(a) Property $2$ of an equivalence relation states that if $a\sim b$ then $b\sim a$. property $3$ states that if $a\sim b$ and $b\sim c$ then $a\sim c$. What is wrong with the following proof that properties $2$ and $3$ imply property $1$? Let $a\sim b$; then $b\sim a$, whence, by property $3$ (using $a=c$), $a\sim a$.
(b) Can you suggest an alternative of property $1$ which will insure us that properties $2$ and $3$ do imply property $1$?
My solution: Let $X=\{0,1,2\}$. Let's consider relation $$R=\{(0,1),(1,0),(0,2),(2,0),(1,2),(2,1)\}\subset X\times X$$ We can check that this relation is symmetric and transitive. Since $(0,1)\in R$ and $(1,0)\in R$ then $(0,0)\in R$. However, we see that it is not true. What is the problem? Can anyone explain it in detail?
The issue is property (2) states that for all $a,b \in X$ we have that $a\sim b$ if and only if $b \sim a$. This is $\textit{not}$ saying that $a \sim b$ and $b \sim a$ for all $a,b \in X$.
So, if you were to let $b = a$. Property (2) simply implies that $a \sim a$ if and only if $a \sim a$, a statement that is true regardless of whether or not $a$ really is related to $a$. And property (3) does nothing to help us out of this predicament.