Image of Intersection under Relation

Theorem

Let $S$ and $T$ be sets.

Let $\RR \subseteq S \times T$ be a relation.

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


Then:

$\RR \sqbrk {S_1 \cap S_2} \subseteq \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$


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 $\RR \subseteq S \times T$ be a relation.

Let $\powerset S$ be the power set of $S$.

Let $\mathbb S \subseteq \powerset S$.


Then:

$\ds \RR \sqbrk {\bigcap \mathbb S} \subseteq \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$


Family of Sets

Let $S$ and $T$ be sets.

Let $\family {S_i}_{i \mathop \in I}$ be a family of subsets of $S$.

Let $\RR \subseteq S \times T$ be a relation.


Then:

$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i} \subseteq \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$

where $\ds \bigcap_{i \mathop \in I} S_i$ denotes the intersection of $\family {S_i}_{i \mathop \in I}$.


Proof

\(\ds S_1 \cap S_2\) \(\subseteq\) \(\ds S_1\) Intersection is Subset
\(\ds \leadsto \ \ \) \(\ds \RR \sqbrk {S_1 \cap S_2}\) \(\subseteq\) \(\ds \RR \sqbrk {S_1}\) Image of Subset is Subset of Image


\(\ds S_1 \cap S_2\) \(\subseteq\) \(\ds S_2\) Intersection is Subset
\(\ds \leadsto \ \ \) \(\ds \RR \sqbrk {S_1 \cap S_2}\) \(\subseteq\) \(\ds \RR \sqbrk {S_2}\) Image of Subset is Subset of Image


\(\ds \leadsto \ \ \) \(\ds \RR \sqbrk {S_1 \cap S_2}\) \(\subseteq\) \(\ds \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}\) Intersection is Largest Subset

$\blacksquare$


Also see


Also see


Sources

  • 1955: John L. Kelley: General Topology ... (previous) ... (next): Chapter $0$: Relations: Theorem $5 \ \text{(c)}$
  • 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{I}$: Sets and Functions: Problem $\text{AA}$: Relations
  • 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.12$: Set Inclusions for Image and Inverse Image Sets: Theorem $12.2 \ \text{(b)}$