Properties of Restriction of Relation

Theorem
Let $$\left({S; \mathcal{R}}\right)$$ be a relational structure.

Let $$T \subseteq S$$ be a subset of $$S$$.

Let $$\left({T; \mathcal{R}\restriction_T}\right)$$ be the restriction of $$\mathcal{R}$$ to $$T$$.

If $$\mathcal{R}$$ on $$S$$has any of the properties:


 * Reflexive
 * Antireflexive
 * Symmetric
 * Antisymmetric
 * Asymmetric
 * Transitive
 * Antitransitive

... then $$\mathcal{R}\restriction_T$$ on $$T$$ has the same properties.

Reflexivity

 * Suppose $$\mathcal{R}$$ is reflexive on $$S$$.

Then $$\forall x \in S: \left({x, x}\right) \in \mathcal{R}$$, and so $$\forall x \in T: \left({x, x}\right) \in \mathcal{R}\restriction_T$$.

Thus $$\mathcal{R}\restriction_T$$ is reflexive on $$T$$.


 * Suppose $$\mathcal{R}$$ is antireflexive on $$S$$.

Then $$\forall x \in S: \left({x, x}\right) \notin \mathcal{R}$$, and so $$\forall x \in T: \left({x, x}\right) \notin \mathcal{R}\restriction_T$$.

Thus $$\mathcal{R}\restriction_T$$ is antireflexive on $$T$$.

Symmetry

 * Suppose $$\mathcal{R}$$ is symmetric on $$S$$.

Then $$\left({x, y}\right) \in \mathcal{R} \implies \left({y, x}\right) \in \mathcal{R}$$.

So, if both $$x$$ and $$y$$ are in $$T$$, $$\left({x, y}\right) \in \mathcal{R}\restriction_T$$ and $$\left({y, x}\right) \in \mathcal{R}\restriction_T$$ and so $$\mathcal{R}\restriction_T$$ is symmetric.


 * Similarly for asymmetry.

Let $$\left({x, y}\right) \in \mathcal{R} \implies \left({y, x}\right) \notin \mathcal{R}$$.

If both $$x$$ and $$y$$ are in $$T$$, $$\left({x, y}\right) \in \mathcal{R}\restriction_T$$ but $$\left({y, x}\right) \notin \mathcal{R}\restriction_T$$ still.

And so $$\mathcal{R}\restriction_T$$ is asymmetric.


 * Now suppose $$\mathcal{R}$$ is antisymmetric.

Then $$\left({x, y}\right) \in \mathcal{R} \and \left({y, x}\right) \in \mathcal{R} \implies x = y$$.

By the above argument, the same applies to $$\mathcal{R}\restriction_T$$.

Transitivity

 * Suppose $$\mathcal{R}$$ is transitive.

Then $$\left({x, y}\right) \in \mathcal{R} \and \left({y, z}\right) \in \mathcal{R} \implies \left({x, z}\right) \in \mathcal{R}$$

Therefore, if $$x, y, z \in T$$, it follows that $$\left({x, y}\right) \in \mathcal{R}\restriction_T, \left({y, z}\right) \in \mathcal{R}\restriction_T \implies \left({x, z}\right) \in \mathcal{R}\restriction_T$$.


 * Suppose $$\mathcal{R}$$ is antitransitive.

Then there are no $$\left({x, z}\right) \in \mathcal{R}$$ such that $$\left({x, y}\right) \in \mathcal{R} \and \left({y, z}\right) \in \mathcal{R}$$.

If $$x, y, z \in T$$, then the same still applies, and $$\mathcal{R}\restriction_T$$ remains antitransitive.

Note
If a relation is:
 * non-reflexive,
 * non-symmetric, or
 * non-transitive

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