Equivalence Relation saturating a subset

Question

Following this question, can someone help me understand (even better with some basic visualization) the meaning of "saturation" when it comes to partition and its counterpart (equivalence relation)? Does saturation mean that there is a union of equivalence classes that make up the entire set S or just part of it?

Answer 1

Let $X$ be a set and let $R$ be an equivalence relation on $X$. Then, $X$ can be partitioned into pairwise disjoint equivalence classes $\{E_\alpha\}_{\alpha \in I}$ where $I$ is some index set. Now, every $E_\alpha$ is by definition a subset of $X$. So, in symbolic terms, we have $$ X = \bigcup_{\alpha \in I} E_\alpha $$ and $E_\alpha \cap E_\beta = \varnothing$ whenever $\alpha \neq \beta$.

Consider an arbitrary (for now) subset $A$ of $X$. Because $A \subseteq X$, it is clear that $$ A \subseteq \bigcup_{\alpha \in I} E_\alpha. $$ However, $A$ might not be the union of equivalence classes, i.e. it may be impossible to write $$ A = \bigcup_{\alpha \in J} E_\alpha $$ for any subset $J$ of $I$. This is because an equivalence class $E_\alpha$ containing a point $a \in A$ may also contain points from $X \setminus A$.

Next let us assume that $A$ is saturated. The definition of saturation prevents the above from occurring. Namely, it states that $A$ can be expressed exactly as the union of some of these equivalence classes. Put otherwise, the saturation of $A$ ensures that for each $\alpha \in I$, \begin{equation}\label{eq:star}\tag{$\star$} A \cap E_\alpha \neq \varnothing \implies E_\alpha \subseteq A. \end{equation} Therefore, if $A$ is saturated, then an equivalence class of $X$ either contains only points from $A$, or none at all.