# Form relation inside set $A = \{1,2,3\}$ so that $\text{R}:\text{A}\leftrightarrow \text{A}$

## Problem

Form relation inside set $$A = \{1,2,3\}$$ so that $$\text{R}:\text{A}\leftrightarrow \text{A}$$

## Attempt to solve

I know that $$\text{R}:\text{A}\leftrightarrow \text{A}$$ is true when relation is reflexive, symmetric and transitive. Which means when

$$\underbrace{(\forall x \in A : xRx)}_{\text{relfexive}} \wedge \underbrace{(\forall x,y \in A : xRy \implies yRx )}_{\text{symmetric}}\wedge \underbrace{(\forall x,y,z \in A : (xRy \wedge yRz \implies xRz))}_{\text{transitive}}$$ $$\implies \underbrace{\text{R}:\text{A}\leftrightarrow \text{A}}_{\text{equivalence relation}}$$

I could try to form binary relation explicitly $$\forall x \in A$$.

Example of symmetric binary relation:

$$R = \{ \langle 1,1 \rangle, \langle 2,2 \rangle, \langle 3,3 \rangle \}$$

Example of symmetric relation, which is also transitive

$$R = \{ \langle 1, 2 \rangle, \langle 2,3 \rangle \rangle, \langle 3 ,1 \rangle \}$$

So by combining these two i get relation that is equivalence relation?

$$R = \{ \langle 1,1 \rangle, \langle 2,2 \rangle, \langle 3,3 \rangle ,\langle 1, 2 \rangle, \langle 2,3 \rangle \rangle, \langle 3 ,1 \rangle \}$$ I'am not quite sure if this is right? If someone could clarify on what is going on since i don't think i quite understand this yet.

For example I don't understand why reflexive + symmetric + transitive would make relation equivalence relation?

Note that your relation is not an equivalence relation because if $$(1,2)\in R$$, then it must be true that $$(2,1)\in R$$ as well (similar for the other elements).
The easiest example of a non-empty equivalence relation over $$A$$ would be $$\{(1,1)\}$$; or the set containing all possible pairs in $$A\times A$$.