A description/proof of that union of indexed subsets minus union of subsets indexed with same indices set is contained in union of subset minus subset for each index
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 for any set and any possibly uncountable indices set, the union of any subsets (the 1st set of subsets) indexed with the indices set minus the union of any subsets (the 2nd set of subsets) indexed with the indices set is contained in the union of the subset from the 1st set minus the subset from the 2nd set for each index.
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 set, \(S\), any possibly uncountable indices set, \(A\), and any sets of subsets, \(\{S_\alpha\vert \alpha \in A\}\) and \(\{S'_\alpha\vert \alpha \in A\}\), \(\cup_{\alpha \in A} S_\alpha \setminus \cup_{\alpha \in A} S'_\alpha \subseteq \cup_{\alpha \in A} (S_\alpha \setminus S'_\alpha)\).
2: Proof
For any element, \(p \in \cup_{\alpha \in A} S_\alpha \setminus \cup_{\alpha \in A} S'_\alpha\), \(p \in S_\alpha\) for an \(\alpha\) and \(p \notin S'_\alpha\) for each \(\alpha\). \(p \in S_\alpha \setminus S'_\alpha\) for an \(\alpha\), \(p \in \cup_{\alpha \in A} (S_\alpha \setminus S'_\alpha)\).
3: Note
Equality does not hold in general, because for any \(p \in \cup_{\alpha \in A} (S_\alpha \setminus S'_\alpha)\), \(p \in S_\alpha \setminus S'_\alpha\) for an \(\alpha\), \(p \in S_\alpha\) and \(p \notin S'_\alpha\) for the \(\alpha\), but if \(p \in S'_\beta\) for a \(\beta \neq \alpha\), \(p \notin \cup_{\alpha \in A} S_\alpha \setminus \cup_{\alpha \in A} S'_\alpha\).
