description/proof of that union of complements of subsets is whole set iff intersection of subsets is empty
Topics
About: set
The table of contents of this article
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that for any set, the union of the complements of any possibly uncountable number of subsets is the whole set if and only if the intersection of the subsets is empty.
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:
\(S\): \(\in \{\text{ the sets }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S_j \subseteq S \vert j \in J\}\):
//
Statements:
\(\cup_{j \in J} (S \setminus S_j) = S\)
\(\iff\)
\(\cap_{j \in J} S = \emptyset\)
//
2: Proof
Whole Strategy: Step 1: see that \(\cup_{j \in J} (S \setminus S_j) = S \setminus \cap_{j \in J} S\); Step 2: suppose that \(\cup_{j \in J} (S \setminus S_j) = S\); Step 3: see that \(\cap_{j \in J} S = \emptyset\); Step 4: suppose that \(\cap_{j \in J} S = \emptyset\); Step 5: see that \(\cup_{j \in J} (S \setminus S_j) = S\).
Step 1:
\(\cup_{j \in J} (S \setminus S_j) = S \setminus \cap_{j \in J} S\), the proposition for any set, the union of the complements of any possibly uncountable number of subsets is the complement of the intersection of the subsets.
Step 2:
Let us suppose that \(\cup_{j \in J} (S \setminus S_j) = S\).
Step 3:
\(S \setminus \cap_{j \in J} S = \cup_{j \in J} (S \setminus S_j)\), by Step 1, \(= S\), by the supposition, which implies that \(\cap_{j \in J} S = \emptyset\).
Step 4:
Let us suppose that \(\cap_{j \in J} S = \emptyset\).
Step 5:
\(\cup_{j \in J} (S \setminus S_j) = S \setminus \cap_{j \in J} S\), by Step 1, \(= S \setminus \emptyset\), by the supposition, \(= S\).
