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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.

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$R$ 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.

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  • $\begingroup$ So because zero does not appear in any other form other than being related to itself, that means that it's "confined" only to $(0,0)$. But if the relation contained $(1,0)$ then the doors would open for both symmetry and transitivity to be tested? $\endgroup$ – Dimitri Nov 25 '13 at 6:07
  • $\begingroup$ @Dimitri: Symmetry and transitivity must be tested even for the relation that you have; it’s just that the tests are passed rather easily. But I think that you have the right idea, yes: if $\langle 1,0\rangle$ were in the relation, then symmetry would require that $\langle 0,1\rangle$ also be present, and since we already have $\langle 2,1\rangle$, transitivity would require that we also have $\langle 2,0\rangle$. $\endgroup$ – Brian M. Scott Nov 25 '13 at 6:09
  • $\begingroup$ I think I'm understanding a bit. What about if the relation was just $(0,0)$ then? This brings back if $x$ and $y$ were equal. In such case transitivity cannot be tested because we don't have a $z$ for which we can test this. But does it hold for symmetry? Must there be at least 2 ordered pairs to determine this? $\endgroup$ – Dimitri Nov 25 '13 at 6:14
  • $\begingroup$ @Dimitri: Transitivity can be tested, and it passes: in that case the only $x,y,z$ such that $\langle x,y\rangle$ and $\langle y,z\rangle$ are $x=y=0$ and $y=z=0$, in which case the pair $\langle x,z\rangle$ is also $\langle 0,0\rangle$, which is in the relation. You also have symmetry: the only $x,y$ such that $\langle x,y\rangle$ is in the relation is $x=y=0$, in which case $\langle y,x\rangle$ is also in the relation, because it’s the same ordered pair: both $\langle x,y\rangle$ and $\langle y,x\rangle$ are $\langle 0,0\rangle$. $\endgroup$ – Brian M. Scott Nov 25 '13 at 6:21
  • $\begingroup$ Great. I think I'm getting the idea here. The $\forall$ portion of the definition confused me a little since I'm trying to somehow relate everything to every variable. It also threw me off because I was expecting $x,y,z$ to be different values, but of course the definition never said anything about that. $\endgroup$ – Dimitri Nov 25 '13 at 6:25

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