Properties of Restriction of Relation
Theorem
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
If $\RR$ on $S$ has any of the properties:
... then $\RR {\restriction_T}$ on $T$ has the same properties.
Proof
Reflexivity
Restriction of Reflexive Relation is Reflexive
Suppose $\RR$ is reflexive on $S$.
So:
- $\forall x \in S: \tuple {x, x} \in \RR$
We are given $T$ is a subset of $S$, so:
- $\forall x \in T: \tuple {x, x} \in \RR$
By definition of restriction:
- $\forall x \in T: \tuple {x, x} \in \RR {\restriction_T}$
Hence $\RR {\restriction_T}$ is by definition reflexive on $T$.
$\blacksquare$
Restriction of Antireflexive Relation is Antireflexive
Suppose $\RR$ is antireflexive on $S$.
Then by definition of antireflexive:
- $\forall x \in S: \tuple {x, x} \notin \RR$
We are given $T$ is a subset of $S$, so:
- $\forall x \in T: \tuple {x, x} \notin \RR$
We are given $\RR {\restriction_T} \subseteq T \times T$, so:
- $\forall x \in T: \tuple {x, x} \notin \RR \restriction_T$
Hence by definition, $\RR {\restriction_T}$ is antireflexive on $T$.
$\blacksquare$
Symmetry
Restriction of Symmetric Relation is Symmetric
Suppose $\RR$ is symmetric on $S$.
Then:
\(\ds \tuple {x, y}\) | \(\in\) | \(\ds \RR {\restriction_T}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {x, y}\) | \(\in\) | \(\ds \paren {T \times T} \cap \RR\) | Definition of Restriction of Relation | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {x, y}\) | \(\in\) | \(\ds T \times T\) | |||||||||||
\(\, \ds \land \, \) | \(\ds \tuple {x, y}\) | \(\in\) | \(\ds \RR\) | Definition of Set Intersection | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {y, x}\) | \(\in\) | \(\ds T \times T\) | |||||||||||
\(\, \ds \land \, \) | \(\ds \tuple {y, x}\) | \(\in\) | \(\ds \RR\) | $\RR$ is symmetric on $S$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {y, x}\) | \(\in\) | \(\ds \paren {T \times T} \cap \RR\) | Definition of Set Intersection | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {y, x}\) | \(\in\) | \(\ds \RR {\restriction_T}\) | Definition of Restriction of Relation |
and so $\RR {\restriction_T}$ is symmetric on $T$.
$\blacksquare$
Restriction of Asymmetric Relation is Asymmetric
Suppose $\RR$ is asymmetric on $S$.
Then:
\(\ds \tuple {x, y}\) | \(\in\) | \(\ds \RR {\restriction_T}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {x, y}\) | \(\in\) | \(\ds \paren {T \times T} \cap \RR\) | Definition of Restriction of Relation | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {x, y}\) | \(\in\) | \(\ds \RR\) | Intersection is Subset | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {y, x}\) | \(\notin\) | \(\ds \RR\) | $\RR$ is asymmetric on $S$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {y, x}\) | \(\notin\) | \(\ds \paren {T \times T} \cap \RR\) | Proof by Contraposition and Intersection is Subset | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {y, x}\) | \(\notin\) | \(\ds \RR {\restriction_T}\) | Definition of Restriction of Relation |
and so $\RR {\restriction_T}$ is asymmetric on $T$.
$\blacksquare$
Restriction of Antisymmetric Relation is Antisymmetric
Suppose $\RR$ is antisymmetric on $S$.
Then:
\(\ds \set {\tuple {x, y}, \tuple {y, x} }\) | \(\subseteq\) | \(\ds \RR {\restriction_T}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \set {\tuple {x, y}, \tuple {y, x} }\) | \(\subseteq\) | \(\ds \paren {T \times T} \cap \RR\) | Definition of Restriction of Relation | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \set {\tuple {x, y}, \tuple {y, x} }\) | \(\subseteq\) | \(\ds \RR\) | Intersection is Subset | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds y\) | $\RR$ is Antisymmetric on $S$ |
Thus $\RR {\restriction_T}$ is antisymmetric on $T$.
$\blacksquare$
Transitivity
Restriction of Transitive Relation is Transitive
Suppose $\RR$ is transitive on $S$.
Then by definition:
- $\set {\tuple {x, y}, \tuple {y, z} } \subseteq \RR \implies \tuple {x, z} \in \RR$
So:
\(\ds \set {\tuple {x, y}, \tuple {y, z} }\) | \(\subseteq\) | \(\ds \RR {\restriction_T}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \set {\tuple {x, y}, \tuple {y, z} }\) | \(\subseteq\) | \(\ds \paren {T \times T} \cap \RR\) | Definition of Restriction of Relation | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {x, z}\) | \(\in\) | \(\ds \paren {T \times T} \cap \RR\) | $\RR$ is transitive on $S$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {x, z}\) | \(\in\) | \(\ds \RR {\restriction_T}\) | Definition of Restriction of Relation |
Therefore, if $x, y, z \in T$, it follows that:
- $\set {\tuple {x, y}, \tuple {y, z} } \subseteq \RR {\restriction_T} \implies \tuple {x, z} \in \RR {\restriction_T}$
and so by definition $\RR {\restriction_T}$ is a transitive relation on $T$.
$\blacksquare$
Restriction of Antitransitive Relation is Antitransitive
Suppose $\RR$ is antitransitive on $S$.
Then by definition:
- $\set {\tuple {x, y}, \tuple {y, z} } \subseteq \RR \implies \tuple {x, z} \notin \RR$
So:
\(\ds \set {\tuple {x, y}, \tuple {y, z} }\) | \(\subseteq\) | \(\ds \RR {\restriction_T}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \set {\tuple {x, y}, \tuple {y, z} }\) | \(\subseteq\) | \(\ds \paren {T \times T} \cap \RR\) | Definition of Restriction of Relation | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {x, z}\) | \(\notin\) | \(\ds \paren {T \times T} \cap \RR\) | as $\RR$ is antitransitive on $S$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {x, z}\) | \(\in\) | \(\ds \RR {\restriction_T}\) | Definition of Restriction of Relation |
Therefore, if $x, y, z \in T$, it follows that:
- $\set {\tuple {x, y}, \tuple {y, z} } \subseteq \RR {\restriction_T} \implies \tuple {x, z} \notin \RR {\restriction_T}$
and so by definition $\RR {\restriction_T}$ is an antitransitive relation on $T$.
$\blacksquare$
Connectedness
Restriction of Connected Relation is Connected
Suppose $\RR$ is connected on $S$.
That is:
- $\forall a, b \in S: a \ne b \implies \tuple {a, b} \in \RR \lor \tuple {b, a} \in \RR$
So:
\(\ds a, b\) | \(\in\) | \(\ds T\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {a, b}\) | \(\in\) | \(\ds T \times T\) | |||||||||||
\(\, \ds \land \, \) | \(\ds \tuple {b, a}\) | \(\in\) | \(\ds T \times T\) | Definition of Ordered Pair and Definition of Cartesian Product | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {a, b}\) | \(\in\) | \(\ds \paren {T \times T} \cap \RR\) | |||||||||||
\(\, \ds \lor \, \) | \(\ds \tuple {b, a}\) | \(\in\) | \(\ds \paren {T \times T} \cap \RR\) | as $\RR$ is connected on $S$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {a, b}\) | \(\in\) | \(\ds R \restriction_T\) | |||||||||||
\(\, \ds \lor \, \) | \(\ds \tuple {b, a}\) | \(\in\) | \(\ds R {\restriction_T}\) | Definition of Restriction of Relation |
and so $\RR {\restriction_T}$ is connected on $T$.
$\blacksquare$
Also see
Properties of Relation Not Preserved by Restriction
If a relation is:
- serial,
- non-reflexive,
- non-symmetric,
- non-transitive or
- non-connected
it is impossible to state without further information whether or not any restriction of that relation has the same properties.