properties of relations

Question

I'm trying to do some chapter problems on equivalence relations. I'm stuck in the second section "properties of relations."

Question: Let $A=\{a,b,c,d\}$. Give an example of a relation $R$ on $A$ that is neither reflexive, symmetric ,or transitive.

What I tried doing was writing out all the pairs and then canceling out the ones that matched with the laws. I was left with $\{(d,b) (d,c)\}$. Does that mean $R=\{(d,b) (d,c)\}$ is not reflexive, transitive, or symmetric ?

Answer 1

Well, the relation you give is indeed neither reflexive nor transitive, but there are many ways to get such a relation. "Cancelling out pairs that don't match the laws" is not a well defined procedure. (Well, it is defined for reflexivity, but it does not do what you want: You would be left with $\{(a,a),(b,b),(c,c),(d,d)\}$ which is reflexive).

I think the problem would like you to find a relation that involves all four of the members of the set. You could, for example, add in $(a,b)$. Or even $(a,a)$; just one element being equivalent to itself does not make the relation reflexive if others are not.

Answer 2

Well, here's a way to do it. Maybe over kill.

Let $S = A \times A$. That is the universal relationship.

As $(x,x) \in S$ for all $x$ it is reflexive. Let's destroy that be removing $(a,a)$.

$T = A \times A \setminus \{(a,a)\}$.

$T$ is not reflexive because $a \not T a$.

But if $x T y$ then $y T x$. So T is symmetric. Let's kill that. Let's remove $(a,b)$.

$W = A \times A \setminus \{(a,a),(a,b)\}$

Not $W$ is not symetric because $b W a$ but $a \not W b$.

Is $W$ transitive? To be honest, I'm not sure. But $a R c$ and $c W d$ and $a W d$. Let's remove $(a,d)$.

Let $R = A \times A \setminus \{(a,a),(a,b),(a,d)\}$

$R$ is not transitive because $a R c$ and $c R d$ but $a \not R d$.

And it's not symmetric as $a \not R a$ nor reflexive as $b R a$ but $a \not R b$.

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Or we could have built from scratch;

$S = \{(a,b)\}$

Not reflexive: $a \not S a$.

Not symmetric: $a S b$ but $b \not S a$.

Transitive? Vacuously so. There are no $x S y$ and $y S z$. So for all zero of those it is vacuously true $x S y; y S z \implies x S z$.

But let's kill this by adding $R = \{(a,b),(b,c)\}$

Then we have $a R b$ and $b R c$ but $a \not R c$ so it is not transitive.

Answer 3

Question: Let $A=\{a,b,c,d\}$. Give an example of a relation $R$ on $A$ that is neither reflexive, symmetric ,or transitive.

I built that from knowing to exclude reflexivity; there's no symmetry, and no transitivity.

Let $$R =\{(a,b ), (b, c), (c, d)\}$$