A description/proof of that disjoint union of complements is disjoint union of whole sets minus disjoint union of subsets
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of set.
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.
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, \(\{S_\alpha\}\), and any subsets, \(S'_\alpha \subseteq S_\alpha\), \(\coprod_\alpha (S_\alpha \setminus S'_\alpha) = \coprod_\alpha S_\alpha \setminus \coprod_\alpha S'_\alpha\).
2: Proof
For any \(p \in \coprod_\alpha (S_\alpha \setminus S'_\alpha)\), \(p \in S_\alpha \setminus S'_\alpha\) for an \(\alpha\), \(p \notin S'_\alpha\) for each \(\alpha\), \(p \notin \coprod_\alpha S'_\alpha\), so, \(p \in \coprod_\alpha S_\alpha \setminus \coprod_\alpha S'_\alpha\); for any \(p \in \coprod_\alpha S_\alpha \setminus \coprod_\alpha S'_\alpha\), \(p \notin \coprod_\alpha S'_\alpha\), \(p \notin S'_\alpha\) for each \(\alpha\), \(p \in S_\alpha \setminus S'_\alpha\) for an \(\alpha\), so, \(p \in \coprod_\alpha (S_\alpha \setminus S'_\alpha)\).
