description/proof of that set minus (set minus set) is union of 1st set minus 2nd set and intersection of 1st set and 3rd 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 set.
Target Context
- The reader will have a description and a proof of the proposition that any set minus (any set minus any set) is the union of the 1st set minus the 2nd set and the intersection of the 1st set and the 3rd 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:
\(S_1\): \(\in \{\text{ the sets }\}\)
\(S_2\): \(\in \{\text{ the sets }\}\)
\(S_3\): \(\in \{\text{ the sets }\}\)
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Statements:
\(S_1 \setminus (S_2 \setminus S_3) = (S_1 \setminus S_2) \cup (S_1 \cap S_3)\)
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2: Natural Language Description
For any sets, \(S_1, S_2, S_3\), \(S_1 \setminus (S_2 \setminus S_3) = (S_1 \setminus S_2) \cup (S_1 \cap S_3)\).
3: Proof
Whole Strategy: Step 1: see that \(S_1 \setminus (S_2 \setminus S_3) \subseteq (S_1 \setminus S_2) \cup (S_1 \cap S_3)\); Step 2: see that \((S_1 \setminus S_2) \cup (S_1 \cap S_3) \subseteq S_1 \setminus (S_2 \setminus S_3)\).
Step 1:
For any \(p \in S_1 \setminus (S_2 \setminus S_3)\), \(p \in S_1\), \(p \notin S_2 \setminus S_3\), \(p \notin S_2\) or \(p \in S_3\), so, \(p \in S_1 \setminus S_2\) or \(p \in S_1 \cap S_3\), so, \(p \in (S_1 \setminus S_2) \cup (S_1 \cap S_3)\).
Step 2:
For any \(p \in (S_1 \setminus S_2) \cup (S_1 \cap S_3)\), \(p \in S_1\), \(p \notin S_2\) or \(p \in S_3\), so, \(p \notin S_2 \setminus S_3\), so, \(p \in S_1 \setminus (S_2 \setminus S_3)\).
4: Note
\(S_1 \setminus (S_2 \setminus S_3) = (S_1 \setminus S_2) \cup S_3\) does not necessarily hold, because for example, if \(S_3\) contains a point not contained in \(S_1\), the left hand does not contain the point but the right hand side contains the point.
\(S_1 \setminus (S_2 \setminus S_3)\) is not necessarily equal to but contains \((S_1 \setminus S_2) \setminus S_3\), as is proved in another article.
