# A simple conceptual doubt related to sets and relations

I have a doubt in a Types of Relations on a Set

Suppose we have a $A = \{1,2,3\}$ and we define some relations as follows:-

$R_1 = \{(1,1),(2,2),(3,3)\}$

$R_2 = \{(1,1),(2,2)\}$

$R_3 = \{(1,2),(1,3),(2,3),(2,1),(3,1),(3,2)\}$

$R_4 = \{(1,2),(2,1)\}$

$R_5 = \{(1,1),(2,2),(3,3),(1,2),(1,3),(2,3),(2,1),(3,1),(3,2)\}$

$R_6 = \{(1,2),(2,3),(1,3)\}$

Now, I know that $R_1$ will be Reflective, $R_3$ will be Symmetric and $R_5$ will be Transitive. But will $R_2,R_4$ and $R_6$ be Reflective, Symmetric and Transitive respectively?

In short: Is it necessary for a Relation to have all the elements of the Set covered, so that it can be any one of those?

Also, will $R_7 = \{(1,1),(2,2),(3,3),(1,2),(2,1)\}$ be an equivalent relation?

Thank You

About reflexivity: a relation on set $A$ is reflective iff it contains the diagonal $\triangle_A:=\{\langle a,a\rangle\mid a\in A\}$ as a subset.

In your question we have $\triangle_A=R_1$ and evidently $R_2,R_4,R_6$ are not reflexive (do not contain $R_1$ as a subset).

A reflexive relation "covers all elements of $A$".

For being symmetric or transitive it is not necessary to "cover all elements of $A$." In fact the empty relation is transitive and symmetric.

To find out whether $R_7$ is an equivalence relation, just check whether it is reflexive, symmetric and transitive. You can do that...

• thanks, now I got it. So in this way, R7 will be an equivalence relation. – Cheapstrike Aug 16 '16 at 13:35
• You are welcome. – drhab Aug 16 '16 at 13:38