Can a partition of a set be constructed from relations that are NOT equivalence relations?

Question

Is it true that only kind of relations that can completely and uniquely define a partition in a set is the equivalence relation ?

I'm having a hard problem believing that.

Given an original set $A = \{a,b,c,d\}$ and certain partition $\{P_1,P_2\}$ where $P_1 = \{a,b,d\}$ and $P_2 = \{c\}$ I could introduce a relation (not equivalence relation) $A \rightarrow A = \{ (a,d),(a,b),(b,d),(c,c) \}$ and from that relation I could still uniquely partition the original set $A$ into $\{P_1,P_2\}$. How? Well the informal process of building subsets out of the ordered pairs, one-by-one.

First $(a,d)$ implies a subset that contains $a,d$.
then $(a,b)$ implies a subset that contains $a,b,d$ (by transitive property).
then $(b,d)$ adds no information
then $(c,c)$ implies a subset containing $c$.
The unique result is the partition $\{a,b,d\}, \{c\}$.

Am I doing something wrong or forgetting something crucial?

Answer 1

It all depends on what exactly you mean by a relation "completely and uniquely defining a partition". In one sense, a partition and an equivalence relation are basically the same thing. That doesn't mean that you can't use other types of relations to define partitions; but it does mean that in doing so you necessarily also define an equivalence relation.

In your example, it seems that you intend to define a partition using the reflexive transitive symmetric closure of a relation $R$. This is the intersection of all equivalence relations containing $R$, or equivalently the unique minimal equivalence relation containing $R$. By defining an equivalence relation, you are at the same time defining a partition.

Answer 2

Suppose you have a partition of $A$. Then the natural equivalence relation that corresponds to it is that $p$ is related to $q$ whenever $q$ is in the same part as $p$.

What you are doing is taking a non-equivalence relation and then constructing the "equivalence closure", which is the smallest equivalence relation that contains it.

Your partition, for example, has $d$ in the same part as $b$, even though the original relation omits the pair $(d, b)$. The relation says that the relationship between $d$ and $b$ is asymmetric: It has $(b,d)$ but not $(d,b)$, so $d$ is related to $b$ but $b$ is not related to $d$. But your partition doesn't capture this asymmetry.

Similarly, your partition has $a$ in the same part as itself, but your original relation says that $a$ should not be related to itself, because your original relation does not include $(a,a)$. So your partition fails to capture this aspect of the relation.

So the partition that you've constructed is derived from the original relation, but not in the special way that we mean when we say that every partition corresponds to an equivalence relation in a special way.