Properties of Relation Not Preserved by Restriction

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Theorem

If a relation is:

it is impossible to state without further information whether or not any restriction of that relation has the same properties.


Proof

Restriction of Serial Relation is Not Necessarily Serial

Proof by Counterexample:

Let $S = \set {a, b}$.

Let $\mathcal R = \set {\tuple {a, b}, \tuple {b, b} }$.

$\mathcal R$ is a serial relation, as can be seen by definition.

Now let $T = \set a$.

Then $\mathcal R {\restriction_T} = \O$.

So $\not \exists y \in T: \tuple {a, y} \in \mathcal R {\restriction_T}$.

That is, $\mathcal R {\restriction_T}$ is not a serial relation on $T$.

$\blacksquare$


Restriction of Non-Reflexive Relation is Not Necessarily Non-Reflexive

Proof by Counterexample:

Let $S = \set {a, b}$.

Let $\RR = \set {\tuple {b, b} }$.

$\RR$ is a non-reflexive relation, as can be seen by definition:

$\tuple {a, a} \notin \RR$
$\tuple {b, b} \in \RR$

Now let $T = \set a$.

Then $\RR {\restriction_T} = \O$.

So:

$\forall x \in T: \tuple {x, x} \notin \RR {\restriction_T}$

That is, $\RR {\restriction_T}$ is an antireflexive relation on $T$.

That is, specifically not a non-reflexive relation.

$\blacksquare$


Restriction of Non-Symmetric Relation is Not Necessarily Non-Symmetric

Proof by Counterexample:

Let $S = \set {a, b}$.

Let $\RR = \set {\tuple {a, b}, \tuple {b, b} }$.

$\RR$ is a non-symmetric relation, as can be seen by definition.

Now let $T = \set b$.

Then $\RR {\restriction_T} \ = \set {\tuple {b, b} }$.

So:

$\forall x, y \in T: \tuple {x, y} \in \RR {\restriction_T} \implies \tuple {y, x} \in \RR {\restriction_T}$

as can be seen by setting $x = y = b$.

So $\RR {\restriction_T}$ is a symmetric relation on $T$.

That is, $\RR {\restriction_T}$ is not a non-symmetric relation on $T$.

$\blacksquare$


Restriction of Non-Transitive Relation is Not Necessarily Non-Transitive

Proof by Counterexample:

Let $S = \left\{{a, b}\right\}$.

Let $\mathcal R = \left\{{\left({a, b}\right), \left({b, a}\right), \left({b, b}\right)}\right\}$.

$\mathcal R$ is a non-transitive relation, as can be seen by definition.

Now let $T = \left\{{b}\right\}$.

Then $\mathcal R \restriction_T \ = \left\{{\left({b, b}\right)}\right\}$.

So:

$\forall x, y \in T: \left({x, y}\right) \in \mathcal R \restriction_T \land \left({y, z}\right) \in \mathcal R \restriction_T \implies \left({y, z}\right) \in \mathcal R \restriction_T$

as can be seen by setting $x = y = z = b$.

So $\mathcal R \restriction_T$ is a transitive relation on $T$.

That is, $\mathcal R \restriction_T$ is not a non-transitive relation on $T$.

$\blacksquare$


Restriction of Non-Connected Relation is Not Necessarily Non-Connected

Proof by Counterexample:

Let $S = \set {a, b}$.

Let $\mathcal R = \set {\tuple {a, a}, \tuple {b, b} }$.

$\mathcal R$ is a non-connected relation, as can be seen by definition: neither $a \mathrel {\mathcal R} b$ nor $b \mathrel {\mathcal R} a$.

Now let $T = \set a$.

Then $\mathcal R {\restriction_T} = \set {\tuple {a, a} }$.

Then $\mathcal R {\restriction_T}$ is trivially connected on $T$.

$\blacksquare$


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