description/proof of that for set, union of power set of set is set
Topics
About: set
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
- 4: Note
Starting Context
- The reader knows a definition of power set of set.
- The reader knows a definition of union of set.
Target Context
- The reader will have a description and a proof of the proposition that for any set, the union of the power set of the set is the set.
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:
\(S\): \(\in \{\text{ the sets }\}\)
//
Statements:
\(\cup Pow (S) = S\)
//
2: Natural Language Description
For any set, \(S\), \(\cup Pow (S) = S\).
3: Proof
Whole Strategy: Step 1: see that each element of \(\cup Pow (S)\) is contained in \(S\); Step 2: see that each element of \(S\) is contained in \(\cup Pow (S)\); Step 3: conclude the proposition.
Step 1:
Let \(p \in \cup Pow (S)\) be any.
\(p\)'s being in the union of \(Pow (S)\) means that \(p\) is in an element of \(Pow (S)\), which (the element) is a subset of \(S\), so, \(p\) is in a subset of \(S\), which implies that \(p \in S\).
So, \(\cup Pow (S) \subseteq S\).
Step 2:
Let \(p \in S\) be any.
\(\{p\} \in Pow (S)\), so, \(p\) is in an element of \(Pow (S)\), so, \(p \in \cup Pow (S)\).
So, \(S \subseteq \cup Pow (S)\).
Step 3:
So, \(\cup Pow (S) = S\).
4: Note
On the other hand, \(Pow (\cup S) \neq S\) in general: while always \(\emptyset \in Pow (\cup S)\), \(\emptyset\) is not in \(S\) in general.
