Talk:Equivalence of Definitions of Saturated Set Under Equivalence Relation

An alternative proof of 2 implies 1 would go as follows:

Let $T = \displaystyle\bigcup_{u\in U} \left[\!\left[{u}\right]\!\right]$ with $U\subset S$.

By Image of Union under Mapping, $q(T) = \displaystyle\bigcup_{u\in U} q(\left[\!\left[{u}\right]\!\right])$.

By Preimage of Union under Mapping, $q^{-1}(q(T)) = \displaystyle\bigcup_{u\in U} q^{-1}(q(\left[\!\left[{u}\right]\!\right]))$.

It then requires something like the fact that equivalence classes are saturated. This could be proved here (which would clutter the page somewhat) or on a separate page. Disadvantage of the latter is that we're, strictly speaking, not allowed to talk about saturations when we're still defining it, so in order to not risk circular reasonings, I decided to put the argument here for who's interested. (Or in case someone wants to write out the details and add its proof here.) Either way, it is a consequence of the 2nd definition that an equivalence class is saturated. --barto (talk) 14:58, 16 March 2017 (EDT)