description/proof of that for group and subsets, inverse of union of subsets is union of inverses of subsets
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of inverse of subset of group.
Target Context
- The reader will have a description and a proof of the proposition that for any group and any subsets, the inverse of the union of the subsets is the union of the inverses of the subsets.
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:
\(G\): \(\in \{\text{ the groups }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S_j \subseteq G \vert j \in J\}\):
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Statements:
\((\cup_{j \in J} S_j)^{-1} = \cup_{j \in J} {S_j}^{-1}\)
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2: Proof
Whole Strategy: Step 1: let \(g \in (\cup_{j \in J} S_j)^{-1}\) be any and see that \(g \in \cup_{j \in J} {S_j}^{-1}\); Step 2: let \(g \in \cup_{j \in J} {S_j}^{-1}\) be any and see that \(g \in (\cup_{j \in J} S_j)^{-1}\).
Step 1:
Let \(g \in (\cup_{j \in J} S_j)^{-1}\) be any.
\(g^{-1} \in \cup_{j \in J} S_j\). There is a \(j \in J\) such that \(g^{-1} \in S_j\). \(g \in {S_j}^{-1}\). So, \(g \in \cup_{j \in J} {S_j}^{-1}\).
Step 2:
Let \(g \in \cup_{j \in J} {S_j}^{-1}\) be any.
There is a \(j \in J\) such that \(g \in {S_j}^{-1}\). \(g^{-1} \in S_j\). \(g^{-1} \in \cup_{j \in J} S_j\). So, \(g \in (\cup_{j \in J} S_j)^{-1}\).
