0
$\begingroup$

Let $R$ be a symmetric, transitive relation. If $(x, y) \in R$ then the symmetric property implies that $(y, x) \in R$. Using the the transitive property upon $(x, y)$ and $(y, x)$ we can conclude $(x, x) \in R$. Is this fair logic or is it flawed?

$\endgroup$
  • 1
    $\begingroup$ Welcome to MSE. Please use descriptive titles. After reading the title of this question, users have no idea of which topic it belongs. $\endgroup$ – jjagmath Oct 14 at 17:09
  • $\begingroup$ This is a common confusion. Showing $A \implies B$ does not mean you have shown $B$. You still need $A$ FIRST "to get at" $B$. In your case, you have IF $(x, y) \in R$, THEN $(x, x) \in R$. But you need $(x, y) \in R$ to begin with. $\endgroup$ – 0XLR Oct 14 at 18:05
  • $\begingroup$ This question is addressed in many posts already on the site: (math.stackexchange.com/questions/440/…, math.stackexchange.com/questions/3802279/…, math.stackexchange.com/questions/2106732/…, etc.) Also, $\mathbb{R}$ is a standard symbol for the set of real numbers, so in general it would be better to use something else in this context. $\endgroup$ – halrankard2 Oct 14 at 18:05
3
$\begingroup$

You may conclude that for all $x$, if $(x, y)\in R$ for some $y$ then $(x,x)\in R$. That's not the same that proving that $(x,x) \in R$ for all $x$.

| cite | improve this answer | |
$\endgroup$