582: Intersection of Union of Subsets and Subset Is Union of Intersections of Each of Subsets and Latter Subset

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description/proof of that intersection of union of subsets and subset is union of intersections of each of subsets and latter subset

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Topics

About: set

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Proof

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Starting Context

• The reader knows a definition of set.

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Target Context

• The reader will have a description and a proof of the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$S^{\prime }$: $\in \{\text{ the sets }\}$
$A$: $\in \{\text{ the possibly uncountable index sets }\}$
$\{S_{\alpha }\}$: $\alpha \in A$, $S_{\alpha }\subseteq S^{\prime }$
$S$: $S\subseteq S^{\prime }$
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Statements:
$({\cup }_{\alpha \in A}{S}_{\alpha })\cap S={\cup }_{\alpha \in A}(S_{\alpha }\cap S)$.
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2: Natural Language Description

For any set, $S^{\prime }$, any possibly uncountable number of subsets, $\{S_{\alpha }\subseteq S^{\prime }|\alpha \in A\}$, where $A$ is any possibly uncountable index set, and any subset, $S\subseteq S^{\prime }$, $({\cup }_{\alpha \in A}{S}_{\alpha })\cap S={\cup }_{\alpha \in A}(S_{\alpha }\cap S)$.

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3: Proof

For any $p\in ({\cup }_{\alpha \in A}{S}_{\alpha })\cap S$, $p\in S_{\alpha }$ for an $\alpha$ and $p\in S$. $p\in S_{\alpha }\cap S$ for an $\alpha$. $p\in {\cup }_{\alpha \in A}(S_{\alpha }\cap S)$.

For any $p\in {\cup }_{\alpha \in A}(S_{\alpha }\cap S)$, $p\in S_{\alpha }\cap S$ for an $\alpha$. $p\in S_{\alpha }$ for an $\alpha$ and $p\in S$. $p\in {\cup }_{\alpha \in A}{S}_{\alpha }$ and $p\in S$. $p\in ({\cup }_{\alpha \in A}{S}_{\alpha })\cap S$.

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References

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