Properties of equivalence relation

Question

(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?

For a given $a$, neither property 2 nor property 3 guarantees that there is any $b$ such that $a\sim b$. — Daniel Fischer, Nov 02 '17 at 16:59
@DanielFischer, I cannot understand. I was interesting why for relation $R$ reflexivity does not hold. — RFZ, Nov 02 '17 at 17:28
Let $R$ a symmetric and transitive relation on $X$. Let $Y = { x \in X : (\exists y)(x R y)}$. Then $R$ is an equivalence relation on $Y$, but it can be that $Y \subsetneqq X$. And then, $R$ is not an equivalence relation on $X$. — Daniel Fischer, Nov 02 '17 at 17:31
It's not transitive, you have $(0,1) \in R$ and $(1,0) \in R$, but not $(0,0) \in R$. — Daniel Fischer, Nov 02 '17 at 17:41
Mm yes! Thanks a lot. But What is wrong with the above argument in proof? — RFZ, Nov 02 '17 at 17:41
You start with "Let $a\sim b$", but such a $b$ need not exist. — Daniel Fischer, Nov 02 '17 at 17:44
@DanielFischer, could you take a look at that problem? https://math.stackexchange.com/questions/2512308/0-1-cannot-be-partititioned-into-two-sets-with-given-properties — RFZ, Nov 09 '17 at 14:42

score 0 · Answer 1 · answered Nov 02 '17 at 17:03

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.

Properties of equivalence relation

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