description/proof of that for set, subset minus intersection of bunch of subsets is union of subset minus bunch of subsets
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, any subset minus the intersection of any bunch of subsets is the union of the subset minus the bunch of 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:
\(S'\): \(\in \{\text{ the sets }\}\)
\(S\): \(\subseteq S'\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S_j \subseteq S': j \in J\}\):
//
Statements:
\(S \setminus \cap_{j \in J} S_j = \cup_{j \in J} (S \setminus S_j)\)
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2: Proof
Whole Strategy: Step 1: see that \(S \setminus \cap_{j \in J} S_j \subseteq \cup_{j \in J} (S \setminus S_j)\); Step 2: see that \(\cup_{j \in J} (S \setminus S_j) \subseteq S \setminus \cap_{j \in J} S_j\).
Step 1:
Let \(s \in S \setminus \cap_{j \in J} S_j\) be any.
\(s \in S\) and \(s \notin \cap_{j \in J} S_j\).
\(s \notin S_j\) for a \(j \in J\).
\(s \in S \setminus S_j\) for that \(j\).
\(s \in \cup_{j \in J} (S \setminus S_j)\).
Step 2:
Let \(s \in \cup_{j \in J} (S \setminus S_j)\) be any.
\(s \in S \setminus S_j\) for a \(j \in J\).
\(s \in S\) and \(s \notin S_J\) for that \(j\).
\(s \notin \cap_{j \in J} S_j\).
So, \(s \in S \setminus \cap_{j \in J} S_j\).
