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I'm studying set theory/equivalence relation.

For an equivalence relation, $3$ conditions are to be met:

$1)$ reflexivity

$2)$ symmetry

$3)$ transitivity

Does the following set and the relation on it fulfill these criteria, even though $(2)$ and $(3)$ are missing? They are not disproven, they're just absent.

$\{0,1,2\}$

$R=\{(0, 0), (1, 1),(2, 2)\}$

thanks ralph

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    $\begingroup$ What makes you think that symmetry and transitivity are 'absent'? $\endgroup$ – Clive Newstead Aug 10 '20 at 11:35
  • $\begingroup$ Actually, this is just the equality relation on the given set, and equality is the main prototype of equivalence relations. $\endgroup$ – Berci Aug 10 '20 at 11:55
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    $\begingroup$ Be careful when you see something like $\forall x, y \in R, \dots$. It means "for all $x$ and for all $y$", not "for all distinct $x$ and $y$". So you don't even have to worry about the empty set stuff in some of the answers, because indeed for the $x$ and $y$ that are both $0$, $R$ is symmetric, for example. $\endgroup$ – Izaak van Dongen Aug 10 '20 at 14:10
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  1. Since $(\forall x \in A): (x,x) \in R$, then $R$ is clearly reflexive.

I think I’m understanding what you meant by “absent”. But remember the definitions:

  1. For symmetry: $R$ is symmetric iff $xRy$ implies $yRx$ for all $x, y \in A$. Note that in the relation $R$ you can shift the numbers and the ordered pairs resulting from this will still be in $R$. Hence $R$ is symmetric.

  2. For transitivity: $R$ is transitive iff $xRy$ and $yRz$ implies $xRz$. How do we check this? Start by picking the first element of $R$, $(0,0)$. What are the elements in $R$ that have $0$ in the first place? It is $(0,0)$. Now check $(0,0)$ (this $(0,0)$ would be the result from combining both the elements mentioned). You note that $(0,0)$ is still in $R$. Doing this for every element, tou deduce that $R$ is transitive.

The point is that in these definitions $x, y$ and $z$ can be equal to each other. As long they appear in the “right places”, the relation will agree with these properties.

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If there is no counter-example for a statement, the statement is necessarily true. This also applies to your example.

To give another example, let $S(A)$ be the statement $\forall a \in A: a>0$. Then $S(\varnothing)$ is a true statement, because there exists no counter-example to it.

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