I know that when handed an equivalence relation it is possible to deifne equilence classes which are disjoint sets that form a partitoin of the original set $A$. My question is, is using this fact to determine whether or not a relation is an equivalence one a good idea? For example, say I'm given a set $A = \{0, 1, 2, 3, 4, 5\}$ And a relation $R = \{(0,0), (1,1), (1,2),(2,2), (3,3),(4,4) (5,5), (2,1), (2,3), (1,3),(3,1),(3,2)\}$ And I'm asked to determine if it's an equivalence relation. Since it's kind of tricky to search for the transitive property when there are so many elements, let's instead find the set: $\{[x] | x \in A\}$ where $[x] =\{y \in A | (x,y) \in R\}$. If this set turns out to be a partition of $A$ can I then safely conlcude that the relaiton $R$ is indeed an equivalence one? Or is it possible for a non equivalence relation to form a partition the way I stated before? Thanks for your help!
Also in case you're wondering, this is what the set would loook like for the above relation:
$$
[0] = {0}
[1] = {1, 2, 3}
[2] = {2, 1, 3}
[3] = {3, 1, 2}
[4] = {4}
[5] = {5}
$$
So I get a partition of the set A, so according to my theory then the orignal set must be an equivalence relation. Is this correct? Can a regular relation produce a partition as well using the described mehtod above? Thanks again.