A relation that is both an equivalence relation and a function. [closed]

The question is: "Suppose R on A is both an equivalence relation on A and a function from A to A. What is R?" I'm not entirely sure where to start, I'm not even sure I understand what a function from A to A represents, can anybody help with this?

Solution 1:

Recall the following definitions for a relation $R$ on a set $A$ (hence $R \subseteq A \times A$). These are adjectives which describe the kind of relation $R$ is.

• Reflexive: For each $a \in A$, $(a,a) \in R$
• Symmetric: Whenever $(a,b) \in R$, then $(b,a) \in R$
• Transitive: Whenever $(a,b),(b,c) \in R$, then $(a,c) \in R$
• Functional/Right-Unique: If $(x,y),(x,z) \in R$, then $y=z$
  • In other words, if $a$ is related to any other two elementsin $A$, then those must be the same element.
  • Or put differently, any $a$ must be related to either zero or one elements.
  • A totality axiom (i.e. that each $a \in A$ is related to one thing) may be required to match the "typical" definition of a function, as usually given.
• Equivalence: Reflexivity, symmetry, and transitivity all apply

So, consider a relation $R$ which is a right-unique equivalence relation.

• From reflexivity, we know $\{ (a,a) \mid a \in A \} \subseteq R$
• From right-uniqueness, any $a \in A$ can only be related to one element. That is, $a$ can be in the left-coordinate of at most one pair in the relation.

In fact, you don't even need symmetry or transitivity to see what results.