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In a practice paper for an exam there is the following relation:

$$ E = \{(1,1),(2,2),(3,3),(4,4)\}\ \text{ on the set }V = \{1,2,3,4,5\} $$

It would appear that because $(5,5)$ is not in $E$, that it would not be a reflexive relation.

What is unclear is how the relation is still transitive. By our definition for Transitivity, for all elements in set $V$, there would have to be $(a,b)$ and $(b,c)\ldots$ and therefore $(a,c)$ in the set $E$.

This works for element $1$ in $V$ because you have $(1,1)$ and $(1,1)$ and therefore $(1,1)$. I would have assumed however that because there is no tuple $(5,5)$ in $E$ the entire relation could not be transitive?

Hope this makes sense, thanks for any input offered.

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    "By our definition for Transitivity, for all elements in set V, there would have to be (a,b) and (b,c)" No. This is not the definition. ( a,b) do *not* have to be in E for all a,b. But ***IF*** (a,b) does exist, ***AND*** (b,c) exist, then (a,c) must also exist. But if (a,b) does not exist... no harm, no foul. [Note: if (a,b) **did** have to exist for a,b in V, then E **has to** be V x V = {(1,1)(1,2)(1,3)..... (5,3),(5,4)(5,5)} all, every single one of the 25 pairs. – 2017-01-11

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Reflexive means for all $a \in V$ then $(a,a) \in E$. That fails because $5 \in V$ and $(5,5) \not \in V$.

Transitive says nothing about items of $V$ but only about items in $E$. The relation is transitive if every time you have an $(a,b) \in E$ and a $(b,c) \in E$ you must also have a $(a,c) \in E$ then the relation is transitive. If you never have an $(5,x)$ or $(y, 5) \in E$ that will in no way affect what you do have in $E$.

This particular relationship if you have $(a,b) \in E$ then $a \ne 5; b\ne 5;$ and $a = b$. So if $(a,b), (b,c) \in E$ then $a = b = c$ and $(a,c) = (a,a) = (a,b) \in E$. So the relationship is transitive. That $a,b,c$ can never be equal to $5$ is utterly irrelevant.

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    I never thought about it, but reflexive is the only condition that pertains to the elements of V. Symmetry pertains only to elements of E in that if you have *this* element in V you must have *that* element too. Transitivity says if you have two elements in V you must also have a third. A function could still be transitive if you have (a,b) but absolutely no (b,x). For transitivity to fail you must have both an (a,b) and an (b,c) but not the (a,c). If you don't the (a,b) or you don't have the (b,c) it doesn't matter whether or not you have the (a,c). So no (5,x) or (x,5) doesn't matter. – 2017-01-11
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    Yes, @fleablood. In your last point, when you do not have some elements a, b, c, in V, such that $(a, b), (b, c) \in E$, it is a [vacuously true](https://en.wikipedia.org/wiki/Vacuous_truth), that transitivity holds regardless of whether or not $(a, c)$ is, or is not, in $E$. Also, as you seem to have gleaned, it is helpful to examine failures of properties of relations, for which the property fails. If it does not fail, the property holds. – 2017-01-11
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    Hmmm, transitivity does not require three elements in some set $V$ of a transitive relation $E$. If $E = \{(a, a)\} $, E is indeed transitive, (and symmetric). – 2017-01-11
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    I didn't mean to imply transitivity required 3 elements. As you point out if there are no (a,b) (b,c) pairs it's vacuously transitive. If it's non-vacuously transitive it must have (a,b) and (b,c) and a cooresponding (a,c). But a, b, and c need not be distinct. So (a,a) (a,a) and (a,a) will do. – 2017-01-11
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Definition: $R$ is transitive if and only if $\forall x,y (R(x,y)$ and $R(y,x))\Rightarrow R(x,z)$

Can you point to where exactly you think this definition fails? For $x=y=5$ the antecedent fails to hold (since $R(5,5)$ is false) and therefore the implication is true, since an implication with a false antecedent is always true.

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Your definition of transitivity is unclear as stated, and you have misunderstood it, there is no requirements on the members of the underlying set, only on the elements of the relation.

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Proof $E$ is transitive: if $E\left(a,\,b\right)\land E\left(b,\,c\right)$ then $a=b\in V\backslash \left\{ 5\right\}$ and similarly $b=c\in V\backslash \left\{ 5\right\}$, so $a=c\in V\backslash \left\{ 5\right\}$ and $E\left(a,\,c\right)$.