I'm currently at the beginning of studying Set Theory. We've studied the basic concepts among them- the equivalence relation. The question:
Assuming we have a non empty set $A$ and an equivalence relation $E$ on $A$.
The definition of a relation is a set of ordered pairs so the relation is a subset of $A\times A$. By definition of subset, the relation's members must be in $A\times A$ but not every member of $A\times A$ must be in the relation.
So if we go back to the specific example: not every ordered pair of $A\times A$ must be part of the relation group. This rule, for my understanding is applying to the equivalence relation $E$: there's a member of $A\times A$, called $(x,y)$, which is not a member of the relation $E$.
If we go along- $(x,y)$ doesn't have an equivalence class- because it is not part of the relation. Therefore not all members of $A\times A$ are part of an equivalence class.
I guess that i'm wrong somewhere on the way and i'm missing something. what is it?
Thank you in advance, Yaron
Recall that $E$ is a relation on $A$, rather than on $A\times A$. This means that we are concerned with relations between two elements of $A$.
While you're absolutely correct, and while the equivalence relation $E$ could be a proper subset of $A\times A$, the equivalence classes—or the partition they induce—are subsets of $A$.
So stating that $(x,y)\notin E$ will simply tell you that the equivalence classes of $x$ and $y$ are disjoint.
But to answer your main concern about covering, if $E$ is an equivalence relation on $A$, then $E$ is reflexive (on $A$), and therefore for every $a\in A$, $(a,a)\in E$. So indeed there will be an equivalence class which includes $a$.