905: Union of Subsets Is Complement of Intersection of Complements of Subsets

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description/proof of that union of subsets is complement of intersection of complements of subsets

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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: Proof

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Starting Context

• The reader knows a definition of set.
• The reader admits 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.

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Target Context

• The reader will have a description and a proof of the proposition for any set, the union of any possibly uncountable number of subsets is the complement of the intersection of the complements of the subsets.

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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:
$S$: $\in \{\text{ the sets }\}$
$\{S_{\beta }\subseteq S|\beta \in B\}$: $B\in \{\text{ the possibly uncountable index sets }\}$
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Statements:
${\cup }_{\beta \in B}{S}_{\beta }=S\setminus {\cap }_{\beta \in B}(S\setminus S_{\beta })$
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2: Proof

Whole Strategy: Step 1: take $S\setminus S_{\beta }$ as $S_{\beta }$ in 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 1:

$S\setminus (S\setminus S_{\beta })=S_{\beta }$.

As $S_{\beta }$ in 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, let us take $S\setminus S_{\beta }$, which makes ${\cup }_{\beta \in B}(S\setminus (S\setminus S_{\beta }))=S\setminus {\cap }_{\beta \in B}(S\setminus S_{\beta })$, but ${\cup }_{\beta \in B}(S\setminus (S\setminus S_{\beta }))={\cup }_{\beta \in B}{S}_{\beta }$, so, ${\cup }_{\beta \in B}{S}_{\beta }=S\setminus {\cap }_{\beta \in B}(S\setminus S_{\beta })$.

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References

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