2025-07-06

1186: For Group and Subsets, Inverse of Intersection of Subsets Is Intersection of Inverses of Subsets

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description/proof of that for group and subsets, inverse of intersection of subsets is intersection of inverses of subsets

Topics


About: group

The table of contents of this article


Starting Context



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 intersection of the subsets is the intersection 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:
\(G\): \(\in \{\text{ the groups }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S_j \subseteq G \vert j \in J\}\):
//

Statements:
\((\cap_{j \in J} S_j)^{-1} = \cap_{j \in J} {S_j}^{-1}\)
//


2: Proof


Whole Strategy: Step 1: let \(g \in (\cap_{j \in J} S_j)^{-1}\) be any and see that \(g \in \cap_{j \in J} {S_j}^{-1}\); Step 2: let \(g \in \cap_{j \in J} {S_j}^{-1}\) be any and see that \(g \in (\cap_{j \in J} S_j)^{-1}\).

Step 1:

Let \(g \in (\cap_{j \in J} S_j)^{-1}\) be any.

\(g^{-1} \in \cap_{j \in J} S_j\). For each \(j \in J\), \(g^{-1} \in S_j\). \(g \in {S_j}^{-1}\). So, \(g \in \cap_{j \in J} {S_j}^{-1}\).

Step 2:

Let \(g \in \cap_{j \in J} {S_j}^{-1}\) be any.

For each \(j \in J\), \(g \in {S_j}^{-1}\). \(g^{-1} \in S_j\). \(g^{-1} \in \cap_{j \in J} S_j\). So, \(g \in (\cap_{j \in J} S_j)^{-1}\).


References


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