Binary relations: transitivity and symmetry

Question

I've been looking at some examples for transitivity and symmetry.

Suppose $A=\{0,1,2 \} $ and the relation $R=\{ (0,0),(1,1),(2,2),(1,2),(2,1) \}$

Well for starters this is clearly reflixe since $\forall x \in A ,xRx$.

As for symmetry, we define it as : $\forall x,y \in A, xRy \rightarrow yRx$. However, when looking at the relation above, well $(1,2) \rightarrow (2,1)$ is there, but what about zero? Aren't you suppose to have a relation for zero as well? What about if $x=y$, does symmetry hold? For example: $(0,0) \rightarrow (0,0)$. Does symmetry hold for when both $x$ and $y$ are the same values?

For transitivity, it needs to satisfy: $\forall x,y,z, (xRy \land yRz)\rightarrow xRz$. Well in the above case, zero is not related to anything...

The reason I'm questioning this is that the question itself claims that this relation is reflexive (yes), symmetric and transitive and therefore an equivalence relation. Though the confusion remains as to why the transitivity is there and why the symmetry is there.

Answer 1

$R$ is indeed both symmetric and transitive. These properties are a bit different in character from reflexivity. Reflexivity requires certain specific ordered pairs to be in the relation; these don’t. These are only conditional requirements: if some ordered pairs are present, then certain others must also be present.

Symmetry says that if $\langle x,y\rangle\in R$, then $\langle y,x\rangle\in R$. Can you find any $x,y\in A$ such that $\langle x,y\rangle\in R$, but $\langle y,x\rangle\notin R$? No, so $R$ is symmetric. Symmetry of $R$ just says that if $x$ and $y$ are distinct elements of $A$, then either both of the pairs $\langle x,y\rangle$ and $\langle y,x\rangle$ belong to $R$, or neither of them belongs to $R$. In this case $A=\{0,1,2\}$, so the only pairs of elements to be considered are $0$ and $1$, $0$ and $2$, and $1$ and $2$.

• $0$ and $1$: Neither $\langle 0,1\rangle$ nor $\langle 1,0\rangle$ is in $R$; this is fine.
• $0$ and $2$: Neither $\langle 0,2\rangle$ nor $\langle 2,0\rangle$ is in $R$; this is fine.
• $1$ and $2$: Both $\langle 1,2\rangle$ and $\langle 2,1\rangle$ are in $R$; this is fine.

Conclusion: $R$ is symmetric.

Transitivity says that if $\langle x,y\rangle\in R$ and $\langle y,z\rangle\in R$, then $\langle x,z\rangle\in R$. it’s a conditional requirement: if certain ordered pairs are in $R$, then certain other ordered pairs have to be in $R$ as well. If the conditions of the if part aren’t met, it doesn’t impose any requirement. In the particular case at hand we have $\langle x,y\rangle\in R$ and $\langle y,z\rangle\in R$ when:

• $x=y=z=0$;
• $x=y=z=1$;
• $x=y=z=2$;
• $x=1,y=2$, and $z=1$; and
• $x=2,y=1$, and $z=2$.

For any other choice of values for $x,y$, and $z$, the condition that $\langle x,y\rangle\in R$ and $\langle y,z\rangle\in R$ is not satisfied, so transitivity says nothing about those choices. And in every one of these five cases the ordered pair $\langle x,z\rangle$ is in $R$, as required for transitivity, so $R$ is transitive.