Preimage of Intersection under Relation

From ProofWiki
Jump to: navigation, search

Theorem

Let $S$ and $T$ be sets.

Let $\mathcal R \subseteq S \times T$ be a relation.

Let $C$ and $D$ be subsets of $T$.


Then:

$\mathcal R^{-1} \left[{C \cap D}\right] \subseteq \mathcal R^{-1} \left[{C}\right] \cap \mathcal R^{-1} \left[{D}\right]$


General Result

Let $S$ and $T$ be sets.

Let $\mathcal R \subseteq S \times T$ be a relation.

Let $\mathcal P \left({T}\right)$ be the power set of $T$.

Let $\mathbb T \subseteq \mathcal P \left({T}\right)$.


Then:

$\displaystyle \mathcal R^{-1} \left[{\bigcap \mathbb T}\right] \subseteq \bigcap_{X \mathop \in \mathbb T} \mathcal R^{-1} \left[{X}\right]$


Family of Sets

Let $S$ and $T$ be sets.

Let $\left\langle{T_i}\right\rangle_{i \in I}$ be a family of subsets of $T$.

Let $\mathcal R \subseteq S \times T$ be a relation.


Then:

$\displaystyle \mathcal R^{-1} \left[{\bigcap_{i \mathop \in I} T_i}\right] \subseteq \bigcap_{i \mathop \in I} \mathcal R^{-1} \left[{T_i}\right]$

where $\displaystyle \bigcap_{i \mathop \in I} T_i$ denotes the intersection of $\left\langle{T_i}\right\rangle_{i \in I}$.


Also see


Proof

This follows from Image of Intersection under Relation, and the fact that $\mathcal R^{-1}$ is itself a relation, and therefore obeys the same rules.

$\blacksquare$