\overline{S} \subseteq N_1 S \subseteq N'_1 S\).
