description/proof of that subset minus union of subsets is intersection of 1st subset minus 2nd chunk 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 union of any subsets is the intersection of the 1st subset minus the 2nd chunk 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:
\(S\): \(\in \{\text{ the sets }\}\)
\(S_1\): \(\subseteq S\)
\(\{S_{2, j} \subseteq S \vert j \in J\}\): \(J \in \{\text{ the possibly uncountable index sets }\}\)
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Statements:
\(S_1 \setminus \cup_{j \in J} S_{2, j} = \cap_{j \in J} (S_1 \setminus S_{2, j})\)
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2: Proof
Whole Strategy: Step 1: see that \(S_1 \setminus \cup_{j \in J} S_{2, j} \subseteq \cap_{j \in J} (S_1 \setminus S_{2, j})\); Step 2: see that \(\cap_{j \in J} (S_1 \setminus S_{2, j}) \subseteq S_1 \setminus \cup_{j \in J} S_{2, j}\).
Step 1:
Let \(p \in S_1 \setminus \cup_{j \in J} S_{2, j}\) be any.
\(p \in S_1\). \(p \notin \cup_{j \in J} S_{2, j}\), so, \(p \notin S_{2, j}\) for each \(j \in J\). So, \(p \in S_1 \setminus S_{2, j}\) for each \(j\). So, \(p \in \cap_{j \in J} (S_1 \setminus S_{2, j})\).
Step 2:
Let \(p \in \cap_{j \in J} (S_1 \setminus S_{2, j})\) be any.
\(p \in S_1 \setminus S_{2, j}\) for each \(j \in J\), so, \(p \in S_1\) and \(p \notin S_{2, j}\) for each \(j\), so, \(p \notin \cup_{j \in J} S_{2, j}\), so, \(p \in S_1 \setminus \cup_{j \in J} S_{2, j}\).
