Proof that there is exactly one equivalence relation that forms a partition

elementary-set-theoryequivalence-relationsset-partition

So I've just started trying to teach myself some topology and i cant quite grasp how an equivalence class forms a partition more specifically i don't understand the proof that there is exactly one equivalence relation that forms the partition.

For example in my book it says that a partition $D$ is formed from an equivalence relation $R$ and that equivalence relation is unique that is there is only one equivalence relation that forms the partition however they do not give a proof of this i myself cant come up with one so is there a proof of this statement?
Thanks in advance.

Best Answer

Suppose $R_1$ and $R_2$ on a set S induce the same partition of S but are not identical.

Since they are not identical there is ( at lest) one couple $(a,b)\in S\times S$ that belongs to one of them but does not belong to the other. [ Relations are sets - of couples - and two different sets must differ by at least one element, by the extensonnality principle]

So, one relation will produce a partition having as "cell" an equivalence class with $a$ and $b$ as elements; while the other relation will not produce such a partition.

The reason is that in the other relation there is no couple $( a,b)$ , meaning that there is no equivalence class having $a$ and $b$ as members. Since the partition is a set of equivalence classes, and since the two partitions differ by ( at least) one equivalence class, they are not identical.

The fact that the partitions induced are not identical contradicts our assumption.

So $R_1 = R_2$.