# Fake proof, symmetric and transitive relation is already reflexive [duplicate]

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?

• Welcome to MSE. Please use descriptive titles. After reading the title of this question, users have no idea of which topic it belongs. – jjagmath Oct 14 '20 at 17:09
• 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. – 0XLR Oct 14 '20 at 18:05
• 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. – halrankard2 Oct 14 '20 at 18:05

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 $$x$$.