2023-03-26

246: Complement of Product of Subsets Is Union of Products of Whole Sets 1 of Which Is Replaced with Complement of Subset

<The previous article in this series | The table of contents of this series | The next article in this series>

A description/proof of that complement of product of subsets is union of products of whole sets 1 of which is replaced with complement of subset

Topics


About: set

The table of contents of this article


Starting Context



Target Context


  • The reader will have a description and a proof of the proposition that the complement of the product of any subsets is the union of the products of the whole sets, 1 of which is replaced with the complement of a subset for each product.

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.


Main Body


1: Description


For any sets, S1,S2,...,Sn, and any subsets, SiSi, (S1×S2×...×Sn)(S1×S2×...×Sn)=((S1S1)×S2×...×Sn)(S1×(S2S2)×...×Sn)...(S1×S2×...×(SnSn)).


2: Proof


For any p=(p1,p2,...,pn)(S1×S2×...×Sn)(S1×S2×...×Sn), pS1×S2×...×Sn, piSi for an i, piSiSi for an i, pS1×...×(SiSi)×...×Sn for an i, p((S1S1)×S2×...×Sn)(S1×(S2S2)×...×Sn)...(S1×S2×...×(SnSn))\); for any p((S1S1)×S2×...×Sn)(S1×(S2S2)×...×Sn)...(S1×S2×...×(SnSn)), pS1×...×(SiSi)×...×Sn for an i, piSiSi for an i, piSi for an i, pS1×S2×...×Sn, p(S1×S2×...×Sn)(S1×S2×...×Sn)).


References


<The previous article in this series | The table of contents of this series | The next article in this series>