description/proof of that for group, intersection of symmetric subsets is symmetric
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of symmetric subset of group.
Target Context
- The reader will have a description and a proof of the proposition that for any group, the intersection of any symmetric subsets is symmetric.
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:
\(G\): \(\in \{\text{ the groups }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S_j \in \{\text{ the symmetric subsets of } G\} \vert j \in J\}\):
\(S\): \(= \cap_{j \in J} S_j\)
//
Statements:
\(S \in \{\text{ the symmetric subsets of } G\}\)
//
2: Proof
Whole Strategy: Step 1: see that each element of \(S\) belongs to \(S^{-1}\); Step 2: see that each element of \(S^{-1}\) belongs to \(S\).
Step 1:
Let \(s \in S\) be any.
\(s \in S_j\) for each \(j \in J\).
\(s^{-1} \in {S_j}^{-1} = S_j\).
So, \(s^{-1} \in \cap_{j \in J} S_j = S\).
So, \(s = {s^{-1}}^{-1} \in S^{-1}\).
Step 2:
Let \(s^{-1} \in S^{-1}\) be any.
\(s \in S\), so, \(s \in S_j\) for each \(j \in J\).
\(s^{-1} \in {S_j}^{-1} = S_j\).
So, \(s^{-1} \in \cap_{j \in J} S_j = S\).
