description/proof of that for set, union of subsets minus subset is union of (each of former subsets minus latter subset) s
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of set.
Target Context
- The reader will have a description and a proof of the proposition that for any set, the union of any subsets minus any subset is the union of (each of the former subsets minus the latter subset) s.
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:
\(S'\): \(\in \{\text{ the sets }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S_j \subseteq S' \vert j \in J\}\):
\(S\): \(\subseteq S'\)
//
Statements:
\((\cup_{j \in J} S_j) \setminus S = \cup_{j \in J} (S_j \setminus S)\)
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2: Proof
Whole Strategy: Step 1: see that for each \(s \in (\cup_{j \in J} S_j) \setminus S\), \(s \in \cup_{j \in J} (S_j \setminus S)\); Step 2: see that for each \(s \in \cup_{j \in J} (S_j \setminus S)\), \(s \in (\cup_{j \in J} S_j) \setminus S\).
Step 1:
Let \(s \in (\cup_{j \in J} S_j) \setminus S\) be any.
\(s \in \cup_{j \in J} S_j\), so, \(s \in S_j\) for a \(j \in J\).
\(s \notin S\).
So, \(s \in S_j \setminus S\) for that \(j\).
So, \(s \in \cup_{j \in J} (S_j \setminus S)\).
Step 2:
Let \(s \in \cup_{j \in J} (S_j \setminus S)\) be any.
\(s \in S_j \setminus S\) for a \(j \in J\).
So, \(s \in S_j\) for that \(j\) and \(s \notin S\).
So, \(s \in \cup_{j \in J} S_j\).
So, \(s \in (\cup_{j \in J} S_j) \setminus S\).
