Is every relation which is transitive and symmetric also reflexive? I have seen a proof that every relation which is symmetric and transitive is also reflexive.
if $A=\{1,2,3\}$ Then if $R=\{(1,2)(2,1)(1,1)\color{blue}{(2,2)}\}$
here $R$ is symmetric and transitive on $A$ but not reflexive right?
Can anyone clear up this confusion for me?
 A: Foremost, this relation you defined is not transitive because of the following:
$2 \sim 1, 1 \sim 2$, but $2$ is not equivalent to $2$, which should be the case under transitivity.
Secondly, every relation that is symmetric and transitive is not necessarily reflexive. Generally the (false) proof proceeds as follows:
$a\sim b$, so then by symmetry $b\sim a$, then by transitivity, $a \sim a$. 
However, this argument is based on the fact that $\exists \, b $ such that $a\sim b$, which does not have to be the case. 
Consider the relation: $A = \{1, 2, 3\}, R = \{(1,2), (2, 1), (1, 1), (2, 2)\}$.
It is symmetric and transitive but not reflexive. Note that there is no such element $b$ where $3 \sim b$.
A: Every relation that is transitive and symmetric is reflexive on its domain, where the domain $dom(R)$ of a relation $R$ is
$$ dom(R) := \{x \mid \exists y\, xRy \}
$$
(and where, as usual, $xRy$ means $(x,y) \in R$).
This is easy to show: if $x\in dom(R)$, then $xRy$ for some $y$, so $yRx$ by symmetry, and then $xRx$ by transitivity.
The domain of the relation $R$ that you exhibit is just $\{1,2\}$, not all of $A = \{1,2,3\}$.
A: if you tack  $A=\{1,2,3\}$ and $R=\{(1,2),(1,1),(2,1),(2,2)\}$ then $R$ is both symetric and transitive relation on $A$ but not reflexive because $(3,3)\not\in R$
