description/proof of that for \(2\) sets, union of sets is exclusive union of 1st set minus intersection of sets, 2nd set minus intersection of sets, and intersection of sets
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 \(2\) sets, the union of the sets is the exclusive union of the 1st set minus the intersection of the sets, the 2nd set minus the intersection of the sets, and the intersection of the sets.
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:
\(S_1\): \(\in \{\text{ the sets }\}\)
\(S_2\): \(\in \{\text{ the sets }\}\)
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Statements:
\(S_1 \cup S_2 = (S_1 \setminus (S_1 \cap S_2)) \cup (S_2 \setminus (S_1 \cap S_2)) \cup (S_1 \cap S_2)\)
\(\land\)
\(\{S_1 \setminus (S_1 \cap S_2), S_2 \setminus (S_1 \cap S_2), S_1 \cap S_2\} \in \{\text{ the disjoint sets of sets }\}\)
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2: Proof
Whole Strategy: Step 1: see that \(S_1 \cup S_2 = (S_1 \setminus (S_1 \cap S_2)) \cup (S_2 \setminus (S_1 \cap S_2)) \cup (S_1 \cap S_2)\); Step 2: see that \(\{S_1 \setminus (S_1 \cap S_2), S_2 \setminus (S_1 \cap S_2), S_1 \cap S_2\}\) is disjoint.
Step 1:
Let us see that \(S_1 \cup S_2 = (S_1 \setminus (S_1 \cap S_2)) \cup (S_2 \setminus (S_1 \cap S_2)) \cup (S_1 \cap S_2)\).
For each \(s \in S_1 \cup S_2\), \(s \in S_1\) or \(s \in S_2\); when \(s \in S_1\), when \(s \in S_2\), \(s \in S_1 \cap S_2\), otherwise, \(s \in S_1 \setminus (S_1 \cap S_2)\), so, anyway, \(s \in (S_1 \setminus (S_1 \cap S_2)) \cup (S_2 \setminus (S_1 \cap S_2)) \cup (S_1 \cap S_2)\); when \(s \in S_2\), \(s \in (S_1 \setminus (S_1 \cap S_2)) \cup (S_2 \setminus (S_1 \cap S_2)) \cup (S_1 \cap S_2)\), by the symmetry; so, anyway, \(s \in (S_1 \setminus (S_1 \cap S_2)) \cup (S_2 \setminus (S_1 \cap S_2)) \cup (S_1 \cap S_2)\).
So, \(S_1 \cup S_2 \subseteq (S_1 \setminus (S_1 \cap S_2)) \cup (S_2 \setminus (S_1 \cap S_2)) \cup (S_1 \cap S_2)\).
For each \(s \in (S_1 \setminus (S_1 \cap S_2)) \cup (S_2 \setminus (S_1 \cap S_2)) \cup (S_1 \cap S_2)\), \(s \in S_1 \setminus (S_1 \cap S_2)\), \(s \in S_2 \setminus (S_1 \cap S_2)\), or \(s \in S_1 \cap S_2\); when \(s \in S_1 \setminus (S_1 \cap S_2)\), \(s \in S_1\), so, \(s \in S_1 \cup S_2\); when \(s \in S_2 \setminus (S_1 \cap S_2)\), \(s \in S_1 \cup S_2\), by the symmetry; when \(s \in S_1 \cap S_2\), \(s \in S_1\), so, \(s \in S_1 \cup S_2\); so, anyway, \(s \in S_1 \cup S_2\).
So, \((S_1 \setminus (S_1 \cap S_2)) \cup (S_2 \setminus (S_1 \cap S_2)) \cup (S_1 \cap S_2) \subseteq S_1 \cup S_2\).
So, \(S_1 \cup S_2 = (S_1 \setminus (S_1 \cap S_2)) \cup (S_2 \setminus (S_1 \cap S_2)) \cup (S_1 \cap S_2)\).
Step 2:
Let us see that \(\{S_1 \setminus (S_1 \cap S_2), S_2 \setminus (S_1 \cap S_2), S_1 \cap S_2\}\) is disjoint.
For each \(s \in S_1 \setminus (S_1 \cap S_2)\), \(s \notin S_2\), because otherwise, \(s \in S_1 \cap S_2\), and \(s \notin S_1 \setminus (S_1 \cap S_2)\), a contradiction, so, \(s \notin S_2 \setminus (S_1 \cap S_2)\) and \(s \notin S_1 \cap S_2\).
For each \(s \in S_2 \setminus (S_1 \cap S_2)\), \(s \notin S_1 \setminus (S_1 \cap S_2)\) and \(s \notin S_1 \cap S_2\), by the symmetry.
For each \(s \in S_1 \cap S_2\), \(s \notin S_1 \setminus (S_1 \cap S_2)\) and \(s \notin S_2 \setminus (S_1 \cap S_2)\).
So, \(\{S_1 \setminus (S_1 \cap S_2), S_2 \setminus (S_1 \cap S_2), S_1 \cap S_2\}\) is disjoint.
