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

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

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Target Context

• The reader will have a description and a proof of 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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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:
${\cup }_{\beta \in B}(S\setminus S_{\beta })=S\setminus {\cap }_{\beta \in B}{S}_{\beta }$
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2: Proof

Whole Strategy: Step 1: see that ${\cup }_{\beta \in B}(S\setminus S_{\beta })\subseteq S\setminus {\cap }_{\beta \in B}{S}_{\beta }$; Step 2: see that $S\setminus {\cap }_{\beta \in B}{S}_{\beta }\subseteq {\cup }_{\beta \in B}(S\setminus S_{\beta })$.

Step 1:

For any element, $p\in {\cup }_{\beta \in B}(S\setminus S_{\beta })$, $p\in S\setminus S_{\beta }$ for a $\beta$, so, $p\notin S_{\beta }$ for a $\beta$, so, $p\notin {\cap }_{\beta \in B}{S}_{\beta }$, so, $p\in S\setminus {\cap }_{\beta \in B}{S}_{\beta }$.

Step 2:

For any element, $p\in S\setminus {\cap }_{\beta \in B}{S}_{\beta }$, $p\notin {\cap }_{\beta \in B}{S}_{\beta }$, so, $p\notin S_{\beta }$ for a $\beta$, so, $p\in S\setminus S_{\beta }$ for a $\beta$, so, $p\in {\cup }_{\beta \in B}(S\setminus S_{\beta })$.

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References

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