906: Subset Minus Union of Subsets Is Intersection of 1st Subset Minus 2nd Chunk of Subsets

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description/proof of that subset minus union of subsets is intersection of 1st subset minus 2nd chunk 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 that for any set, any subset minus the union of any subsets is the intersection of the 1st subset minus the 2nd chunk of 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:
$S$: $\in \{\text{ the sets }\}$
$S_1$: $\subseteq S$
$\{S_{2,j}\subseteq S|j\in J\}$: $J\in \{\text{ the possibly uncountable index sets }\}$
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Statements:
$S_1\setminus {\cup }_{j\in J}{S}_{2,j}={\cap }_{j\in J}(S_1\setminus S_{2,j})$
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2: Proof

Whole Strategy: Step 1: see that $S_1\setminus {\cup }_{j\in J}{S}_{2,j}\subseteq {\cap }_{j\in J}(S_1\setminus S_{2,j})$; Step 2: see that ${\cap }_{j\in J}(S_1\setminus S_{2,j})\subseteq S_1\setminus {\cup }_{j\in J}{S}_{2,j}$.

Step 1:

Let $p\in S_1\setminus {\cup }_{j\in J}{S}_{2,j}$ be any.

$p\in S_1$. $p\notin {\cup }_{j\in J}{S}_{2,j}$, so, $p\notin S_{2,j}$ for each $j\in J$. So, $p\in S_1\setminus S_{2,j}$ for each $j$. So, $p\in {\cap }_{j\in J}(S_1\setminus S_{2,j})$.

Step 2:

Let $p\in {\cap }_{j\in J}(S_1\setminus S_{2,j})$ be any.

$p\in S_1\setminus S_{2,j}$ for each $j\in J$, so, $p\in S_1$ and $p\notin S_{2,j}$ for each $j$, so, $p\notin {\cup }_{j\in J}{S}_{2,j}$, so, $p\in S_1\setminus {\cup }_{j\in J}{S}_{2,j}$.

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References

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