description/proof of that union of set minus set and set is not necessarily but contains union of 1st set and 3rd set minus union of 2nd 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 the union of any set minus any set and any set is not necessarily but contains the union of the 1st set and the 3rd set minus the union of the 2nd 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:
not necessarily \((S_1 \setminus S_2) \cup S_3 = (S_1 \cup S_3) \setminus (S_2 \cup S_3)\)
\(\land\)
\((S_1 \cup S_3) \setminus (S_2 \cup S_3) \subseteq (S_1 \setminus S_2) \cup S_3\)
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2: Natural Language Description
For any sets, \(S_1, S_2, S_3\), \((S_1 \setminus S_2) \cup S_3\) is not necessarily \((S_1 \cup S_3) \setminus (S_2 \cup S_3)\), but \((S_1 \cup S_3) \setminus (S_2 \cup S_3) \subseteq (S_1 \setminus S_2) \cup S_3\).
3: Proof
Whole Strategy: Step 1: see an example that \((S_1 \setminus S_2) \cup S_3 \neq (S_1 \cup S_3) \setminus (S_2 \cup S_3)\); Step 2: see that \((S_1 \cup S_3) \setminus (S_2 \cup S_3) \subseteq (S_1 \setminus S_2) \cup S_3\).
Step 1:
For the 1st part, a counterexample suffices.
Let \(S_1 = \emptyset\), \(S_2 = \emptyset\), \(S_3 \neq \emptyset\). \((S_1 \setminus S_2) \cup S_3 = S_3 \neq \emptyset\), but \((S_1 \cup S_3) \setminus (S_2 \cup S_3) = S_3 \setminus S_3 = \emptyset\).
Step 2:
For any \(p \in (S_1 \cup S_3) \setminus (S_2 \cup S_3)\), \(p \in S_1\) or \(p \in S_3\), but \(p \notin S_3\), so, \(p \in S_1\), \(p \notin S_2\), so, \(p \in S_1 \setminus S_2\), so, \(p \in (S_1 \setminus S_2) \cup S_3\).
4: Note
\((S_1 \setminus S_2) \cap S_3 = (S_1 \cap S_3) \setminus (S_2 \cap S_3)\) holds, as is proved in another article.
