Image of Intersection under Relation

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Theorem

Let $S$ and $T$ be sets.

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

Let $S_1$ and $S_2$ be subsets of $S$.


Then:

$\mathcal R \sqbrk {S_1 \cap S_2} \subseteq \mathcal R \sqbrk {S_1} \cap \mathcal R \sqbrk {S_2}$


That is, the image of the intersection of subsets of $S$ is a subset of the intersection of their images.


General Result

Let $S$ and $T$ be sets.

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

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

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


Then:

$\displaystyle \mathcal R \left[{\bigcap \mathbb S}\right] \subseteq \bigcap_{X \mathop \in \mathbb S} \mathcal R \left[{X}\right]$


Family of Sets

Let $S$ and $T$ be sets.

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

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


Then:

$\displaystyle \mathcal R \left[{\bigcap_{i \mathop \in I} S_i}\right] \subseteq \bigcap_{i \mathop \in I} \mathcal R \left[{S_i}\right]$

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


Proof

\(\displaystyle S_1 \cap S_2\) \(\subseteq\) \(\displaystyle S_1\) Intersection is Subset
\(\displaystyle \implies \ \ \) \(\displaystyle \mathcal R \sqbrk {S_1 \cap S_2}\) \(\subseteq\) \(\displaystyle \mathcal R \sqbrk {S_1}\) Image of Subset is Subset of Image


\(\displaystyle S_1 \cap S_2\) \(\subseteq\) \(\displaystyle S_2\) Intersection is Subset
\(\displaystyle \implies \ \ \) \(\displaystyle \mathcal R \sqbrk {S_1 \cap S_2}\) \(\subseteq\) \(\displaystyle \mathcal R \sqbrk {S_2}\) Image of Subset is Subset of Image


\(\displaystyle \implies \ \ \) \(\displaystyle \mathcal R \sqbrk {S_1 \cap S_2}\) \(\subseteq\) \(\displaystyle \mathcal R \sqbrk {S_1} \cap \mathcal R \sqbrk {S_2}\) Intersection is Largest Subset

$\blacksquare$


Also see


Also see


Sources