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

Question

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.

Answer 1

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 $R_1 = R_2$.

Answer 2

Let $\Omega$ be a set. We say that a partition $P$ of $\Omega$ is compatible with an equivalence relation $R$ on $\Omega$ if for all $x, y \in \Omega$, $$ xRy\iff \exists S\in P \text{ such that } x \in S \text{ and } y \in S. $$ Claim: For any equivalence relation $R$ on $\Omega$, there exists exactly one partition $P$ of $\Omega$ that is compatible with $R$.

If I understand correctly, you are asking about the proof that there is at most one $P$ compatible with $R$. Suppose $P$ and $P'$ are both compatible with $R$; we shall show that $P = P'$.

Consider an arbitrary $S \in P$. Since $P$ is a partition, $S$ is nonempty; say $x \in S$. Since $P'$ is a partition, there is some set $S' \in P'$ such that $x \in S'$. For any $y \in S$, since $P$ is compatible with $R$, we have $x R y$. Since $P'$ is compatible with $R$, this implies that $y \in S'$. Therefore, $S \subseteq S'$. Similarly, $S' \subseteq S$, i.e., $S = S'$. Thus, $S \in P'$. Since $S$ was an arbitrary element of $P$, this shows that $P \subseteq P'$. By symmetry, we also have $P' \subseteq P$, i.e., $P = P'$.

Answer 3

Let $\sim$ and $\sim'$ be equivalence relations on $S\ne\emptyset.$ For $x\in S$ let $[x]_{\sim}=\{y\in S:y\sim x\}$ and let $[x]_{\sim'}=\{y\in S:y\sim' x\}.$ Let $P=\{[x]_{\sim}:x\in S\}$ and let $P'=\{[x]_{\sim'}:x\in S\}.$

Now for any $x\in S,$ the only member of $P$ that contains $x$ (as a member) is $[x]_{\sim}$ and the only member of $P'$ that contains $x$ is $[x]_{\sim'}\,.$

So if $P=P'$ then $\forall x\in S\,([x]_{\sim}=[x]_{\sim'}).$ That is, if $P=P'$ then $\forall x\in S\,(\{y\in S:y\sim x\}=\{y\in S:y\sim' x\}).$

So if $P=P'$ then $\sim$ and $\sim'$ are the same thing.

Equivalently, if $\sim$ and $\sim'$ are not equal then $P\ne P'.$