# If R is a symmetric and transitive relation on the non-empty set A, then R is reflexive.

I am extremely confused about a question I have been trying to find the answer to for a while now. The full question is:

Use $A=\{1,2\}$ to give an example which shows that the statement "If R is a symmetric and transitive relation on the non-empty set A, then R is reflexive." is false.

My answer was the following but after a bit of research i seem to be finding conflicting answers:

Suppose the $R=\{(1,2),(2,1),(1,1)\}$ on the set $A$. $R$ is symmetric because $(1,2) \Rightarrow (2,1)$ and $(2,1)\Rightarrow (1,2)$ which meets the definition of symmetric. $R$ is also transitive because $(x,y)\land (y,z) \Rightarrow (x,z)$. $R$ is not reflexive on $A$ because for any $a\in A$, $(a,a)\in R$ does not exist. This proves the statement false.

I've been told that $R=\{(1,2),(2,1),(1,1)\}$ is not transitive because it does not include $(2,2)$. Is this true? If it is true then what other possible counter examples could there be?

• You claim $R$ is transitive, but did you show it? Consider $x = z = 2$ and $y=1$. – mlg4080 Feb 23 '16 at 3:58

Clearly $R= \{(1,2),(2,1),(1,1)\}$ is not transitive since $(2,1) \in R$, $(1,2)\in R$ but $(2,2) \notin R$.
But if you consider $R=\{(1,1)\}$, then clearly $R$ is symmetric and transitive (vacuously true) but not reflexive.(since $(2,2)\notin R$).
$R$ is not transitive. (2,1) and (1,2) implies that (2,2) is in the set, supposing transitivity. Also, in your second paragraph, it is not for any a, it is for a specific $a$ that $(a,a) \not \in R$.
Suggestion: take the symmetric and transitive relationship (1,1), which is not a reflexive relationship on $\{1,2\}$. Or even take the relation $\emptyset$ !