description/proof of that for indexed set of sets and disjointed union of indexed set, intersection by disjointed union is intersection by index after intersections by elements of indexed set
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of disjointed union of indexed set of sets.
Target Context
- The reader will have a description and a proof of the proposition that for any indexed set of sets and the disjointed union of the indexed set, the intersection by the disjointed union is the intersection by the index after the intersections by the elements of the indexed 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:
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{L_j \in \{\text{ the possibly uncountable index sets }\}\}_{j \in J}\): \(\in \{\text{ the indexed sets }\}\)
\(\{S_{j, l_j} \in \{\text{ the sets }\}\}_{(j, l_j) \in \cup_{j \in J} \{j\} \times L_j}\):
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Statements:
\(\cap_{(j, l_j) \in \cup_{j \in J} \{j\} \times L_j} S_{j, l_j} = \cap_{j \in J} \cap_{l_j \in L_j} S_{j, l_j}\)
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2: Proof
Whole Strategy: Step 1: see that for each \(p \in \cap_{(j, l_j) \in \cup_{j \in J} \{j\} \times L_j} S_{j, l_j}\), \(p \in \cap_{j \in J} \cap_{l_j \in L_j} S_{j, l_j}\); Step 2: see that for each \(p \in \cap_{j \in J} \cap_{l_j \in L_j} S_{j, l_j}\) , \(p \in \cap_{(j, l_j) \in \cup_{j \in J} \{j\} \times L_j} S_{j, l_j}\); Step 3: conclude the proposition.
Step 1:
Let \(p \in \cap_{(j, l_j) \in \cup_{j \in J} \{j\} \times L_j} S_{j, l_j}\) be any.
\(p \in S_{j, l_j}\) for each \((j, l_j) \in \cup_{j \in J} \{j\} \times L_j\).
\(p \in S_{j, l_j}\) for each \((j, l_j) \in \{j\} \times L_j\) for each \(j \in J\), because \((j, l_j) \in \cup_{j \in J} \{j\} \times L_j\).
So, \(p \in \cap_{l_j \in L_j} S_{j, l_j}\) for each \(j \in J\), because \(p \in S_{j, l_j}\) for each \((j, l_j) \in \{j\} \times L_j\), so, \(p \in S_{j, l_j}\) for each \(l_j \in L_j\).
So, \(p \in \cap_{j \in J} \cap_{l_j \in L_j} S_{j, l_j}\).
So, \(\cap_{(j, l_j) \in \cup_{j \in J} \{j\} \times L_j} S_{j, l_j} \subseteq \cap_{j \in J} \cap_{l_j \in L_j} S_{j, l_j}\).
Step 2:
Let \(p \in \cap_{j \in J} \cap_{l_j \in L_j} S_{j, l_j}\) be any.
\(p \in \cap_{l_j \in L_j} S_{j, l_j}\) for each \(j \in J\).
\(p \in S_{j, l_j}\) for each \(j \in J\) and each \(l_j \in L_j\).
So, \(p \in S_{j, l_j}\) for each \((j, l_j) \in \{j\} \times L_j\).
So, \(p \in S_{j, l_j}\) for each \((j, l_j) \in \cup_{j \in J} \{j\} \times L_j\).
So, \(p \in \cap_{(j, l_j) \in \cup_{j \in J} \{j\} \times L_j} S_{j, l_j}\).
So, \(\cap_{j \in J} \cap_{l_j \in L_j} S_{j, l_j} \subseteq \cap_{(j, l_j) \in \cup_{j \in J} \{j\} \times L_j} S_{j, l_j}\).
Step 3:
So, \(\cap_{(j, l_j) \in \cup_{j \in J} \{j\} \times L_j} S_{j, l_j} = \cap_{j \in J} \cap_{l_j \in L_j} S_{j, l_j}\).
