Equivalence relations and partially ordered sets

Question

An equivalence relation is a binary relation that fulfills reflexivity, symmetry and transitivity.

A partially ordered set is a set equipped with a binary relation that fulfills reflexivity, antisymmetry and transitivity.

For a partially ordered set, some elements of the set might be incomparable, i.e. $x\leq y$ or $y \leq x$ may both not hold. That's why we use the term"partially". An example would be $\subset$ as a binary relation and the set $M:= \{\{1\}, \{2\}, \{1, 2\}\}$. Obviously, $\{1\}$ and $\{2\}$ are incomparable. (The example is taken from here.)

Question: Can also elements of a set, equipped with an equivalence relation, be incomparable?

Answer 1

The word "incomparable" in the statement of your question is a bit misleading. Technically speaking, the question does not make sense because for an equivalence relation the notion of "comparable" is not defined. Of course, you may wish to say that if two elements of the set are related by the given equivalence relation, then they are "comparable", but that does not do justice to the notion of equivalence relation. Moreover, if you choose to do so, then the answer to the question is in general 'yes', because if you have more than one equivalence class you already have "incomparable" (i.e. non-related) elements.

It is best to think about an equivalence relation as a tool to identify elements of a set, whereas an order relation, partial or linear, is a tool to create a hierarchy. Identification and Hierarchy are two essentially different notions.

Answer 2

For every non-trivial equivalence relation (that is, more than one equivalence class exists), there exist two incomparable elements.

For instance, if you equip $\mathbb Z$ with the equivalence relation $x\sim y:\!\iff x-y\text{ is even}$, $0$ and $1$ are incomparable: neither $x\sim y$, nor $y\sim x$.