2022-07-17

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

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

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

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 for any set, the union of the complements of any possibly uncountable number of subsets is the complement of the intersection of the subsets.

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: Structured Description


Here is the rules of Structured Description.

Entities:
\(S\): \(\in \{\text{ the sets }\}\)
\(\{S_\beta \subseteq S \vert \beta \in B\}\): \(B \in \{\text{ the possibly uncountable index sets }\}\)
//

Statements:
\(\cup_{\beta \in B} (S \setminus S_\beta) = S \setminus \cap_{\beta \in B} S_\beta\)
//


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, \(p \in \cup_{\beta \in B} (S \setminus S_\beta)\).


References


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