241: Disjoint Union of Complements Is Disjoint Union of Whole Sets Minus Disjoint Union of Subsets

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A description/proof of that disjoint union of complements is disjoint union of whole sets minus disjoint union 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: 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 the disjoint union of any complements is the disjoint union of the whole sets minus the disjoint union 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: Description

For any sets, $\{S_{\alpha }\}$, and any subsets, $S_{\alpha }^{\prime }\subseteq S_{\alpha }$, $\coprod _{\alpha }(S_{\alpha }\setminus S_{\alpha }^{\prime })=\coprod _{\alpha }{S}_{\alpha }\setminus \coprod _{\alpha }{S}_{\alpha }^{\prime }$.

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2: Proof

For any $p\in \coprod _{\alpha }(S_{\alpha }\setminus S_{\alpha }^{\prime })$, $p\in S_{\alpha }\setminus S_{\alpha }^{\prime }$ for an $\alpha$, $p\notin S_{\alpha }^{\prime }$ for each $\alpha$, $p\notin \coprod _{\alpha }{S}_{\alpha }^{\prime }$, so, $p\in \coprod _{\alpha }{S}_{\alpha }\setminus \coprod _{\alpha }{S}_{\alpha }^{\prime }$; for any $p\in \coprod _{\alpha }{S}_{\alpha }\setminus \coprod _{\alpha }{S}_{\alpha }^{\prime }$, $p\notin \coprod _{\alpha }{S}_{\alpha }^{\prime }$, $p\notin S_{\alpha }^{\prime }$ for each $\alpha$, $p\in S_{\alpha }\setminus S_{\alpha }^{\prime }$ for an $\alpha$, so, $p\in \coprod _{\alpha }(S_{\alpha }\setminus S_{\alpha }^{\prime })$.

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References

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