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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?

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  • $\begingroup$ Is $(3,3)$ in it? Or $(2,2)$? This also shows it's not transitive. $\endgroup$
    – user223391
    Jul 5, 2016 at 16:32
  • $\begingroup$ yes its not there so it is not reflexive...that is my confusion $\endgroup$
    – Umesh shankar
    Jul 5, 2016 at 16:33
  • $\begingroup$ What's your confusion? It's not reflexive, you're right. It's also not transitive. $\endgroup$
    – user223391
    Jul 5, 2016 at 16:35
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    $\begingroup$ You may find this question of interest $\endgroup$
    – MPW
    Jul 5, 2016 at 16:35
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    $\begingroup$ What IS your question? $\endgroup$
    – user223391
    Jul 5, 2016 at 16:39

3 Answers 3

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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 $A = \{1,2,3\}$.

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  • $\begingroup$ ok it is pretty clear now $\endgroup$
    – Umesh shankar
    Jul 5, 2016 at 16:41
  • $\begingroup$ so finally does it mean that if a Relation R from Set $A$ to $A$ is symmetric and transitive then it is reflexive if $Dom(R)=A$ $\endgroup$
    – Umesh shankar
    Jul 5, 2016 at 16:52
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    $\begingroup$ @Umeshshankar Yes, exactly. $\endgroup$
    – BrianO
    Jul 5, 2016 at 19:27
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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 $3 \sim b$.

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    $\begingroup$ A small quibble: the argument doesn't require that $b \ne a$, just that some $b\sim a$ exists at all. $\endgroup$
    – Paul Sinclair
    Jul 5, 2016 at 17:01
  • $\begingroup$ Whoops, thank you for pointing that out! I suppose I was thinking there would be no need to prove reflexivity from symmetry and transitivity if it is already known that $ a \sim a$, but it is a mistake on my part. Edited and corrected. $\endgroup$
    – Christian
    Jul 6, 2016 at 3:29
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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 $(3,3)\not\in R$

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